guys this looks like a fun one it was

posted by my math Solutions so let's

solve it we have two rectangles in a

triangle that we know the area and we're

trying to find the area of this last

triangle I think there's a lot of

different ways to solve this but I want

to share with you what I ended up doing

I called this side x this side y this

side Z and this side W let's start with

this rectangle x times Z is equal to 15.

and after we divide both sides by X we

get Z is equal to 15 over X and if we go

to this rectangle y times W is equal to

12. divide both sides by y now we get W

is equal to 12 over y so now let's

update the Z to make it 15 over X and

let's update the W to make it 12 over Y

and then let's get rid of this stuff now

let's look at this triangle the area of

a triangle is one half base times height

so this is going to be one half x times

y equals nine and after we multiply both

sides by 2 on the left hand side we get

x y and on the right hand side we get

18. this looks important so let's put a

box around it now we're ready for the

question mark since it is the area of a

triangle it's going to be equal to one

half base times height if we turn our

heads sideways we can think of the base

is equal to 15 over X and the height is

12 over Y and now we're just multiplying

fractions 1 times 15 times 12 is equal

to 180 and 2 times x times Y is equal to

2xy and we could simplify this 180

divided by 2 is equal to 90 and then we

still have the X Y earlier we figured

out that x times y was equal to 18 so

I'm going to change this into an 18 and

then 90 divided by 18 is equal to 5 and

it's going to be centimeters squared and

that's the answer to our question the

area of this blue triangle is equal to 5

centimeters squared let's make it bigger

and put a box around it how exciting

Area of Blue Triangle

I thought this was a fun one. There might be a better way to solve it, this was my immediate first attempt.


