hey guys this looks like a fun one it's

got a whole bunch of stuff going on and

it's asking for the area of this blue

triangle if you want to try it on your

own pause it right now CU I'm going to

solve it in 3 2 1 so first let's look at

this square right here this side is

equal to four so this side right here is

also equal to four next I want to look

at this rhombus right here let's verify

that it's a square we know that if this

is a right angle this is also a right

angle in a rhombus opposite angles are

congruent so these two would be

congruent and these two would also be

congruent so we know that rhombus is a

square and then I want to focus on this

triangle this is a right triangle so we

know that we can use the Pythagorean

theorem so A2 plus b^2 is going to equal

c^2 and these two are equal to each

other so let's call them both X so we

can say x^2 + x^2 is equal to 4^2 X2 +

X2 is equal to 2x^2 and 4^2 is equal to

16 after we divide both sides by two we

get x^2 is equal to 8 and after we

square root both sides we get X is equal

to theare < TK of 8 so let's update both

of these X's to s < TK of 8 next let's

go back to this square right here we

know these two sides are four this side

is also equal to four and now let's look

at this right angle again this diagonal

is going to BCT that right angle so both

of these will be 45° and then in this

Square we also have a right angle and if

this is 45 that means this last angle

must also be 45° so we have each of

these angles equal to 45° if we focus on

this triangle we can figure out this

obtuse angle is equal to the sum of

these three which is

135° and now since this is an obtuse

triangle we have to use the law of

cosiness so if we call the sides a b and

c and this angle Theta the law of

cosiness tells us c^2 is = to a 2 + b^2

- 2 * a * B cosine of that angle Theta

so we're going to use this to solve this

side C so let's give ourselves some room

and a is equal to the square < TK of 8

so in the place of each of these A's

we'll plug in square < TK of 8 and then

B is equal to four so in the place of

these B's we'll plug in four and then

Theta is equal to 135 deg

< TK of 82 is equal to 8 + 4 2 is equal

to 16 and then we're going to subtract

all this 2 * 4 is equal 8 and then we

still have the < TK of 8 and to get

cosine of 135 we can look at the unit

circle to get < tk2 over2 and then 8 +

16 is equal to 24 and then this negative

time this negative will give us a

positive and all this is going to be a

fraction we can copy down the 8un 8 * <

tk2 will give usun 16 and this two will

still be in the denominator this otk 16

is equal to 4 * 4 is equal to 32 and 32

/ 2 is 16 24 + 16 is equal to 40 and

after we square root both sides we get C

is equal to the squ < TK of 40 so we can

update this side length C to be square <

TK of 40 and now let's focus on this

square right here since this side isqu <

TK of 40 this side will also be squ < TK

of 40 and then we're ready to focus on

this right triangle and because it's a

right triangle we can do Pythagorean

theorem we'll assign a b and c to the

sides C is equal to < TK of 40 so we can

plug in < TK of 40 here and then B is

equal to > of8 so we'll plug in < TK of

8 and then we'll just copy down the a^ S

the < TK of 402 is 40 and then squ < TK

of 8^2 is equal to 8 and then let's copy

down the A2 and then we can subtract 8

from both sides and we get a^2 is equal

to 32 and then after we square root both

sides we get a is equal to square < TK

of 32 so we can update this side a to be

square < TK of 32 and now since we have

these two sides we're ready to find the

area of this triangle the area is equal

to 1/2 base time height where the base

is the < TK of 32 and the height is the

< TK of 8 we can combine the < TK of 32

and the < TK of8 so we end up with the <

TK of 256 and then theun of 256 is 16

and 1 12 of 16 is 8 we'll give it a

label of centimet squar and that is the

answer to our question let's put a box

around it the area of this blue triangle

is 8 cm squared how exciting

Area of Blue Triangle

I hope you guys like this one!

Let me know if you see an easier way to solve it!

Any requests for other things I should work on?

Thank you!


hey guys this is a fun one it's not

super difficult but I kind of like the

imagery we got a palm tree and a penguin

and we're trying to figure out how tall

is this penguin together they're 230 and

the difference between the two is 200 so

let's give the height of the penguin the

variable X and let's give the height of

the palm tree the variable Y and now we

can write these equations we have y + x

will be equal to 230 and then y - x will

be equal to 200 in order to solve for

this we're going to do something called

elimination we're going to add the two

equations together y + y will give us 2

y the X +x goes away and 230 + 200 is

equal to 430 and then after we divide

both sides by 2 we get Y is equal to 215

and then next let's rewrite this y + x =

230 Y is equal to 215 in the place of

this y I want to plug in 215 and then we

can subtract 215 from both sides we have

X is equal to 15 so we now have the

height of the penguin he is equal to 15

units how exciting

Quick Penguin Math

Someone shared it with me and I liked the picture.


hey guys welcome to this rotation video

we're gonna rotate this figure by 270

degrees these this way to do rotations

in my opinion is using the notes these

are notes for the different rotations

we're gonna use the one for 270 degrees

so first step we want to write down the

3 points of our preimage the 3 points of

the image that we're given it's a

negative 3 to negative 6 2 and negative

3 9 they were going to apply the

transformation given to us from above so

for 270 it says list the Y so we're

switching them so we're gonna take the Y

value and put it in the X spot so I took

the Y from here and put it into X 4

there then it says do negative X or

whatever the opposite of the first one

is now going to become our new second

one so the negative negative 3 becomes 3

this becomes a positive 6 this becomes a

positive 3 now we just plot the points

that we just mapped into so 2 3 2 6 and

9 3 then we're gonna connect the dots

and now you can see that the figure was

rotated 270 degrees so change the

orientation and the quadrant that it's

in so this makes sense if not have a

link down below 2 more practice in

videos and problems on Andy math com

feel free to check it out please like

this video and subscribe to my channel

now sod you guys soon

[Music]

Rotate 270 Degrees

For more visit http://andymath.com.

Subscribe here: https://www.youtube.com/channel/UC6KhU3AMLHC-qv7tj0ZJduw?sub_confirmation=1


hey guys this looks like a fun one it's

a no calculator challenge it's 3 x = 69

and 23 to the Y = 69 and it wants to

know what is 1x + 1 over y if you want

to try it on your own pause it right now

because I'm going to solve it in 3 2 one

first let's copy these down and let's

put everything in parentheses for the

first equation let's exponentiate both

sides to 1 /x power and on the bottom

one let's exponentiate both sides to the

1/ Y power let's go back to the top one

these two exponents are going to

multiply by each other so x * 1X is XX

and X overx is equal to 1 so this is 3

to the 1 or in other words three and

bottom same thing these two exponents

are going to multiply so y * 1 over Y is

y over Y and that's equal to 1 and 23 to

the 1 is 23 so what I like about this is

we now have a 1 /x right here and a one

over y right here we just got to figure

out some way to add these together we

could use logarithms but I want to try

something else let's take this bottom

equation and multiply both sides by

three and since we know 3 is equal to 69

1 /x in the place of this three let's

plug in 69 1 /x on the left hand side 3

* 23 is equal to 69 and on the right

hand side these two have the same base

so we can use these notes right here

anytime two powers are being multiplied

with the same base you can just add the

exponents so these both have a base of

69 so we can just add these two

exponents it'll be 1 y + 1X and on the

left hand side let's give this 69 an

exponent it' be 69 to the 1 now from

here since they both have a base of 69

the only way these two equations can be

equal to each other is if the exponents

are equal to each other so we can set 1

equal to 1 y + 1 /x and now we're

basically done we can rearrange them

like this and we can put a box around it

now we can move it up here in this

challenge 1x + 1 Y is equal to 1 how

exciting

Fun Math Challenge

I hope you guys like this one!


hey guys this looks like a fun one it's

another Katrina a puzzle and it's day

four of our Advent calendar we're going

to do 31 Katrina a puzzles in 31 days so

far we're still on Pace on day four so

for this one there are two circles with

the radius four and two squares and it

once to know what's the Shaded area

which is the area of this Square here if

you want to try it on your own pause it

right now because I'm going to solve it

in three 2 1 since we're trying to solve

for this Square let's give it side

lengths of X and that means the area of

this Square the Shaded area will be

equal to x^2 and this will be the answer

to our question so let's put a box

around it and let's move it down here

next let's focus on the top of the

larger square if we line up with the

center of this circle and drop this line

here this piece is the same length as

the radius of four and this piece is the

same length as the radius of four so one

side of our larger Square will be x + 4

+ 4 and we can do the same thing for the

left hand side if we draw this line

right here parallel with the base that

I'll cut it so that this will be equal

to the radius of four and this will be

equal to the radius of four so once

again the side of the square is x + 4 +

4 now let's go to this side this portion

right here will be equal to four because

it's a radius of the circle and this

portion will be equal to 4 that means

that this last piece from here to here

has to be equal to X because the side

length is x + 4 + 4 and then on bottom

let's draw this down from the center

this piece would be four and this piece

would be four and that leaves X for that

piece so now we can verify our Square

every side is x + 4 + 4 now for the next

step let's draw this right triangle

inside of here the height of our right

triangle will be the same thing as this

right here so it'll be equal to X and

the base of our triangle would be the

same thing as this right here so it'll

be equal to X and then the hypotenuse is

going to be made up of this radius which

is 4 plus this radius which is four so

the whole thing is going to be equal to

eight and now let's focus on that

triangle it's a right triangle so we can

use the Pythagorean theorem it'll be x^2

+ x^2 is equal to 8^2 and if we bring

everything back x^2 that was the area of

our shaded Square so all we have to do

is solve for x^2 x^2 + x^2 is 2 x^2 and

8^2 is equal to 64 and after we divide

both sides by two we get x^2 is equal to

32 and x^2 is what we were trying to

solve for let's change this x^2 into a

32 let's make some room for a label and

there's no units mentioned so we'll just

put unit squared and this is the answer

to our question the Shaded area is 32

unit squared how exciting this was day

four of our Advent calendar this one is

going to be day five it says how far

apart are the corners of these two

regular hexagons so they're both

hexagons they're different sizes and it

wants to know how far apart are these

guess we'll find out tomorrow how that

works give me one second okay I know

what going to happen so tomorrow's going

to be fun how exciting

Aggvent Calendar Day 4

https://andymath.com/agg-vent-calendar/

This is the link to the Aggvent Calendar if you want to participate.


hey guys let's solve for x we have x +

theun x + thek x plus the X going on

forever equal 441 so what value of x

would make this true step number one

let's subtract X from both sides on the

left hand side these two x's cancel so

we just have the square root of all this

stuff it's going to be equal to 441 - x

now this inside of the square root it is

also going to keep going on forever that

means that this stuff inside the square

root is the same thing as this stuff

right here this is equal to 441 that

means this right here is also equal to

441 and then the < TK of 441 is 21 let's

add X to both sides which will give us

21 + x = 441 and then let's subtract 21

from both sides and then we get X is

equal to and 441 - 21 is 420 and that is

the answer to our question that we can

put a box around it so next time you see

420 +un of 420+ of 420 going on forever

you'll be able to say that is exactly

441 how exciting

Fun Math Problem

I hope you guys love it!


hey guys this looks like a fun one it

says the equilateral triangle sits in

the center of the semicircle and it

wants to know what is the area of the

square this problem is day seven of our

agvent calendar for the month of

December we're going to solve 31 Katrina

a puzzles in 31 days and I'm doing my

best to keep us on Pace if you want to

try this one on your own pause it right

now because I'm going to solve it in

three two 1 we're trying to find the

area of this orange Square let's label

the sides of the square s and the area

will be equal to s^2 next they're giving

us the diameter of the semicircle is

equal to 6 that means that the radius

would be equal to 3 so if we connect the

center of the semicircle to this corner

of the square that would also be equal

to three and from here let's finish up

this right triangle this side of our

triangle will be equal to S and the base

we don't know yet let's call it X and

now let's solve for x in the given

equilateral triangle the x is this p

right here it's half of the base let's

take this radius and rotate it over here

that means that the height of this

equilateral triangle is equal to three

and if we connect this vertex to this

vertex we have another right triangle

the hypotenuse of our new right triangle

is the same as a side of our equilateral

triangle and since X is half the

distance the whole distance would be 2X

and now we have three sides of a right

triangle let's do Pythagorean theorem

it'll be X 2 + 3^2 is equal to the

quantity 2x^2 3^2 is equal to 9 and this

exponent will distribute to both of

these to give us 2^2 x^2 and 2^2 is

equal to 4 next let's subtract x^2 from

both sides on the left hand side these

will cancel each other out and on the

right hand side 4x^2 - x^2 is 3x^2 next

we can divide both sides by three on the

left hand side 9 / 3 is equal to 3 and

on the right hand side these two threes

will cancel each other out so we're left

with x^2 is equal to 3 next let's square

root both sides on the left hand side

the square root and the square will

cancel each other out and now we've

solved for x x is equal to the < TK of 3

and that is what we were trying to do up

here in the place of this x let's plug

in < tk3 and now let's take this radius

and bring it back over here and we now

have another right triangle let's do

Pythagorean theorem it'll be < tk3 ^2 +

s^2 is equal to 3^2 over here the square

root and the square will cancel each

other out and 3^2 is equal to 9 after we

subtract three from both sides we end up

with s^2 is equal to 9 - 3 which is 6

and S2 is what we were looking for so in

the place of this s squ let's plug in

six this is the area of the orange

Square let's give it a label of unit

squared and put a box around it how

exciting and this is the puzzle for day

eight when I place two three four five

triangles like this what's the

overlapping area can't wait to see

tomorrow's video how exciting

Aggvent Calendar Day 7

https://andymath.com/agg-vent-calendar/

Here is a link to the Aggvent Calendar if you want to check it out!


hey guys so many people requested this

problem my DMs were flooded with this

one so I guess we got to do it it's

asking for the area of this shaded

region this region is a rectangle so the

area is going to be base time height

where the base is this right here and

the height is equal to 10 we don't know

the length of the base yet so let's call

it X so if we copy down the arrow we

know this entire distance right here is

equal to 14 and since this piece is X

that means this remaining piece is going

to be 14 - x and we have the same thing

going on right here this whole entire

distance is equal to 8 this is X which

leaves 8 - x for this piece the length

of the base of this larger rectangle is

equal to 20 so we can say 8 - x + x + 14

- x is equal to 20 and now we can solve

for x so first we'll combine like terms

this negative X and this positive X can

cancel each other out and we can bring

down thisx we can combine the 8 plus the

14 to give us 22 and this is all going

to be equal to 20 we can subtract 22

from both sides on the left hand side we

havex and on the right hand side 20 - 22

is -2 so we havex is equal to -2 that

means that X is equal to pos2 so let's

update this x to 2 and now we're ready

to find the area of our region it's

going to be the base times the height

where the base is equal to 2 and the

height is equal to 10 and that's going

to give us 20 square units and that's

the answer to our question let's put a

box around it how exciting

Popular Math Problem

https://andymath.com/geometry-challenges/

For more math challenges, check out the above link!


hey guys welcome to this function's

video it says let F ofx = 5x + 3 find F

of 3 so the F here is referring to the F

here and notice how there's an X here

but there's a three here since this x

says there's an X right there you're now

going to put a 3 where there once was an

X so it's 5 * 3 + 3 5 * 3 is 15 + 3 and

15 + 3 is 18 so the answer to F of3 is

18 and this is why we just plugged in

the three for the X and that's how to

evaluate functions comment below on what

you would like me to solve next and I

will talk to you soon bye Andy maath

Evaluating Functions #andymath  #math  #maths  #sat  #tutorials  #tiktokmath  #mathhelp

Please Subscribe, and don't forget to hit the bell icon so you don't miss next video! Also, please feel free to share this video with your friends and family. Let learn Math together with me, Charlie.
Here is the playlist of the videos. [ There is so much more to discover in this playlist ]


okay so let's solve a limit this is the

limit as X approaches 69 of 6X squared

minus 408x minus 414 over x minus 69.

since we want to know the limit as X

approaches 69 we just substitute 69 in

for each of these x's and we can see

what we get so after you plug it into a

calculator all this on top gives us 0

and 69 minus 69 is also zero so this is

an indeterminate form we don't actually

know what number this is so we have to

find a different way to solve this so

let's try to simplify the stuff inside

the parentheses another way you can

think of a fraction is as a division

problem we're going to take this and

we're going to divide by x minus 69.

we're going to do some long division and

this is called polynomial long division

first step we want to know how many

times does X go into 6X squared well

it'll go in 6X times and that's because

6X times x gives us 6X squared and then

we also want to do 6X times the negative

69 which gives us negative 414x now

after we multiply this we want to

subtract these and this negative is

going to distribute both to the negative

6X and it's going to make the 414x

positive 6X squared minus 6X squared

cancel each other out and negative 408x

plus 414x is 6X and now that we're done

subtracting we can bring down the

negative 4 14. now we want to ask

ourselves how many times does X go into

6X well that's going to be 6 times and

that's because 6 times x equals 6X and

we also want to do 6 times negative 69

which is negative 414. and now we want

to subtract this from this well this

negative will distribute to both terms

making the first one negative and the

second one positive these cancel and

these cancel so we're just left with

zero so the answer to this divided by

this is 6X plus 6. so we can change this

whole thing into 6X plus 6. let's smush

this all together and now we're ready to

apply the limit so we want to know what

is this inside thing equal to when X

approaches 69 or we can plug in 69 for

the X at 6 times 69 is 414 so we have

414 plus 6 and that's equal to 420. so

the limit as X approaches 69 of this

right here is is equal to 420. now let's

put a box around it how exciting

Limit using Polynomial Long Division

This is not the most efficient way to solve the problem, but I think it's a fun review of limits and Polynomial Long Division.


hey guys this looks like a fun one we're

trying to find the area of

ACF so that's the area of this reddish

triangle we're also given that this is a

rectangle and that ABC which is this

triangle is congruent to a e c which is

this triangle up here so these two are

congruent as well if you want to try it

on your own pause it right now because

I'm going to solve it in 3 21 so how do

we approach this I'm just going to go

freestyle and try to find everything I

can first thing I'm noticing is this

right triangle so we can make this side

C and do Pythagorean theorem we have 3^2

+ 4^2 is equal to c^ 2 32 is equal to 9

4 2 is equal to 16 and then we can bring

down the c^ 2 9 + 16 is equal to 25 so

we have 25 is equal to c^2 to solve for

C we can square root both sides square >

of 25 is 5 and square < TK of c^2 is C

so now we figured out that this C is

equal to 5 and we can give it a label of

meters to match these other ones what

should we do do next okay I know what to

do next in a rectangle opposite sides

are congruent so this side is also equal

to four ABC is congruent to a e c this

order of these letters is very important

since the first two letters of this are

ab and the first two letters of this are

AE we know side AB is congruent to side

AE so we know that this AB is equal to

AE which means we can label that as 4 M

and we also know since the second letter

is B and this second letter is e that

angle B is equal tole angle e so since

angle B is a right angle that means this

angle e is also a right angle these

congruent triangles gave us quite a bit

of info next let's focus on these two

triangles this angle and this angle are

both congruent because they're vertical

angles and now we know that angle AF is

congruent to cfd that's what we just

found with the vertical angles we know

that angle e is equal to angle D because

they're both right and we know that AE

is congruent to CD because they're both

equal to four therefore we can conclude

that these two triangles are congruent

and that's using the angle angle side

theorem any two triangles where we have

a corresponding angle angle and a side

those two triangles have to be congruent

and now we have another congruent

statement so looking at these first two

letters and these first two letters we

know the segment AF is equal to segment

CF that means this AF is equal to this

CF and since we know CF is five that

means we know this is five and now we're

pretty much done let's get rid of this

stuff and if you look at our red

triangle we have a base and we have a

height and give me a second I'll make

this easier to see so there we go now we

can see the base is five and the height

is four and if it's easier we can flip

this so here's the base of five and the

height of four the area of a triangle is

equal to 1/2 base time height we can

plug in five for the base and plug in

four for the height and 1/2 * 5 * 4 is

equal to 10 so the area of this red

triangle is 10 m squar this is the

answer to our question let's put a box

around around it and here's the next

problem I'm going to work on I'll post

the video as soon as I finish how

exciting

Geometry Challenge from Social Media

I hope you guys like this one!


hey guys so let's solve for x in this

equation it has an absolute value inside

of an absolute value this should be fun

the way to get rid of this outer

absolute value is we can take the stuff

inside and set it equal to 16 because if

all this stuff inside is equal to 16

then this will be a true statement we

can also take this inside stuff and set

it equal to negative 16. because if this

stuff inside here is negative 16 this

will also be true so now we have two

equations but we've managed to get rid

of that outer absolute value that's how

we get rid of absolute value signs we

take whatever was inside of them and set

equal to the positive or the negative of

what it's equal to so to get this alone

I'm going to subtract 3x from both sides

so left hand side is just going to be

negative absolute value of x and this

right hand side is going to be 16 minus

3x then I'm going to divide both sides

by negative 1. left hand side is just

the absolute value of x right hand side

will be negative 16 plus 3x let's do the

same thing to this one down here let's

subtract 3x from both sides the

left-hand side is negative absolute

value of x equals and this will be

negative 16 minus 3x divide both sides

by negative 1 and we get absolute value

of x on this side on this side we get 16

plus 3x so we're going to take both of

these equations and split each of them

up into two different equations take

this inside X and we're going to set it

equal to this negative 16 plus 3x we're

also going to take this inside X and set

it equal to the negative of this which

would be positive 16 minus 3x same thing

here we take the X out and set it equal

to that 16 plus 3x and we take the X out

and set equal to the negative of that

which is negative 16 minus 3x for both

these two equations let's subtract 3x

from both sides so the left-hand side is

going to be negative 2X and then for

this side we just bring down the 16 or

bring down the negative 16. divide

everything by negative 2 and this one

becomes x equals 8 and this one becomes

x equals negative 8. and on this side we

can add 3x to both sides of both these

equations X Plus 3x is equal to 4X these

cancel so we bring down the 16 and these

cancel so we bring down the negative 16.