you guys haven't seen this all over the

place someone even emailed it to me

asking me to solve it so let's just

solve it so this thing is a square

they're telling us that right there now

we have three triangles that we know the

areas and then we have this triangle we

don't know the area since it's a square

all the sides are congruent so I'm going

to call each of the sides X so this

green portion right here I'm going to

call it Y which means this yellow

portion would be the whole thing x minus

y but let's call this blue portion Z and

that would make this yellow portion

equal to x minus Z so let's look at this

green triangle the area of a triangle is

one half base times height so we can say

one half x times Y is equal to the area

which is four and then if we look at

this blue one it'd be one half the x

times Z is equal to two and then for

this last triangle it'd be one half this

thing times this thing equals three I

don't really like fractions so let's

multiply both sides of all three of

these equations by two all these

one-halves will go away and the right

sides are just going to double we'll

have eight four and six so now let's

focus on these top two equations let's

divide both sides of both equations by X

this this X over X cancels this X over X

cancels and these just become 8 over X

and 4 over X so now let's do some

substitution for this Z I want to plug

in 4 over X and for this y I want to

plug in 8 over X so now on the left hand

side let's distribute x times x is x

squared x times negative 8 over X is

negative x times 8 over X these two x's

will cancel so it's just negative 8. and

then this times this is negative x times

4 over X but these X's cancel again to

give us negative 4. and this times this

is a negative times a negative which is

positive it's going to be a fraction or

4 times 8 is 32 and x times x is x

squared and this is all still equal to

6. well negative 8 negative 4 can

combine to negative 12. I don't really

like fractions so let's get rid of this

x squared in the denominator by

multiplying both sides by x squared x

squared times x squared is X to the

fourth x squared times negative 12 is

negative 12x squared x squared times 32

over x squared is just going to be 32

and then 6 times x squared is 6X squared

and then we get subtract 6X squared from

both sides which will make this side 0

and this side would be negative 18x

squared and now we want to factor this

since it's set equal to zero so x to the

fourth here means we want an x squared

and an x squared and then we want two

numbers that add to negative 18 and

multiply to 32. that's going to be

negative 16 and negative 2. this plus

this is negative 18 and this times this

is positive 32. now since this times

this equals zero we can set each of

these equal to zero we can add 16 to

both sides or add 2 to both sides

negative 16 and positive 16 cancel and 0

plus 16 is 16. negative 2 plus 2 cancel

and 0 plus 2 is 2. let's smush things

together and take the square root square

root of x squared is the absolute value

of x and square root of 16 is 4 square

root of x squared is the absolute value

of x and the square root of 2 is square

root of 2. since X represents the length

it must be positive so these absolute

values are not necessary so we know X is

either 4 or square root of 2 and let's

check them so if x is equal to 4 the

area of the square would be equal to 4

Square squared and if x is square root

of 2 then the area would be square root

of 2 squared 4 squared is equal to 16

and square root of 2 squared is equal to

2. so the area of this green triangle

right here is four there's no way the

area of the whole Square can be just two

so we can disregard that value so we

know that the area of the entire square

is equal to 16. so we can add this 2

plus the 3 plus the 4 plus the question

mark and set it equal to 16. and after

we subtract 9 from both sides we get the

question mark is equal to seven that

means the area of this red triangle is

equal to seven centimeters squared we

can blow it up and put a box around it

how exciting

Area of Red Triangle

This is a popular one I've seen in a couple different places. I am not sure where it originated. If anyone knows the creator, let me know!


hey guys this looks like a fun one it

says what's the angle we want to know

this angle right here this is day 11 of

our Advent calendar for the month of

December we're going to solve 31 Katrina

ad puzzles in 31 days I just remembered

something somebody said it was hard to

see the yellow against white so let's

make this thing black so now you can see

there's two circles and the angle I

think it's safe to assume that this is

tangent here here and here 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

let's copy this down here and I'm going

to leave the circle yellow here because

I'm going to be writing in here these

two circles are tangent right here so

let's construct the tangent line between

the two circles and then I'm also

assuming this is tangent right here so

let's draw this tangent line there's a

special property if you have two

tangents to a circle if the tangents

intersect these two distances are the

same so if we go up here let's shorten

this tangent line and shorten this

tangent line and then since these are

both tangent lines to the same Circle

there the same length and then if we

take this line and make it all red we

now have a nice looking red triangle and

it's an isoceles triangle since these

two sides are congruent that means these

two base angles are congruent let's call

them both a degrees let's clean this up

and let's give ourselves some room

because we're going to look at this

other tangent line and we're going to

extend it all the way up here this red

distance and this red distance will also

be equal to each other and if we

complete this right here we have another

cool looking red triangle that's also is

oses since these two sides are congruent

the base angles are congruent let's give

them both a measure of B de and now

let's clean this up and then this angle

right here is made up of b and a we can

make it a single angle of B+ a de and

this B+ a degrees that is our question

mark so we can say question mark is

equal to B+ a de this looks important

let's put a box around it and let's move

it down here next let's focus on this

quadrilateral right here this angle will

be a right angle because of the

rectangle and let's focus on our new

quadrilateral the sum of the interior

angles of any quadrilateral is

360° so we can say 90 + a plus a plus b

plus b equal 360 then let's subtract 90

from both sides on the left hand side

these are gone a plus a is 2 a B plus b

is 2b and then on the right hand side

360 - 90 is 270 then we can divide both

sides by 2 on the left hand side this is

A + B and on the right hand side 270 / 2

is 135 so now we know a + b is equal to

135 and our question mark was B plus a

we can switch these and now in the place

of B plus a we can plug in 135 so this

is the answer to her question this angle

is going to be

135° let's put a box around it how

exciting this is the answer to day 11

this is the puzzle for day 12 it says

what's the area of the semi semicircle

inside this equilateral triangle so

we're given that this circle six we're

given that's an equilateral triangle

that's super helpful and then we're

given this semicircle this will be a fun

one how exciting

Aggvent Calendar Day 11

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

Here is the link to the page!