divide everything by 4 and we get x

equals 4 or x equals negative 4. now we

have four possible solutions to try

let's get rid of all this work let's

make these larger and then bring them up

so now I'm going to plug each of these

values in for X the H the 4 the negative

8 and the negative 4. 3 times 8 is 24

absolute value of 8 is 8 24 minus eight

is sixteen absolute value of 16 is equal

to 16 so this one's good 3 times 4 is 12

absolute value 4 is 4 12 minus 4 is 8

absolute value of 8 is 8 8 does not

equal 16 this one's no good 3 times

negative 8 is negative 24 absolute value

of negative 8 is positive eight negative

24 minus 8 is negative 32. absolute

value of negative 32 is positive 32 32

does not equal 16. 3 times negative 4 is

negative 12 absolute value of negative 4

is equal to four negative 12 minus 4 is

negative 16 absolute value of negative

16 is positive 16. 16 16 does equal 16

so x minus 4 is another solution so

these are the two solutions and here's

all the work again how exciting if you

guys want to try more of these I have

this page on andymath.com the link is

down below in the description check it

out and let me know if you have any

other questions I'll happily take

requests

Absolute Value Challenge Problem


hey guys now we're going to add up some

vectors the first one is asking for it

in component form this right here is

component form if three is and four J's

so when adding vectors all you do is

just add the corresponding component so

3 plus 7 is 10 I and negative 4 plus 3

is negative 1 and there's the answer to

a now if we want to show it as a

magnitude in a direction the way that

you find the magnitude is you do the

square root of this squared plus this

squared

and that's the magnitude square root of

101 then to find the direction so you do

tangent inverse of negative 1 over 10

which is equal to negative five point

seven degrees and then it right after

you do this you want to check your

quadrant so 10 i minus J is down here so

negative 5.7 makes sense so then this

would be written as agne tude equals

square root of 101 and the direction is

negative five point seven degrees that's

it hope that helped talk to you later

bye

Adding Vectors

Vector Addition GREAT EXAMPLE! For more math made easy visit andymath.com.

Subscribe here: https://www.youtube.com/channel/UC6KhU3AMLHC-qv7tj0ZJduw?sub_confirmation=1


hey guys so we have this video of a

Pikachu Professor making a calculus

problem for their

students let's see if we can solve it

I'm pretty sure this is the equation

right here this is an integral from 2 Pi

to 10 of a whole bunch of trig stuff DX

if you want to try this on your own

pause it right now because I'm going to

solve it in three two one first let's

copy it down and let's give ourselves

some space I want to pull some stuff out

of this one right here on top I want to

pull out this cosine Cub 3x on bottom

the six and the eight have a two in

common so I want to pull that two out

which will turn the six into a three and

the eight into a four and let's also

pull a negative 1 out of both of these

so in the end we're factoring out a -2

now I'm ready to focus on this middle

piece right here let's clean it up a

little bit this looks very familiar to

me and the fact some of the inputs for

these are 3x I'm thinking we should

check the triple angle formulas I don't

have them memorized but I found them

here they are right here and this is

identical to this one right here so all

of this is equal to seeking of 3x and

secant means 1/ cosine so we can rewrite

this as 1/ cosine

3x now let's look at this cosine cubed

of 3x that Cube means that there are

three of them being multiplied by each

other and I'd like to take one of them

and put it over here and then for these

two let's recombine them into a cosine

squar now we have a cosine 3x over

cosine 3x they cancel to give us one and

then we're just multiplying by this one

so all of this can disappear and then

let's smush everything together right

now we're subtracting this second term

another way can think of it take the

subtraction symbol make the denominator

negative and add the two terms and now

both of our terms are being divided by

-2 that means they're both being

multiplied by - one2 so let's pull out

that - one2 and now we're just left with

cosine 2 3x+ sin^2 3x well this is a

Pythagorean identity sin s of theta plus

cosine sare thet is equal to 1 Theta can

be anything including 3x so sin s 3x +

cosine 3x is just going to be equal to 1

so using the Pythagorean identities we

just changed all of this stuff into one

and now we can smush everything together

and we're going to take this integral

the integral from 2 Pi to 10 of 1 that's

just going to be X evaluated from 2 Pi

to 10 and then we still have the - 1/2

in front so all this means is we're

going to take our - 1/2 and multiply it

by x - x where the first X is equal to

this top number 10 and this x is equal

to the bottom number 2 pi and now we're

ready to distribute the -22 * 10 is -5

and -2 * -2 piun is pi this is the

answer to Pikachu's challenge question

let's put a box around it how exciting

Tricky Calculus Integral

https://andymath.com/trigonometric-identities/
https://andymath.com/indefinite-integrals-trigonometric-functions/


what is the area and perimeter of this

rectangle there is going to be base

times height so we can just say a equals

then the base is this right here times

the height which is this right here and

we can see that this is an A minus B

times an A plus b that is a difference

of two squares so it's just going to be

equal to a squared minus B squared or in

other words 105 squared minus 2 root

2739 squared 105 squared is 11 25 minus

and then 2 squared is 4 and root 2739

squared is 2739 and 1125 minus 10 956 is

69 nice and next let's find the

perimeter we're going to do 2 of this

side here plus 2 of this side here these

twos can distribute which would give us

210 minus 4 root two seven three nine

plus two ten plus four root two seven

three nine these two match but one's

positive one's negative so they cancel

each other out so we're just left with

the perimeter equal to 210 plus 210

which is 420. so the area is 69 and the

parameter is 420. nice

Area and Perimeter Challenge

Collab with Geeklyhub


hey guys this is one of the most

requested videos I've ever had and for

good reason this is such a cool problem

I did not expect it to have the outcome

that it had so the problem gives us a

semicircle right here and a quarter

Circle right here and then this red line

with a length of five that goes from the

vertex of this rectangle tangent to the

semicircle if you want to try this on

your own pause it right now cuz I'm

going to solve it in 3 2 1 so first up

let's assign a radius to this semi

circle a lot of people wanted to assign

this radius to match but there really is

not evidence that this radius is the

same as this radius so let's call it

something else like X using our method

we're not going to need this radius to

equal this radius and that right there

is one of my favorite things about this

problem we don't care if the radi are

the same so this last distance right

here will also be R because it's a

radius of the same semicircle and this

height will also be X because it's a

radius of this quarter Circle and now we

can come up with a formula for the area

of the rectangle the area of a rectangle

is base time height the base is going to

be equal to x + 2 R and the height is

going to be

Full Video 👆

This was a popular one. I wanted to share it with more people on shorts. I hope you love it!


hey guys this looks like a fun one 2

squared of x equals square root of 2

plus X so for the very first step we can

do a domain restriction we know that

under these square roots we can't have a

negative number so that means that the X

has to be greater than or equal to zero

if x were less than zero this would be

undefined on the left hand side now we

could put a box around that so we're off

to a pretty good start we've already

gotten rid of all the negative numbers

next to get rid of both these square

root symbols let's Square both sides of

the equation on this right hand side the

square root and the square cancel

leaving us with Just 2 plus X on the

left hand side this Square distributes

to both terms 2 squared and the square

root of x squared 2 squared is equal to

4 the square root and the two cancel

each other out leaving us with just 4X

and we were able to do this step because

X is greater than or equal to zero next

we can subtract X from both sides 4X

minus one X is 3x these X's cancel

leaving us with just two we can divide

both sides by three and that leaves us

with x equal to two-thirds and that's

the answer to our question how exciting

Quick Square Root Challenge

I thought this was a fun one.


hey guys let's solve some trigonometric

equation it says sine of X Plus root 3

cosine of x equals zero so first I

copied it down this has a sine X and A

cosine X in it so this can be kind of

tricky to work with let me show you a

trick for this one if you divide

everything by cosine X sine X over

cosine X is equal to tangent X root 3

cosine X over cosine X is equal to just

root 3 and then 0 over cosine X is still

just zero then we can subtract root 3

from both sides and then we have tangent

of x equals negative root 3. so from

here we can use tangent inverse on our

calculator or we could use the unit

circle if we look at a unit circle we

can see these two values right here are

answers so the x equals 2 pi over 3 or 5

pi over 3. and so this is periodic you

can keep going on forever where we keep

adding Pi a better way to express this

would be 2 pi over 3 plus pi times n and

that is the answer to our question and

we can even click show answer and it

gives us the same answer right there if

you guys want to try more of these I

have this page on andymath.com link is

down in the description how exciting

How to solve this Trig Equation #shorts

https://andymath.com/solving-trigonometric-equations/

#shorts


hey guys this looks like a fun one it's

two equilateral triangles with a

distance of two between the Peaks and

through those two peaks it draws a

quarter Circle and it wants to know what

is the area of that quarter Circle if

you want to try it on your own pause it

right now cuz I'm going to solve it in 3

2 1 so the question is asking us to find

the area of this quarter Circle the area

of a quarter circle is equal to < R 2 /

4 so the only thing we got to do is

figure out what is the radius of this

quarter Circle I'm thinking let's

reflect the whole image down here so now

we have a semicircle and four of these

equilateral triangles in equilateral

triangles all the angles are equal to

60° I'm specifically interested in this

angle right here and we can cut this 60

and half by bringing down this segment

right here and it cut this angle into

30° and now let's focus on this 30

degree angle since the vertex is on the

circle this is called an inscribed angle

and inscribed angles are always half the

inscribed Arc so the measure of this Arc

right here is 60° and if we shift the

angle up here it's now called a central

angle and that is equal to The subtended

Arc so this angle is 60° and both of

these sides of the angle are equal to

the radius now let's focus on this

triangle right here since these two

sides are equal to each other it is an

isoceles triangle and every Isles

triangle the base angles are congruent

so both these angles are equal to each

other since every triangle has 180° this

is only possible if both of those are

60° and now since all three angles are

equal we know this is an equilateral

triangle so that means all the sides

have to be equal so both of these sides

are equal to two so now let's let's

bring back our original figure and let's

focus on the quarter Circle the radius

of this quarter circle is equal to 2 and

now we're ready to use our area formula

in the place of the radius let's plug in

two 2^ 2 is equal to four and these

fours will cancel each other out so the

area is equal to Pi square units and

this is the answer to our question let's

put a box around it the area of this

quarter circle is equal to Pi square

units how exciting

Fun Geometry Puzzle

Here's another Catriona Agg puzzle. I hope you guys like it!


hey guys it says no numbers are given

how many degrees is the red angle if you

want to try this on your own posit right

now because I'm going to solve it in

three two one let's focus on this

triangle and the semicircle anytime you

have a triangle inscribed in a circle

where one side is the diameter the

triangle will always be a right triangle

so this angle ACB has to be 90 degrees

and this is always true here's a nice

little visualization to show it if you

notice if we rotate the visual one side

will always stay horizontal and one side

always stays vertical that would only be

possible if it was a right triangle

since there's 180 degrees total in a

triangle these other two angles must add

to 90 degrees now once again that's

because we need all three of these

angles to add up to 180 degrees every

triangle has a special Point called the

in Center there are two very important

details about the in Center first it's

the center of the inscribed Circle and

second it's the intersection of the

angle bisectors so for this figure right

here Point m is the in Center it is the

the center of the inscribed Circle and

since these two lines intersect at that

point that means that these lines are

bisecting these two angles to make it

easier let's number these angles one two

and three so the measure of angle two

right here has to be half of the measure

of angle BAC now the same thing for

angle three it's cutting angle ABC in

half so we'd say the measure of angle

three is equal to measure of angle ABC

divided by two let's look at this

equation again and let's divide

everything by two the measure of angle

BAC divided by two that is the same

thing as the measure of angle two and

then we're going to add to that ABC

divided by 2 which is angle three and

that's going to be equal to 90 divided

by 2 which is 45 degrees and I'll smush

all this together and add angle one to

both sides of the equation angles one

two and three are the three angles of

this triangle so we know they're all

going to add up to 180 degrees and if we

subtract 45 degrees from both sides 180

degrees minus 45 degrees is 135 degrees

and on the right hand side we just have

the measure of angle one so the measure

of this red angle is 135 degrees and

that is the answer to the question let's

put a box around it how exciting

No Number Geometry Challenge

I thought this was a fun one. I hope everyone like it.


hey guys this looks like a fun one it's

made up of four squares and it wants to

know what's this shaded area if you want

to try it on your own pause it right now

cuz I'm going to solve it in 3 2 1 first

step I want to focus on this triangle

right here so let's make it blue and

then I want to zoom in on this region

right here looking at this blue triangle

this angle is a right angle and that's

because we know all of these are squares

and then this angle right here is also a

right angle and that's because it's one

of the angles of this larger Square next

if you ever have two parallel lines and

a line cutting through them it will

always be true that these two angles are

equal to each other these are called

alternate interior angles and this is

the alternate interior angle theorem we

have this exact situation going on here

these two lines are parallel and this

line is cutting through them that means

these two angles are equal to each other

if we look at this pink triangle and

this blue triangle these have two pairs

of corresponding congruent angles that

means the third angles must also be

congruent and now let's pull out these

two triangles and we'll look at those

later next I want to focus on this

square with an area of five that means

that the side lengths are each going to

be square OT of five and if we look at

this square with an area of 20 the side

lengths are each going to be square < TK

of 20 next I want to look at this square

right here the base of the square is <

tk2 +un 5 that means that the side is

also going to beunk 20 +un 5 and let's

move this < tk2 label from the bottom to

the top of this square and now we're

ready to mess around with this blue

right triangle if we label the

hypotenuse C we can do the Pythagorean

theorem we can do < TK 20 2 +un 5^2 is

equal to c^2 and for this first term the

square root and the square will cancel

each other out to give us 20 and same

thing here otk 5^ 2 is going to be 5 20

+ 5 is equal to 25 and we can take the

square root of both sides and we end up

with 5 is equal to C so let's update

this C up here to be five and now that

we have the measurements of the blue

triangle let's update it down here next

I want to focus on the top of this

Square it's going to be equal to < TK 5

+un 20 this is the exact same thing as

the hypotenuse of this pink triangle it

will also beunk 5 + < tk2 and < tk2 can

be split up as < TK 4 * 5 and then this

can be split up into < TK 4 * < TK 5 and

< TK 4 is equal to 2 so it ends up being

2un 5 so now we can simplify this side

length if we add up < TK 5 + 2un 5 it

ends up being 3un 5 and now we're almost

done they want to know what's the Shaded

area we called this length down here s

that is the same as the base of our pink

Square so once we know s we know the

answer to our question the area of this

pink square is going to be s s so let's

solve for S I want to give this pink

triangle the same orientation as the

blue triangle all three of these

corresponding pairs of angles are

congruent which means these two

triangles are similar that means we can

set up a proportion for their

corresponding sides we have 3un 5 over 5

is going to be equal to S over < tk2 and

then we can cross multiply which gives

us 5S is = to 3un 5 * < tk2 and then

this < TK 5 < tk2 can be combined into <

TK 5 * 20 that's going to be < TK of 100

which is equal to 10 and 3 * 10 is equal

to 30 and after we divide both sides by

5 we get S is equal to 6 now we're ready

to plug this in we can plug in 6 for

this s and 6 s is equal to 36 and let's

give it a label of unit squared and that

is the answer to our question the area

of this shaded region is 36 units

squared let's put a box around it how

exciting

Pink Square Puzzle

This is another Catriona Agg Puzzle. I am hoping to solve all of them I can find this year!


you guys this looks like a fun one it

says find the area of the red region so

we want the area of the region contained

by these three circles and these circles

have a radius of one two and three

meters so for the first step let's

connect this Center to this Center and

then this Center to this Center and then

this Center to this Center this forms a

triangle and let's look at the triangle

down here so this is a radius of this

circle which is one and this is a radius

of this circle which is three which

makes this entire top side one plus

three which is four and same here this

is a radius of three and this is a

radius of two which would make this

whole bottom a length of five and then

this radius is two and this radius is

one which makes this side a length of

three so this is a three four five

triangle that is a Pythagorean triple

that means this has to be a right

triangle and here's the Pythagorean

theorem that verifies it all three four

five triangles are right triangles let's

find the area of this whole triangle the

area of a triangle is one half base

times height so if we rotate this thing

we can see that the base is four and the

height is three four times three is 12

and one half of 12 is 6. so the area of

this triangle is equal to 6. so now that

we have that let's look at these

individual pieces of the triangle we

have this yellow sector right here this

orange sector and the green sector let's

start by finding this orange piece if

this central angle is equal to D the

area of this sector is going to be equal

to D over 360 times pi r squared and we

know this is a right angle so that means

we can plug in 90 for the D and the

radius of the circle is equal to one so

we can plug in 1 for r 1 squared is

equal to one and multiplying by one we

don't need this one anymore 90 over 360

is equal to 1 4 so we end up with 1 4 Pi

or pi over four next we have the area of

this yellow sector let's make this angle

D degrees so we can use this formula

right here the radius of this circle is

equal to 3 and 3 squared is equal to

nine now we got to find the D let's copy

this triangle down again and this right

here is the angle D so now we have to

use some trigonometry we know that sine

of D is equal to the opposite over the

hypotenuse the side opposite of the D is

equal to 3 and hypotenuse is equal to

five so now to solve for D we need to do

the inverse of sine this sine inverse

gets rid of the sign so we just have D

on the left hand side and for the sine

inverse of three-fifths we can use a

calculator to get the approximate value

of this and we get 36.9 degrees and so

we can change this D into 36.9 so

fortunately this isn't exact anymore so

we got to make this an approximately

equal after we do all this math we get

the areas approximately 2.9 and last

let's do this green sector so we'll make

this angle right here D which means we

can use this formula again for the r the

radius of this circle is equal to 2 and

the 2 squared is equal to 4. and now to

find the D we know that the three angles

of a triangle add up to 180 degrees so

90 plus 36.9 plus d will equal 180. 90

plus 36.9 is 126.9 and after we subtract

that from both sides we get D is equal

to 53.1 and we can change this D into

53.1 and then once again we've got to

change this to approximate and after we

do this the area is approximately 1.9

and now we have everything we need to

find the area of this red region it's

going to be equal to this large triangle

minus the yellow region which was

approximately 2.9 but here's the exact

value right here minus the green region

which is approximately 1.9 but this is

the exact value for the green region and

then minus the pi over 4 which this is

the exact value and after we do all this

math we get approximately

0.464256 meters squared and these are

the answers to the question this is the

exact area and this is the approximate

area of the red region how exciting

Find Area of Red Region

I thought this was a fun one, and it includes a little bit of trigonometry!


hey guys you've heard of the quadratic

formula it's a great way to solve ax^2 +

BX + Cal 0 but did you also know there

was a cubic formula if you want to solve

for a x Cub + bx2 + cx+ d = 0 x is going

to be equal to all of this stuff and

this is so long that it requires two

rows I thought it'd be fun to test this

out on X Cub is equal to 8 so first

let's copy down the formula and we need

to rewrite X Cub equal 8 in this format

First Step let's subtract 8 from both

sides and that gives us X Cub minus 8

equals 0 and let's give ourselves some

space so that we can include the 0 x^2

and the 0x and then this x Cub needs a

coefficient so let's give it a one and

now we're ready to use this formula I

think the zeros could help us simplify

this quite a bit so let's start with b

equals 0 in the place of all of these

B's let's plug in zero and then now

everywhere where there's a

multiplicative zero is just going to be

zero so this whole term can vanish so

you can see just by the fact that b was

Zero we were able to get rid of a lot of

these terms and then we can smush

everything everything together and we

can even move this up here next let's do

the other zero the C equals z so we can

change this C right here into zero and

this C right here into zero and both of

those are going to vanish as well next

let's focus on the a is equal to one so

we can plug in one for both of these A's

and we can plug in one for both of these

A's and all four of these denominators

are 2 * 1 so they're all just going to

become two and now we can focus on that

D it's equal to8 so all four of these

D's are going to be8 and now we can just

simplify things this negative and this

negative will cancel each other out and

8 / 2 is 4 this squared on the outside

will distribute to the top and the

bottom so we're going to have 8^ 2 and

2^2 the 82 is equal to POS 64 and 2^2 is

equal to 4 and then 64 / by 4 is 16 squ

< TK of 16 gives us 4 and 4 + 4 is equal

to 8 and then the cube root of 8 is

equal to 2 and now we can do the second

cube root the negative and negative will

cancel each other out 8 / 2 is 4 and

this squar will go to the top and bottom

and 8^2 is 64 and 2^2 is equal to 4 64 /

4 is 16 and the < TK of 16 is equal to 4

4 - 4 is equal to 0 and then the cube

root of 0 is just 0 so then we end up

with x = 2 + 0 which is just 2 so the

cubic formula gave us the real solution

to X Cub = 8 it's xal 2 let's put a box

around it how exciting

Cubic Formula Part 1

https://youtu.be/XswsrgTQAWI
Part 2 is here.