hey guys this looks like a fun one we're

given a rectangle and a semicircle and

this distance of six right here and it

wants to know what is the area of the

rectangle this is day 10 of our Advent

calendar for the month of December we're

going to do 31 Katrina a puzzles in 31

days if you want to try it on your own

pause it right now because I'm going to

solve it in three two one the puzzle

asking for the area of the rectangle so

let's write out that formula the area of

a rectangle is base time height so we

just got to figure out what is the base

and what is the height first let's find

the center of the semicircle and draw

this radius that's part of the base and

then we can draw this radius of the

semicircle that'll be the height of our

rectangle and to finish up the base of

our rectangle it's not quite a full

radius so let's just give it a variable

X so now we can express the area of the

rectangle the base is x + r and the

height is equal to R and then we can

distribute that x * R is XR and r * R is

R 2 so the area of our rectangle is XR +

r 2 this looks important let's put an

ellipse around it and I don't think we

need this stuff anymore and let's move

this down here next let's swing this

radius down here let's label this other

unknown distance as y this YX and R make

a nice looking right triangle let's do

Pythagorean theorem it'll be x^2 + y^2

is equal to R 2 and then I see another

right triangle let's get rid of of this

radius and focus on this right triangle

and we can do Pagan theorem again it'll

be the quantity x + r^ 2 + y^2 = 6^ 2

now let's play around with this a little

bit x + r 2 is the same thing as x + r *

x + r and after we multiply this out

we'll have x^2 + 2xr + r 2 these two

things are the same so let's get rid of

this I'm noticing up here we have an x^2

+ y^2 so let's bring this X2 over here

so now these look the same and let's

bring this stuff up here so we know that

x^2 + y^2 is equal to r^ 2 so in the

place of this x^2 + y^2 let's substitute

r^ 2 and then r^ 2 + r 2 is 2 R 2 let's

smush all of this stuff together and

they both have a two in there let's

factor out that two 2 * what gives us

2xr that's XR and 2 * what gives us 2 R

2 that'll be R 2 and if you didn't fully

follow that just imagine Distributing

this two to both of these it bring us

back up here so this is the same thing

as this let's get rid of this and move

this up here next on the right hand side

this 6 squar can be changed into 36 and

now both sides of our equation can be

divided by two on the left hand side

these twos will cancel each other out

and on the right hand side 36 / 2 is 18

so now we have XR plus r 2 is equal to

18 but that's the same thing as our area

XR + r 2 so the place of this XR + r^ 2

let's substitute in 18 let's give it a

label of unit squared and this is the

answer to our question the area of this

rectangle is 18 units squared how

exciting this was day 10 of our Advent

calendar and this will be day 11 it's

asking us what's this angle but it

doesn't really seem to provide very much

other information how exciting

Aggvent Calendar Day 10

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

10 days in, 21 to go!


hey guys this looks like a fun one it

says solve for x and it gives us XM + Dy

= a n + xth we got to get both X's to

the same side of the equation so I'm

going to subtract xth from both sides

and then while we're at it this Dy does

not have an X in it so we want to move

it to the other side and we can do that

by subtracting Dy from both sides this

positive Dy and negative Dy will cancel

each other out and the positive XT and

the negative XT will cancel each other

out so we're left with X Maus xth = a

minus Dy now all the X's are on the same

side and every term has an X in it so

let's factor out that X when we Factor

the X out of this first term it's an MA

that's remaining and when we Factor the

X out of this term it's a th that's

remaining next to get the X all by

itself we're going to divide both sides

by ma minus th the ma minus th's on top

and bottom will divide each other out

and that leaves us with X = a nus dy

over ma minus th this is the answer to

our question let's put a box around it

how exciting

How exciting!

I hope you guys like this one!


hey guys I thought this was a fun

problem and I thought of a fun way to

demonstrate it let's focus right here

this X in the numerator can be broken

out into < TK x * squ root of x now

these two square root of x's match so

they can cancel each other out and now

we're stuck with just the square root of

x now let's focus on the square root of

square root of x the square root of a

square root of a number is the same

thing as the fourth root of a number now

let's focus on this right here we can

break this x into four fourth roots of x

and then one of these in the numerator

will cancel with this one in the

denominator that leaves us with of these

fourth root of x we can write these

three fourth roots of x as 4th root of x

cubed and now we have a square root of

the fourth root of x cubed that's the

same thing as the eth root of x cubed

and then looking at this x up here we

can break this down into 8 eth roots of

x and this eth root of x cubed this

three here is telling us there are three

of those now we can cancel one of these

with one of those another one with

another one and the third one with the

third one and we're left with five of

these eth roots of x or in other words

the eth root of x to the 5th now we have

the square root of the 8th root of x 5th

that's the same thing as the 16th root

of x 5th is the same thing as the 16th

root of x to the 5th and now to get rid

of this five we can take the fifth root

of both sides of the equation this fifth

root and this five are going to cancel

each other out and the fifth root of 32

is 2 and now to get X all by itself we

want to get rid of this 16th root we're

going to do both sides of the equation

to the 16th power this 16 exponent and

the 16th root will cancel each other out

so we just have X and 2 to the 16th

power is 2 multiplied out 16 times

that's going to be

65,536 and that is the answer to our

question let's put a box around it how

exciting

Radical Math Problem


an area of an equilateral triangle is

root 3 over 4S squared where s is the

length of each of the sides so how do we

figure this out first let's draw out an

equilateral triangle and if we drop down

a perpendicular line that will be our

height so the area of this triangle is

one half base times height let's bring

down half of this triangle so this is

one half s because it's half of the side

and it's a right triangle because this

is the height so we can do a squared

plus b squared equals c squared one half

squared is one fourth and then s squared

is s squared and all these are just H

squared and S squared if we subtract 1 4

S squared from both sides on the left

hand side the H squared is now alone and

on the right hand side we have

three-fourths of s squared we can square

root both sides on the left hand side we

just have H and on the right hand side

the square root can be distributed to

all three things we'll bring down the

root 3 square root of four is two and

square root of x squared is s now we

have H in terms of s so let's plug that

in up here in place of this H I'm going

to plug in root 3 over 2s and S times s

is s squared and then in the denominator

2 times 2 is 4 and that gives us the

same area that we have up here how

exciting

Equilateral Triangles Area #shorts

https://andymath.com/
#shorts


guys it says solve for x x squared

equals one so I think I want to use the

logarithms to solve this so let's copy

down the x squared equals one and let's

make some room so we can plug in the

natural log so before we move on we have

to verify that X does not equal zero and

if we check up here 0 is not a solution

to this so we know that X does not equal

zero so now we can move on one of the

properties of logarithms is this

exponent can be brought to the front so

we end up with two natural log of x

equals natural log of one the input of a

natural log must be positive so let's

make this absolute value of x it wasn't

an issue before because it was being

squared but now it's no longer being

squared next let's divide both sides by

two these twos cancel so on the left

hand side we just have the natural log

of x and on the right hand side another

way to rewrite this is one half natural

log of one so just like before when we

brought the two to the front this one

half can become the exponent so the

one-half exponent means the same thing

as the square root so one to the

one-half is the same thing as the square

root of one and let's bring down the

rest of this stuff so natural log of

square root of 1 is just natural log of

1 one and next for both sides of the

equation we're going to do e to the

power of e to the power of a natural log

or inverse operation so the left-hand

side is just going to be absolute value

of x and on the right hand side same

thing so we're just left with one and

now we have the absolute value of x is

equal to one in order for this to be

true the X inside the absolute value

would have to be either one or negative

one and that's the answer to our

question let's put a box around it so if

you ever see x squared is equal to one

we know that X is equal to one or

negative one how exciting

Solve x^2=1 with Logarithms

I think there is probably and easier way to solve this.