hey guys there's this interesting

property of logarithms where you can

take this exponent and bring it to the

front it's something I think most people

tend to memorize but it feels kind of

weird doesn't it I wanted to see if we

can figure out why we can do it this is

a really nice animation here but

typically in books it looks more like

this on one side of the equation the a

is in front on the other side of the

equation the a is the exponent so we're

going to find this equation first let's

talk about log form versus exponential

form both of these equations mean the

exact same thing B the Y = X is the same

thing as y = log base B of X and this

will always be true the base of the

exponent will always be the base of the

logarithm and the exponent will always

be the thing that the logarithm is equal

to and the thing that the exponent is

equal to will always be the argument of

the logarithm and here's some colorcoded

circles to help us remember let's copy

down both of these equivalent equations

and I want to play around with them a

little bit first let's do this one b y =

x let's exponentiate both sides of the

equation to the power of a on the left

hand side these exponents will multiply

so we have B to the power of y * a and

we can change the order to get B to the

a * Y and then on the right hand side

this will be x to the A and now we're

back in exponent form let's identify

these three parts again and let's

rewrite this in logarithmic form this

green circle will be a y the base of the

logarithm will be B and the argument of

the logarithm will be x to the A and

this looks important let's put a box

around it and then let's go down to our

yal log base B of X I'm thinking we

should multiply both sides of this by a

on the left hand side we'll have a * Y

and on the right hand side we'll have a

* log Bas B of X this also looks

important let's put a box around it now

let's compare our two boxes since this

and this are both equal to a y we can

set them equal to each other so now we

have a * log Bas B of X is equal to log

base B of x to the A and that's what we

trying to find and here's all the work

right here how exciting

Why can you move the exponent in logarithms?

I thought this was an interesting one. I hope you guys like it!


hey guys this looks like a fun one we

have a square right here split up into

three regions and all three of them have

the same area and then we have a quarter

circle with a radius of 13 and the

question is asking what is the area of

the square if you want to try this on

your own pause it right now cuz I'm

going to solve it in 3 2 1 Let's copy

down and let's focus on the Square let's

label the sides of the square X so

ultimately we're trying to find X2 I

want to focus on this blue triangle

right here it has a base of X and let's

label the height Y and since the blue

area equals the red area equals the

green area we know that this triangle is

1/3 of the entire square area of this

triangle is 1/2 base time height where X

is the base and Y is the height for the

right hand side 1/3 of the square is

going to be 1/3 of x^2 next we can

multiply both sides by two on the left

hand side the 2 * 1/2 cancel so we just

have XY and on the right hand side the

two gets multiplied by the 1/3 to give

us 2/3 and then we can bring down the

x^2 next we we can divide both sides by

X and we can do that because we know X

is not equal to zero on the left hand

side we have y and on the right hand

side we have 2/3 of X so let's change

this y up here to 2/3 x and now let's

play around with a quarter Circle the

radius of this quarter circle is 13 and

that's going to be true all the way up

here so this length is equal to 13 and

that means the hypotenuse of our

triangle is also 13 and now we can do

Pythagorean theorem so we have A2 + B2 =

c^2 or in this case it's x^2 + 2/3 x^2

is equal to 132 we can bring down the

x^2 and then this squar is going to

distribute to both the 2/3 and the x

squared and then this squar is going to

distribute to both the two and the three

and that'll give us four nths and let's

smush everything together and set that

equal to 13 s which is 169 nice and now

let's combine like terms I want to give

these a common denominator so let's

multiply the x^2 by 9 over 9 so now

we're doing 9 9 + 4 9 which is 13 9 x^2

and that's still equal to 169 let's

multiply both sides by the reciprocal

9913 on the left hand side we just have

x^2 and on the right hand side we'll

have 117 and that's the answer to our

question the area of the square is 117

cm squared let's put a box around it how

exciting

Cool Geometry Challenge

https://andymath.com/geometry-challenges/

For more geometry challenges try the above link!


hey guys this looks like a fun one the

small circles each have an area of one

and I want to know what's the area of

the larger Circle if you want to try it

on your own pause it right now cuz I'm

going to solve it in 3 2 1 first from

this tangent line let's draw this radius

and call it little r and we're given

that the area of this circle is 1 so

that means PI Little R 2 will be equal

to 1 this looks important let's put a

box around it now for the larger Circle

let's get rid of the question mark and

let's draw the radius here that touches

the tangent line and let's call it Big R

since we're trying to find out what is

the area of this larger Circle we

ultimately want to know what is pi Big R

2 and this is also important let's put a

box around it and to save room let's put

these side by side next let's connect

all the centers of the circles it's

going to be composed of a big radius a

little radius another little radius and

a third little radius and if we darken

this line right here we now have two

similar right triangles and the reason

we know the right triangles is because

radi are always at right Ang to the

tangent lines let's copy this down here

and let's focus on the triangles this

piece right here is going to be little r

plus little r which is two little r and

this piece right here is going to be Big

R plus little r and now we're ready to

set up a proportion let's do this little

triangle first we're going to do one

little r over 2 little r and then we can

focus on this larger triangle this piece

is Big R and this bottom piece is going

to be 2 little r plus Big R plus little

r which is the same thing as Big R + 3

little r so the other half of the

proportion is going to be be Big R over

Big R + 3 little r on this side the

little r and the little r are going to

cancel each other out so that leaves us

1/ 12 is equal to Big R over the

quantity Big R + 3 little r next we can

cross multiply 2 * Big R is 2 Big R and

then 1 * this is just going to be this

next we can subtract Big R from both

sides of the equation on the left hand

side 2 Big R minus 1 Big R is 1 Big R

and on the right hand side these cancel

each other out to give us three little r

so we end up with Big R is equal to 3

little r let's copy down here and let's

Square both sides on the left hand side

Big R S stays as Big R S but on the

right hand side this Square will

distribute to both of these terms giving

us 3^ 2 * little R 2 and 3^2 is equal to

9 let's copy it down again and let's

multiply both sides by pi on the left

side we have Pi Big R 2 and on the right

hand side we can move this over and

we'll have 9 pi little R 2 and we know

that PI Little R 2 is equal to 1 so in

the place of this PI Little R squ we can

plug in one and this multiply by 1 isn't

doing anything so we can get rid of it

and we have now solved for pi Big R 2 so

we can update the question mark up here

to be nine and now we have the answer to

the question I want to know what's the

area of the large Circle so that'll be 9

units squared let's put a box around it

how exciting

Fun Geometry Challenge

Puzzle by Catriona Agg

Such a cool one!


hey guys I made this video the other day

and I showed a shortcut to solve it a

bunch of people asked me to solve it the

hyper complicated way so let's do it so

it gives us some links here and it wants

to know what is the length of a b so

first let's find this length right here

this is a right triangle so if we call

this side C we can use Pythagorean

theorem we have 2 squared plus 4 squared

equals c squared 2 squared is four four

squared is sixteen four plus sixteen is

twenty and after we square root both

sides we get C is equal to square root

of 20. so let's change the C into square

root of 20. so next let's find this side

let's call it C and we could do 2

squared plus 2 squared equals c squared

both of these two squareds are four four

plus four is eight and we could square

root both sides to get C is equal to the

square root of eight and the exact same

thing is going to happen right here so

this will also be square root of 8.

let's focus our energy right here in

fact enhance let's focus our energy

right here this is a 90 degree angle

this right here is an isosceles right

triangle which means that's 45 degrees

and then for this last angle there's no

shortcut so let's call it d degrees and

let's use some trigonometry we know that

the tangent of D is equal to the

opposite over the adjacent well this

opposite side is 4 and the adjacent side

is 2 so this is going to become four

over two and four over two is equal to

2. then we can tangent inverse both

sides on the left hand side we have D

and it'll be equal to the tangent

inverse of two which is approximately

63.4 degrees so let's change this D to

63.4 degrees now we can get this last

angle let's call it n all four of these

will add up to 360. we can combine the

45 to 90 and the 63.4 to give us 198.4

after we subtract 198.4 from both sides

we get n is equal to 161.6 so let's

change that that was good so we can zoom

back out now we can connect this last

side and focus on this triangle right

here so this side right here is made up

of 2 square root of eights so let's call

it 2 square root of 8. this side is root

20 and this side is unknown let's call

the unknown side C and in fact this

unknown side is the a b that we wanted

to solve from before or so we're going

to use a thing called law of cosines

here it is right here you'll recognize a

portion of it as a squared plus b

squared equals c squared that's the

Pythagorean theorem but since it's not a

right triangle we have this little extra

portion so for both of these A's I'm

going to plug in 2 square root of 8 and

for both of these B's I'm going to plug

in square root of 20. and then for this

angle Theta we're going to plug in 161.6

this Square goes to both the 2 and the

square root of 8. 2 squared is equal to

4 and square root of 8 squared is equal

to 8 and then 4 times 8 is equal to 32

and then square root of 20 squared will

give us 20. next we can plug all of this

stuff into a calculator and it will give

us a negative 48. there it is right

there and then minus negative 48 is

positive 48 and 32 plus 20 plus 48 is

equal to 100 and after we square root

both sides we get C is equal to 10. so

we can update this C to be 10. and then

returning back to the original problem

we have that right here the C is equal

to 10. how exciting

Hyper Complicated Method

https://youtube.com/shorts/fmDUseL-YUA
I solved this with a shortcut last time. Hopefully you guys enjoy this hyper complicated method!


hey guys let's solve for x first we have

this equilateral triangle right here all

triangles have 180 degrees so if all

three angles are the same each angle is

going to be 60 degrees next let's look

at this regular pentagon the formula for

the sum of the interior angles of a

polygon is n minus 2 times 180 where n

is the number of sides of the polygon

for a pentagon it has five sides and

this will give us the sum of all the

interior angles if we want to know how

big each angle is we're going to divide

by 5 and this will give us 108 degrees

so this one's 108 this one's 108 this

one's 108. this one down here we already

have the 60s so we're going to need

additional 48 and same thing for this

one an additional 48. but next let's

look at this right triangle right here

if we added these three angles they're

all going to add up to 180 degrees the

48 Plus 90 is 138 and if we subtract 138

from both sides we get x equals 42 and

that is our answer for X how exciting

Fun Polygon Math #shorts

https://andymath.com/angles-of-polygons/

#shorts


hey guys this looks like a fun one it

says find the area of the blue rectangle

and the rectangle is twice as wide as it

is tall and it hits the semicircle in

these various places if you guys want to

try this on your own pause it right now

cuz I'm going to solve it in 3 2 1 first

let's connect this point to this point

and I want to focus on this triangle

right here first off it's a right

triangle because this is a rectangle and

second it's isos because these two sides

are congruent that means it's a 45 4590

triangle and then since this angle is 45

we now know this angle this plus this is

a linear pair which means they have to

add to 180° so this angle will be

135° next let's finish up the circle and

I want to focus on this

135° this is called an inscribed angle

and if we look at it it's cutting the

circle into two parts this red Arc and

this blue arc a cool property about

inscribed angles is they are exactly

half of the arc that they subtend so

this red Arc is going to be double the

135 de so it'll be 270° and then since

the red Arc is 270° that leaves us 90°

for the blue arc for the next step let's

draw this radius here and this radius

here and let's focus on this 90° Arc

this angle right here is called a

central angle and central angles are

equal to the subtended arcs so this is

going to be a 90° angle and we usually

notate that as a box in the corner and

since both of these are radi they both

have a length of five so if we connect

this here we have another isoceles right

triangle if we use a shortcut for 45

4590 triangles we can see this green

side is equal to 5 < tk2 and here's the

steps of the Pythagorean theorem to

verify it and we just did all of this

work just so we could get this green

length right here we can go back to here

and let's bring back our rectangle

looking at our rectangle if we call the

shorter side x this longer side is going

to be 2X and we have formed another

right triangle let's focus on that and

now we can do Pythagorean theorem to

solve for x we have x^2 + 2x^2 = 5 <

tk22 we can bring down the x^2 and then

this squar will go to both the 2 and the

X and this two will go to both the 5 and

the square root so we have x^2 + 4 x^2 =

25 * 2 x^2 + 4x^2 is equal to 5 x^2 and

25 * 2 is equal to 50 after we divide

both sides by 5 we get x^2 is equal to

10 and after we square root both sides

we get X is equal to the squ < TK of 10

and now we know the shorter side of the

rectangle this x can change into < TK 10

and this 2x can change into 2un 10 and

now we can find the area of the

rectangle it's equal to base time height

where the base is equal to 2un 10 and

the height is equal to < TK 10 and

that's equal to 2 * 10 which is equal to

20 let's give it a label of square units

and put a box around it the area of the

blue rectangle is 20 square units how

exciting

Tricky Geometry Challenge

https://andymath.com/geometry-challenges/

For more geometry challenges, try the above page!


hey guys this looks like a fun one we're

given two blue squares and it wants to

know what is the area of the red

triangle and I'm noticing something it's

not giving us anything about the

distance between the two squares so it

could realistically be any one of these

scenarios so I'm guessing every single

one of these red triangles have the

exact same area if you want to try it on

your own pause it right now because I'm

going to solve it in 3 2 1 so first

let's look at the formula for the area

of a triangle area equals 12 base time

height looking at this triangle we don't

know the base so let's call it X and

then for the height of this triangle we

need something at right angles this

right here would be at right angles

because this is a square so this is also

right angles this right here will be the

height of our triangle we don't know

that either so let's call it y so now

let's update this formula the base of

our triangle's X so let's call this x

and the height of our triangle is y so

let's call this y so this is what we're

trying to figure out so next let's use

the areas of the blue squares if this

one has an area of 16 each side will be

equal to four and if this one has an

area of 36 each side will be equal to

six and then from here just like this

was a right angle right here this one

would also be a right angle and then if

we look at this angle right here let's

call it Theta this angle will also be

Theta because these are vertical angles

and vertical angles are always congruent

so let's take this whole thing and copy

it down here and I want to focus on

these white triangles Let's Line them up

like this these are similar right

triangles we know that because because

they have two pairs of congruent angles

so that means we can set up a proportion

we can say the base of this triangle 4

over the base of this triangle X is

equal to the height of this triangle y

ided by the height of this triangle 6

and from here we can cross multiply x *

Y is equal to XY and on the other side 4

* 6 is equal to 24 and now we have x * y

that's all we really needed up here so

we can scoot this up here and in the

place of x y let's plug in 24 1 12 * 24

is equal to 12 and let's give it a label

of me squared this is the answer to our

question the area of the red triangle is

12 M squar and that'll be true for all

of these red triangles they all have an

area of 12 M squar how exciting now I'm

trying to put together an end card

here's a button to subscribe and here's

a nice video you can check out how

exciting

2 Squares 1 Triangle

I hope you guys love this one. It's neat how the areas are all the same!

If anybody is reading this description, feel free to comment "It Looks like a panda." or "Is that the panda from 2016?" Or something along those lines. If enough people do it, I think it could be funny and confusing. :)


how do we solve for x in this right

triangle we're going to use pythagorean

theorem pythagorean theorem says a

squared plus b squared equals c

squared a and b are the two smaller

sides and c would be the largest side

so to plug in the values we get 3

squared plus 4 squared equals c squared

3 squared is 3 times 3 which is 9

4 squared is 4 times 4 which is 16

and we'll leave c squared as c squared 9

plus 16 is 25

and that still equals c squared and then

to solve for c we will square root both

sides

square root of 25 is 5 so we get c

equals 5.

that's the answer to the question for

more math videos please check out

andymath.com

have a good one i'll talk to you soon

bye

Pythagorean Theorem Fast Example #Shorts

𝐋𝐈𝐍𝐊𝐒 𝐓𝐎 𝐑𝐄𝐋𝐀𝐓𝐄𝐃 𝐓𝐎𝐏𝐈𝐂𝐒
𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐞𝐚𝐧 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/pythagorean-theorem/
𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐞𝐚𝐧 𝐓𝐫𝐢𝐩𝐥𝐞𝐬: https://andymath.com/pythagorean-triples/

𝐄𝐪𝐮𝐢𝐥𝐚𝐭𝐞𝐫𝐚𝐥 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬: https://andymath.com/equilateral-triangles/

𝐈𝐬𝐨𝐬𝐜𝐞𝐥𝐞𝐬 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬: https://andymath.com/isosceles-triangles/

𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐀𝐧𝐠𝐥𝐞 𝐒𝐮𝐦 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/triangle-angle-sum-theorem/

𝐒𝐢𝐝𝐞-𝐒𝐩𝐥𝐢𝐭𝐭𝐞𝐫 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/side-splitter-theorem/

𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐏𝐫𝐨𝐩𝐨𝐫𝐭𝐢𝐨𝐧𝐚𝐥𝐢𝐭𝐲 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/triangle-proportionality-theorem/

𝐏𝐚𝐫𝐚𝐥𝐥𝐞𝐥 𝐋𝐢𝐧𝐞𝐬 𝐏𝐫𝐨𝐩𝐨𝐫𝐭𝐢𝐨𝐧𝐚𝐥𝐢𝐭𝐲 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/parallel-lines-proportionality-theorem/

𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐌𝐢𝐝𝐬𝐞𝐠𝐦𝐞𝐧𝐭 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/triangle-midsegment-theorem/

𝐄𝐱𝐭𝐞𝐫𝐢𝐨𝐫 𝐀𝐧𝐠𝐥𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/exterior-angle-theorem/

𝐒𝐩𝐞𝐜𝐢𝐚𝐥 𝐑𝐢𝐠𝐡𝐭 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬 𝟑𝟎-𝟔𝟎-𝟗𝟎 𝟒𝟓-𝟒𝟓-𝟗𝟎: https://andymath.com/special-right-triangles/

𝐇𝐢𝐧𝐠𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/Hinge-theorem/

𝐒𝐢𝐦𝐢𝐥𝐚𝐫 𝐑𝐢𝐠𝐡𝐭 𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞𝐬: https://andymath.com/geometric-mean-similar-right-triangles/

Visit any of the the above links for more practice problems, notes, solutions and videos for all these related topics. For even more math topics, visit https://andymath.com/.

#Shorts


hey guys so we have a circle i want to

find the area of this red region of the

circle so let's call that x in order to

solve for x first i'm going to find this

blue sector right here and then i'm

going to subtract this orange triangle

right here and if we call the large

sector a and the triangle b we know that

a minus b is going to give us x first

let's solve for a if this is 60 degrees

right here we know this angle right here

is going to be 120 degrees the area of

an entire circle is pi r squared and

this is only 120 out of 360 degrees of

an entire circle so we'll say the

proportion 120 over 360. and 120 over

360 is the same thing as one third and

then the radius here is six so i'm going

to plug in 6 for r 6 squared is 36 and

then 1 3 of 36 is 12. so 12 pi is our

answer and we can even move it up here

to the a next let's look at this orange

triangle this is the same angle from

above so it's still 120 and now we don't

really need the 60 anymore and this side

of the triangle right here is a radius

as well so it's also going to be six

because they're both coming from the

center to the edge of the circle now we

don't even need the circle anymore

enhance and let's flip it around and

bisect this 120 degrees that's going to

give us 60 degrees for both sides of

this let's focus on this left triangle

and then we know it's a right angle

anytime you bisect that top angle in an

isosceles triangle it always meets at

right angles and this last angle is

going to be 30 degrees now we have a 30

60 90 triangle this is a special type of

triangle here's the notes right here on

andymath.com the side opposite the 30

degrees is always half the hypotenuse

this hypotenuse is 6 so this side right

here is going to be 3 and then the side

opposite the 60 is always square root of

3 times the smallest side so this right

here is gonna be three root three since

these two are similar triangles this is

also three root three and when we

combine them we get six root three let's

clean up the stuff we don't need now we

need to find the area of this triangle

it's gonna be one half base times height

the base is this six root three and the

height is three so six times one half is

three and then three root three times

three is nine root three so that's our

value for the triangle so let's change

these up here we'll make the a into the

twelve pi and we'll make the b into nine

root three and now we have the answer to

the question the area of this red region

is 12 pi minus 9 root 3.