you guys this looks like a fun one one

over one plus one over one plus one over

x equals 420 over 834. so for the first

step we want to do this one plus one

over X in order to give them a common

denominator we're going to multiply this

one on the left hand side by X over x

and x over x times 1 is just X over X

and now that they have a common

denominator we can write them as a

single fraction so it becomes X plus one

over X so now let's focus on this right

here this is basically saying 1 divided

by X Plus 1 over X the division of a

fraction is the same thing as

multiplying by the reciprocal so if we

change this divide into a multiply we

can flip this fraction and one times

this is just this so we want to rewrite

this whole thing as X over X plus one

now let's get rid of this stuff next we

want to add these together and we need

to give this a common denominator with

this so we're going to multiply this one

by X Plus 1 over X plus one well this

times 1 is just this thing and now that

these two fractions have a common

denominator we can write them as a

single fraction so all we did was add

the numerators and last X Plus X is

equal to 2 x with this right here we're

doing the division thing again we're

doing one divided by this fraction down

here anytime you do that we can change

the Divide to a multiply and flip the

fraction and then one times something is

just the something so now we want to

rewrite this entire left side as this

fraction right here or in other words

you can think of it as bringing this X

Plus 1 to the top now let's get rid of

this stuff I'm noticing on the side we

can simplify this fraction both the top

and the bottom are divisible by six on

top 420 divided by 6 is 70. and on

bottom made 34 divided by 6 is 139. in

order to solve this proportion Let's

cross multiply we have the 139 times X

plus one equals 70 times 2x plus 1 and

then if we distribute 139 times x is

139x 139 times 1 is 139 70 times 2x is

140x and 70 times 1 is 70. now we have

an x on both sides so let's subtract

139x from both sides these cancel so on

the left hand side we only have 139 and

on the right hand side 140x minus 139x

is just X now we still have the plus 70.

now on the left hand side 139 minus 70

is 69. nice that is the answer to our

question so we can put a box around it

how exciting

Nice Rational Equation

https://andymath.com/rational-equations/

Thank you for watching these videos! I hope you have a nice week!


hey guys so many people sent me this one

it was posted by math love and it got

thousands of likes so let's do it first

step I want to focus on this triangle

right here if we call this angle a we

know that a + 45 + 43 has to equal 180

and then after we subtract 45 and

subtract 43 from both sides that'll give

us a is equal to 92 so we can update

this to be

92° and then next let's focus on this

relationship right here this is called a

linear pair if we let this angle be n°

it we know that n + 92 is equal to 180°

and after we subtract 92 from both sides

we get n is equal to 88 so we can update

this there and then this angle right

here is also 88 you can either think of

it as a linear pair with the 92 or you

can think of it as vertical angles with

this 88 next let's focus on this

triangle right here let's let this angle

be D de then we know that D plus the 88

+ 24 needs to equal 180 and after we

subtract 88 and 24 from both sides we

get D is equal to 68 so we can update

this to be

68° and then if we extend this here we

have another linear pair and if we make

this y Dees then the y + 68 needs to

equal 180 and then we can subtract 68

from both sides and we get Y is equal to

112 and next let's focus on this

triangle right here we can call this

angle M and then once again we can do

all this math and we find that m is

equal to

66 so then we have this linear pair

right here so we know that t + 66 = 180

or in other words T is equal to 114 and

now we're down to this convex

quadrilateral right here we know that

all four of these angles are going to

add to

360° well 112 + 114 + 92 is equal to 318

and then after we subtract 318 from both

sides we get X is equal to 42 and that

is the answer to the question let's put

a box around it how exciting

Relaxing Math Problem

This is wasn't super difficult, but I found it relaxing to go through. Kinda like math ASMR.


guys this looks like a fun one we're

giving some triangles with some angles

this is congruent to this and they give

us the values of these two angles and

they're asking us to solve for x let's

do this if you want to try this on your

own pause it because I'm going to solve

it in three two one for the first step

let's focus on this linear pair right

here I want to find out what is this

other angle combine these form 180

degrees so we know that the question

mark plus 45 will equal 180. that'll be

true if the question mark is 135 so we

know that this angle is 135 degrees next

let's focus on this triangle right here

what is the value of this angle the sum

of the interior angles of a triangle is

always equal to 180 degrees and this

will be true for 30 so we can change

this angle to 30 degrees and now that we

know this 30 degrees we won't need this

135 anymore starting from the vertex of

this angle let's draw out this blue

segment and construct a circle with that

as the radius now let's find where this

circle intersects this let's connect

this intersection point to the center of

our Circle this is another radius of the

circle so it's equal to this segment

right here and now we don't need this

circle anymore next let's focus on this

triangle right here since there are two

sides congruent in this triangle we know

it's an isosceles triangle and the angle

is opposite the congruent sides are also

congruent so we know that this angle

right here is also 30 degrees and now to

find the last angle of the triangle we

can set them all equal to 180 and this

will be true for 120. so this angle

right here is 120 degrees next I want to

find this angle right here and we're

going to use these three angles to get

it this angle plus this angle plus this

angle will all add up to 180 degrees and

this statement will be true for 15. so

we know that this angle right here is 15

degrees next I'm thinking let's focus on

this triangle right here since these

angles are both 15 degrees we know it's

an isosceles triangle and the sides

opposite the congruent angles are also

congruent next let's connect this

intersection point with this vertex

right here now let's focus on our new

triangle to find this angle down here we

need to add the 15 degrees and the 45

degrees to get 60 degrees and now let's

get the last two angles of this triangle

this is another isosceles triangle so we

know that these two angles are congruent

to each other and we also know that if

we add up all three of these angles it's

going to be 180 degrees this will be

true if both of these angles are 60

degrees and now since all three angles

are 60 degrees we know it's an

equilateral triangle so all three sides

are going to be congruent so now from

here I'm interested in this angle right

here so let's focus on these three

angles and find that angle right there

these three angles are also going to add

to 180 degrees and this will be true if

the angle is 90 degrees so let's put one

of those little boxes right there to

show that's 90 degrees for our next step

let's focus on this right triangle right

here it's also an isosceles triangle so

we know that these two angles are going

to be congruent to each other and if we

add up all three of these angles we can

find out that each of these are 45

degree angle let's look at this x

degrees in this 15 degrees let's make

them red we know that the sum of these

two angles will be equal to this larger

angle so we have X plus 15 is equal to

45. so we subtract 15 from both sides we

find that X is equal to 30. and that is

the answer to our question they wanted

us to solve for x so let's clean things

up a little bit and put a box around it

x equals 30 how exciting

Geometry Challenge Problem

I thought this was a fun math problem. I hope you guys like the video!


guys let's find this limit it's the

limit as X approaches 0 of 1 minus the

square root of 1 minus X over X first

let's copy it down and let's try just

plugging in zero see what happens so in

the place of both these X's I just

plugged in zero well this minus zero

isn't doing anything and the square root

of one is just one and then one minus

one is zero by using direct substitution

we just got zero over zero which is an

indeterminate form this isn't giving us

any information given 0 over 0 the

answer could still be anything so we

have to find another way to solve this

looking at this I'm seeing a binomial

that includes a square root so usually

you're going to use the conjugate so in

this case 1 minus the square root the

conjugate would be one plus the square

root and then we're going to multiply

that to top and bottom so we're not

actually changing the value of the

rational expression so let's see what

happens when we use the conjugate on

bottom we're just going to leave it as x

times this we're not going to need to

distribute the X usually is the one that

isn't the conjugate you can just leave

it as the product of the two so now on

top let's multiply this stuff out so we

have 1 times 1 which will give us one

and we have one times the square root

which is the square root and then we

have the negative square root times 1

which is the negative square root and

then we have the negative square root

times the positive square root a

negative times a positive is going to be

negative and then the square root of 1

minus x times the square root of 1 minus

X when you multiply those the square

root goes away and we have 1 minus X so

before we simplify let's distribute this

negative to both of these terms so this

positive one will become a negative 1

and the negative X will become a

positive X so now we have the positive

of this square root and the negative of

this square root they're going to cancel

each other out and then we have positive

one and negative one those will also

cancel each other out and now let's

smoosh everything together and this X on

top and bottom will also cancel each

other out and now there's nothing on top

we should put the implied one now let's

do some direct substitution what happens

if we plug 0 into the X well the zero

isn't doing anything so this just

becomes one over one plus square root of

one and one plus square root one is one

plus one which is two and that is the

answer to our question let's put a box

around it so the limit as X approaches 0

of this is equal to one half and here it

is in Desmos and you can see that it

does approach one half from both

directions

if you guys want to try more of these

limit problems I have this page on

animat.com it's got a whole bunch of

problems with the answers the link is in

the description how exciting

Limit with Conjugate

https://andymath.com/intro-to-limits/
For more limit practice, check out the above link!