Geometry Challenge Problem

For more math fun, check out andymath.com!


hey guys this looks like a fun one it

has two circles inside of a square and

it wants to know what is this angle this

angle is created with these three points

of tangency if you want to try it on

your own postive right now cuz I'm going

to solve it in 3 2 1 first I want to get

this question mark out of the way and

then we'll construct the common tangent

line of these two circles let's label

this angle X De and this angle y de now

our goal is to find what is x + y next

let's label the centers of both of these

circles this radius will be at right

angles to this side right here and

that's because the tangent line is

always at right angles to the radius

that intersects at that point here's the

theorem right here and then let's

connect this other radius and that'll be

at right angles to this tangent line

let's find the size of this angle right

here we know that y plus this angle has

to equal 90° that means this angle is

the same thing as 90 - y de and let's

look at this triangle more closely this

side and this side are both radi of the

circle since these two sides are equal

that means this is an isoceles triangle

and when you an isoceles triangle these

base angles are also equal so since this

one is 90- y that means this one will

also be 90- Y and now looking up here

since this is 90 minus y and this whole

thing is 90 that leaves y for this angle

right here so we've shown that both of

these angles are equal to Y de and now

let's go back up here and we're going to

do basically the same thing if we

connect this radius it'll be at right

angles to this tangent line and now to

find this angle if this is X that leaves

90- X for this angle and if we complete

complete this radius this makes another

Isles triangle and that's because this

radius is equal to this radius and in

all Isles triangles the base angles are

congruent so if this is 90 - x that

means this one is also 90 - x since this

is 90 - x that leaves X for this angle

right here next let's get rid of this

piece of the tangent line and we're

going to combine the angle X and the

angle y to get X Plus Y and then since

this whole thing is a square that means

this angle right here is a 90° Angle now

we have a really cool looking

quadrilateral the sum of the interior

angles of any quadrilateral is equal to

360° that means if we add up all these

angles we'll get

360° so x + x + y + y + 90 is equal to

360 and then from here we can subtract

90 from both sides of the equation we

end up with x + x + y + y is equal to

270 and then y + y is equal to 2 y and x

+ x is equal to 2x next we can divide

both sides of the equation by two on the

left hand side we have X + y and on the

right hand side 270 / 2 is 135 so now in

the place of this x + y we can plug in

135° and this 135 is in the same place

as the question mark so we know that the

question mark is equal to

135° this is the answer to our question

let's put a box around it how exciting

hey guys it's a couple days later and I

was editing the video and I realized I

missed something that I wanted to share

with you it doesn't matter how big the

circles are the circle can go all the

way up to here till it hits the top or

it can go all the way down here till

this one hits the side the angle will be

135° in all these scenarios and there

was another detail I wanted to share

there's another reason why these two

angles match and these two angles match

if you have a circle and you extend two

tangent lines until they intersect these

two sides will always be congruent so

that means this forms an isoc triangle

and these two base angles would be

congruent so if you isolate this

triangle right here these two sides are

equal making these two base angles equal

or on this other side if you isolate

these in extend them these two sides

would be equal making these base angles

equal I hope you guys like these two

bonus insights how exciting

What's the angle? (With Bonus Footage)

Watch to the end, the bonus footage is super cool!


hey guys this looks like a fun one this

is another Katrina a puzzle it's given

us that this is made up of five squares

and it wants to know what's the total

shaded area if you want to try this on

your own pause it right now cuz I'm

going to solve it in 3 2 1 first I want

to look at these three angles they're

going to add up to

180° and that's the same thing as a

triangle so if we look at this triangle

on the left here this right angle can be

brought here and the purple angle can be

brought there these three angles will

also add up to 180° and then we can do

the same thing for this triangle over

here this blue right angle can go here

and then that means that this last angle

would have to be the green angle so

these two triangles have three congruent

angles that means the two triangles are

similar next let's focus on this literal

triangle let's label this side a this

side B and this hypotenuse C well this C

is also one of the sides of these

squares so another side of the square

would also be C in fact all of these

sides would be equal to C I don't think

there's anything else I know for now so

let's pull out these triangles and if we

rotate this one now the two similar

triangles are oriented the same let's

change these four C's into 4 C and now

since these are similar triangles we can

find the scale factor from C to 4 C that

means our scale factor is four to get

the base we're going to multiply a * 4

to get 4 a and to get this height we're

going to multiply B * 4 to get 4 B and

now we know these sides so we can update

this figure let's rotate this triangle

back and you can see that the 4B goes

there and the 4 a goes there so that

looks pretty good now we have three

different ways to measure the side of

this larger Square let's set them equal

to each other so we'll have this side a

+ 4 b equal to this side 4 a and then we

can also do this side b + 13 equal to

this side 4 a and now we have two

equations and two variables so we can

solve for both A and B let's start with

this one we can subtract a from both

sides and that gives us 4 B is equal to

3 a and next we can divide both sides by

four and that will give us B is equal to

34s a and this looks important so let's

put a box around it next let's look at

this equation let's copy it down and in

the place of the B we're going to plug

in 3/4s a and then we want to combine

like terms so let's subtract 34s a from

both sides of the equation on the right

hand side these are going to cancel each

other out and on the left hand side

we're going to do 4 A minus 34s a in

order to subtract these let's give them

a common denominator let's rewrite this

4 as 164s and then we're going to do

164s a - 3 4 a which is equal to 134s a

now to get the a all by itself we'll

multiply both sides by the reciprocal

413 on the left hand side these are

going to cancel each other out and on

the right hand side these two 13s will

cancel each other out so we end up with

a is equal to four and this looks

important so let's put a box around it

now going back to our previous box we

know B is equal to 34s of a a is equal

to four so in the place of this a let's

plug in four and then when we multiply

this 34s * 4 four the fours are going to

cancel each other out we end up with B

is equal to 3 in the place of both of

these A's we can plug in four and then 4

* 4 is equal to 16 and in the place of

both of these B's we can plug in three

and 4 * 3 is equal to 12 so now let's

focus on this little triangle right here

this is a special triangle it's called a

Pythagorean triple anytime a right

triangle has legs three and four the

hypotenuse will always be five and since

this side is five that means this side

is also five and the area of this Square

would be 5 * 5 which is 25 and 25 would

be the area of all these squares so if

we want the area of the Shaded region

we're going to do 4 * 25 and 4 * 25 is

100 and that is the answer to our

question let's put a box around it how

exciting

Fun Math Problem

Another Catriona Agg Problem. I thought this was a fun one. Let me know if you know another way to solve it!


hey guys this looks like a fun one we

have 12 to power of x - 4 is equal to

the < TK of 2 to^ 2x - 4 if you want to

try this on your own pause it right now

cuz I'm going to solve it in 3 2 1 so

first let's copy down the exponents and

let's leave the parentheses blank we

want to express these two with the same

base 1/2 and S > of two both contain a

two so for both of them we're going to

do two to some power for this on the

left we're going to use a negative

exponent so if you have X to 1 that's

equal to 1 /x that takes the X on top

and puts it on bottom or if you have 1X

to the1 that takes the X on bottom and

puts it on top so regardless of which

one you start with it takes the

reciprocal to make it look like our

situation we can change these X's into

twos and now we have 2 the1 is equal to2

well that means that we can rewrite this

1/2 as 2 to the1 so we haven't changed

anything on the left hand side we've

only changed how we expressed it so now

on the right hand side side we're going

to use rational exponents it says x 1 /

n is equal to the N root of x well in

this particular example it's the square

root so we're going to do the square

root of x and the square root has an

implied index of two so that means this

n is going to be equal to two so the

square < TK of two is the same thing as

2 to the 1/2 power that means this

Square < TK of two is the same thing as

2 to the 1/2 power and now we can move

on to the next step we're doing the

power of a power these are the notes

right here if you have x to the a print

pareses to the B that's the same thing

as X the a * B so for the left hand side

2 -1 x - 4 is the same thing as 2-1 * x

- 4 and then if we distribute this 1 to

both terms this x - 4 becomes x + 4 and

then we're going to do the same thing on

the right hand side this 1/2 is going to

multiply by the 2x - 4 and that 1/2 will

cut both of these terms in half so 2x -

4 is going to change into 1 x - 2 so we

have 2 to this power is equal to 2 to

this power now that we've made the bases

match we can drop the base Drop the

Base and we can set the exponents equal

to each other and solve for x we have a

negative X and a positive X let's move

this x to this side we'll do that by

adding X to both sides and I want to

move this two to the other side so we'll

do that by adding two to both sides on

the left hand side these two x's will

cancel each other out leaving us with

six and on the right hand side these two

twos will cancel each other out leaving

us with X+ X or 2x after we divide both

sides by two we get 3 is equal to X and

that is the answer to our question xal 3

let's put a box around it how exciting

Exponential Drop the Base

I hope you guys love it!


hey guys this looks like a fun one I

think we would call it a radical

equation first step to solve this I want

to get rid of this radical right here so

let's copy this down and square both

sides on the right hand side this square

and square root are going to cancel each

other out giving us just the inside

stuff and we could just copy down the

left hand side normally with these we

would multiply this out but this has a

matching one right here so we're just

going to subtract that from both sides

the left hand side is just going to be 0

and the right hand side we bring down

this other stuff and smush them together

let's bring this to the other side and

add 8 to both sides these cancel so the

left hand side is just this and the

right hand side is going to be eight now

to get rid of this square root we're

going to square both sides again the

left hand side is just going to be this

Inside Stuff x minus 4 and the right

side is going to be 64. add 4 to both

sides and we get x equals 68 how

exciting

Radical Equation Challenge #shorts

#shorts


guys this looks like a fun one this was

posted by my math Solutions we've got

three rectangles and it wants us to know

the area of this rectangle in the middle

so first let's focus on this gray

rectangle area of it is going to be base

times height this two centimeters only

goes to here so let's make the rest of

this X and since the bottom is X plus 2

the top would also be X plus two and we

could change the base into X plus two

now for the height we know this part is

three and for the unknown part let's

call it Z and since this is three and

this is only two we need to make this Z

plus one so that both Heights are equal

to Z plus three and now let's bring

these rectangles back let's get rid of

the question mark and bring this

rectangle inside here and let's focus on

this rectangle here this 26 centimeter

squared refers to this backwards L shape

but we don't know the gray area so we'll

call that a we know that a plus 26 is

going to be equal to the area of the

whole rectangle which is 7 times 6. 7

times 6 is 42 and after we subtract 26

from both sides we find out that a is

equal to 16. and a is this rectangle

made up of x times e so really we found

out that x times Z equals 16. and that's

really what's most important here and

now let's extend this other rectangle

and focus on it if we call the area of

this region B we know that b plus 36 is

going to be equal to 7 times 8. 7 times

8 is 56 and after we subtract 36 from

both sides we get D is equal to 20. and

this B is also equal to the base times

the height or in other words x times Z

plus one and that's really what's

important here now we have a system of

equations we have two variables and two

equations let's get X by itself for both

of these for this top one we'll divide

both sides by Z these Z's will cancel

each other out leaving us with just x

equals 16 over Z and for this one let's

divide both sides by Z plus one and then

these will cancel leaving us with just x

equals a 20 over Z plus one so we have

found that X is equal to 16 over Z well

in the place of this x down here I'm

going to plug in 16 over Z and now we

have a proportion we can solve this by

cross multiplying we have 16x times Z

plus 1 is equal to this 20 times Z and

then we can distribute the 16. 16 times

Z is 16z and then 16 times 1 is 16. and

on this side we have 20 Z and then we

can subtract 16z from both sides on the

left hand side these are gone so we just

have 16 and on the right hand side we

have 4z we can divide both sides by 4

and we have 4 equals z and then we can

rewrite that as Z equals four so since

we know the Z equals 4 I'm going to plug

in 4 for this Z right here and 16

divided by 4 is 4. so we know x equals 4

so we can plug in 4 for this X and we

know Z equals 4 so we can plug in 4 for

this Z here 4 plus 2 is equal to 6 and

here 4 plus 3 is equal to 7 and 6 times

7 is equal to 42. we have now found the

value of our question mark it's 42. so

for this question mark let's put in 42

centimeters squared that is the answer

to our question and it's already got a

box around it how exciting

Area Challenge Problem

I thought this was a fun one. I hope you guys like it!


hey guys this looks like a fun one it

wants us to solve for x and it gives us

this triangle that is not to scale we

have angle measures of a de 2 a de and a

de it gives us the base is 5 cm and it

wants us to find this longest side if

you want to try on your own pause it

right now because I'm going to solve it

in 3 2 1 so first thing we know the sum

of the interior angles of any triangle

adds to 180° so a + 2 a + a will equal

180 and on the left hand side a + 2 a +

a simplifies to 4 a and that's still

equal to 180 now to get a by itself we

can divide both sides by four 4 a ID 4

is equal to a and 180 / 4 is 45 so we

now know that a is equal to 45 and every

single time I do one of these videos

someone tells me that a is actually

equal to

45° but if you notice with the degree

symbol up here this is a degrees so if a

was 45° that would make this 45° de so

we know it's not 45°

so a is only 45 and that would make this

angle 45° I wonder if someone will still

comment that in the comments even after

I said this but a is for sure equal to

45 and then this a degrees up here will

change into 45° and this 2 a degrees

will change into 90° so we know for sure

this image is not to scale because 90°

would be perpendicular right here in

fact we can fix this and this is what it

should look like and now that we fix the

triangle let's get rid of the knot and

now this is image to scale that looks

nice now in any isos triangle that's any

triangle with two angles the same the

sides opposite those congruent angles

will be congruent since this side is 5

cm this side will also be 5 cm so right

now there's two different ways we can

solve this we can use the notes for a 45

4590 triangle and this tells us that the

hypotenuse is whatever this side is

times the < tk2 so if this side is 5 cm

the hypotenuse will be 5 < tk2 2 cm or

the other method is to do Pythagorean

theorem it's a^ 2 + b^2 is equal to c^2

so we can do 5^ 2 + 5^2 is = to x^2 on

the left hand side 5^2 is equal to 25

and 5^2 is equal to 25 and 25 + 25 is

equal to 50 and then to solve for x we

can square root both sides of the

equation the rooot 50 can be broken down

into rooot 25 * 2 and then we can give

each of these their own square root and

the square root of 25 is equal to 5 and

when taking the square < TK of x^2 since

we know this is a distance it must be

positive so this simplifies into X and

now we have X is equal to POS 5 < tk2

this is the answer to our question let's

put a box around it how exciting hey

guys I was just editing this and I

realized right at the end here I want to

point out that X does not equal 5 < tk2

cm it's 5 tk2 without the centim label

and that's because that hypotenuse would

end up being 5 < tk2 cm s I know this

one isn't as crazy as some of my other

videos but I really like the fact that

the image is not to scale and I really

like the discussion of the labels for

the variables how you don't want to make

it 45 degrees degrees so I thought this

would make a fun video I hope you guys

liked it and if you like to see more

math content I'm going to put a

subscribe button somewhere up here how

exciting

Triangle Not to Scale


so we're want to find the area of this

shaded sector so this is a circle that's

been cut up and we're finding how much

area is this shaded stuff so first what

is the area of the entire circle well

the area of a circle is < R 2 but this

isn't the entire circle how much of the

circle is this an entire circle would be

360° and this is 172 of those 360° so

we're just going to multiply these two

things by each other now we still need

to plug in for R for the P pi r 2qu and

this is the r the 10 is the radi

10^ 2 is 100 * < * the 172 over 360 so

now if we plug this into a calculator

get approximately 1

15.1 and they don't give us a unit so we

can say unit squared CU they don't tell

us feet or inches or anything like that

and that's the answer to the question

comment below on what I should solve for

next I'll talk to you soon bye Andy

maath

Area of a Sector #andymath  #math  #maths  #sat  #tutorials  #tiktokmath  #mathhelp

Please Subscribe, and don't forget to hit the bell icon so you don't miss next video! Also, please feel free to share this video with your friends and family. Let learn Math together with me, Charlie.
Here is the playlist of the videos. [ There is so much more to discover in this playlist ]


hey guys you got to watch this video

this guy hits a home run and then throws

the bat

up and it stays up for really long

time I got a timer going right

here and it stays up for 11.83 seconds

so let's figure out how high did the bat

reach we can visualize the path of the

bat as a prao where the horizontal axis

can be t for time in seconds and the

vertical axis can be H for the height

that the bat goes so we're interested in

this point right here how high did the

bat reach so this is the formula we're

going to use the height is equal to -4.9

T ^2 plus the initial velocity time time

plus the initial height so we'll call

the initial height zero because the top

of the bat is really close to the ground

as he throws it and we don't know the

initial velocity but we do know at the

time 11.83 seconds the height once again

was zero so we can use this information

to find the initial velocity let's copy

down our equation let's give our our El

some space let's plug in zero and in the

place of these T's let's plug in 11.83

and in the place of the H let's plug in

zero 4.9 * 11.83 2 is-

68575 and let's copy down everything

else let's add 68575 to both sides we

end up with 68575 is equal to 11.83 *

the initial velocity after we divide

both sides by 11.83 we get the initial

velocity is

57.97 and that's measured in meters per

second and if you're curious that's

about

28.6n km/ hour or 129. 68 mph so that's

pretty fast we can update the initial

velocity up here to

57.97 and now we don't need this stuff

anymore now that we know our initial

velocity here's our new equation

representing the height so now to find

this maximum height we first need to

plug in this point right here which we

know is

11.83 the axis of symmetry will go

through that maximum point we can find

this Axis of Symmetry by taking the aage

aage of these two x intercepts and

that's going to give us

5915 and so now we take our equation

again plug in 5915 for T and we get our

height of 171.5 M and that is the answer

to our question how exciting

How high was the throw?

https://andymath.com/quadratic-word-problems/

For more math involving parabolas, check out this page!


hey guys welcome to this probability

question you flip a coin

two times what is the probability you

get two heads the best way to do this is

to

show all the possible outcomes we're

gonna flip the coin twice so we have our

first toss and our second toss so our

first toss can be a head and our second

toss can be a head or our first toss

could be a head

and our second toss could be a tail our

first toss could be a tail and the

second could be a head

or the first toss could be a tail and

the second could be a tail this

should represent all possible outcomes

each of these has an equal probability

of happening

so there's a one and four chance so now

was it probably of getting two heads

well this

is the one scenario where we get two

heads so the answer is one

fourth as a percent that would be 25 and

that's the answer to the question if you

flip a coin twice the problem of you

getting two heads in a row

is 25 comment below on what you'd like

me to solve next and i'll talk to you

soon

bye

Probability 2 Coins Fast Example #Shorts

𝐋𝐈𝐍𝐊𝐒 𝐓𝐎 𝐑𝐄𝐋𝐀𝐓𝐄𝐃 𝐓𝐎𝐏𝐈𝐂𝐒
𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲- 𝐂𝐨𝐢𝐧 𝐓𝐨𝐬𝐬𝐞𝐬: https://andymath.com/probability-coin-tosses/
𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲- 𝐌𝐚𝐫𝐛𝐥𝐞𝐬: https://andymath.com/probability-marbles/

𝐅𝐚𝐜𝐭𝐨𝐫𝐢𝐚𝐥𝐬: https://andymath.com/factorials/

𝐂𝐨𝐦𝐛𝐢𝐧𝐚𝐭𝐢𝐨𝐧𝐬 𝐚𝐧𝐝 𝐏𝐞𝐫𝐦𝐮𝐭𝐚𝐭𝐢𝐨𝐧𝐬: https://andymath.com/combinations-and-permutations/

𝐄𝐱𝐩𝐞𝐫𝐢𝐦𝐞𝐧𝐭𝐚𝐥 𝐚𝐧𝐝 𝐓𝐡𝐞𝐨𝐫𝐞𝐭𝐢𝐜𝐚𝐥 𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲: https://andymath.com/experimental-and-theoretical-probability/

𝐂𝐨𝐦𝐩𝐥𝐞𝐦𝐞𝐧𝐭 𝐨𝐟 𝐚𝐧 𝐄𝐯𝐞𝐧𝐭: https://andymath.com/complement-of-an-event/

𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲 "𝐀𝐭 𝐋𝐞𝐚𝐬𝐭 𝐎𝐧𝐞": https://andymath.com/probability-at-least-one/

𝐌𝐮𝐭𝐮𝐚𝐥𝐥𝐲 𝐄𝐱𝐜𝐥𝐮𝐬𝐢𝐯𝐞 𝐄𝐯𝐞𝐧𝐭𝐬 (𝐃𝐢𝐬𝐣𝐨𝐢𝐧𝐭): https://andymath.com/mutually-exclusive-events/

𝐂𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧𝐚𝐥 𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐢𝐞𝐬: https://andymath.com/conditional-probabilites/

𝐌𝐚𝐫𝐠𝐢𝐧𝐚𝐥 𝐅𝐫𝐞𝐪𝐮𝐞𝐧𝐜𝐢𝐞𝐬 𝐚𝐧𝐝 𝐃𝐢𝐬𝐭𝐫𝐢𝐛𝐮𝐭𝐢𝐨𝐧𝐬: https://andymath.com/marginal-frequencies-and-distributions/

𝐓𝐰𝐨-𝐖𝐚𝐲 𝐓𝐚𝐛𝐥𝐞𝐬 (𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲): https://andymath.com/two-way-tables-statistics/

𝐓𝐫𝐞𝐞 𝐃𝐢𝐚𝐠𝐫𝐚𝐦𝐬: https://andymath.com/tree-diagrams/

𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲 𝐰𝐢𝐭𝐡 𝐃𝐢𝐜𝐞: https://andymath.com/probability-with-dice/

𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲- 𝐑𝐨𝐮𝐧𝐝 𝐓𝐚𝐛𝐥𝐞: https://andymath.com/probability-round-table/

𝐏𝐫𝐨𝐛𝐚𝐛𝐢𝐥𝐢𝐭𝐲- 𝟓 𝐜𝐚𝐫𝐝 𝐏𝐨𝐤𝐞𝐫 𝐇𝐚𝐧𝐝𝐬: https://andymath.com/probability-poker-hands/

𝐕𝐞𝐧𝐧 𝐃𝐢𝐚𝐠𝐫𝐚𝐦𝐬: https://andymath.com/venn-diagrams/

Visit any of the the above links for more practice problems, notes, solutions and videos for all these related topics. For even more math topics, visit https://andymath.com/.

#Shorts