guys a bunch of people mentioned me in

this one so let's do this x squared plus

y squared equals 25 and the 2x plus y

equals 10 and we want to evaluate X Plus

y so first let's focus on the 2x plus y

equals 10. let's get the Y all by itself

let's subtract 2x from both sides on the

left hand side we just have y and on the

right hand side we have 10 minus 2X and

this is important so let's put a box

around it next let's focus on the x

squared plus y squared equals 25. let's

bring it down and in the place of this y

right here I want to plug in 10 minus 2x

so now this is 10 minus 2x being squared

so we're going to multiply 10 minus 2x

times 10 minus 2x 10 times 10 is 100 10

times negative 2x is negative 20x

negative 2x times 10 is negative 20x and

negative 2x times negative 2x is

positive 4x squared and these two can

combine to become negative 40x let's

bring down the x squared and bring down

the equals 25. anytime I do one of these

several people always mention why don't

I do this shortcut well I did try it out

and it ended up not being much of a

shortcut so it low as nicely as this

method in these videos so now let's

combine like terms x squared plus 4x

squared is 5X squared we can bring down

the negative 40x and if we subtract 25

from both sides 100 minus 25 is 75 and

25 minus 25 is zero from here all three

of these terms are divisible by 5 so

let's divide both sides by 5. each of

these terms will divide by 5 to give us

x squared minus eight X plus 15 and it's

still equal to zero and now to solve

this we want to factor we want two

numbers that will add to negative 8 and

multiply to positive 15. well negative 3

plus negative 5 will give us negative 8

and negative 3 times negative 5 will

give us positive 15. so these are the

factors of this so from here if this

equals 0 or if this equals 0 the whole

thing will equal zero because anything

times 0 equals zero so we have this or

this and if we add 3 to both sides we

get x equals 3 and if we add 5 to both

sides we get x equals five so X can be

either 3 or 5. so we don't really need

any of this work anymore and let's bring

these up here since our goal to figure

out what is X Plus y we want to find out

our corresponding y values so let's use

this 2x plus y equals 10. we'll copy it

down solve for y in the place of this x

let's plug in 3. 2 times 3 is 6. let's

subtract 6 from both sides and we get y

equals 4. and same thing for this one if

we plug in 5 for the X 2 times 5 is 10

subtract 10 from both sides and we get y

equals zero these are our two possible

scenarios and we want to evaluate X Plus

y for each of these scenarios for the

first scenario X Plus y ends up being 3

plus 4 which is equal to 7 and for this

scenario X Plus Y is 5 plus 0 which is

equal to five and these are the two

possible answers to the question X Plus

y equals seven or X Plus y equals five

and if you want to see what this looks

like graphically here it is right here

the x squared plus y squared equals 25

is the red circle and the 2x plus y

equals 10 is this blue line and here's

our two intersection points three four

and five zero how exciting

Fun Math Warmup Problem


guys I think this is a pretty cool one

even though this is in terms of X you

can still find the exact value of this

red region if you want to try this on

your own posit because I'm going to

solve it in three two one first let's

copy this down here and let's look only

at the large triangle the base of the

triangle is the two pieces added

together so it's going to be 9x plus 6.

and the height is going to be 2x plus 6

and this hypotenuse will still be 13x

this is a right triangle which means we

can use Pythagorean theorem the quantity

9x plus 6 squared plus the quantity 2x

plus 6 squared is equal to the quantity

13x squared so the right hand side's

pretty easy both the 13 and the X get a

squared and 13 squared is equal to 169.

nice and then on the left hand side

we're going to add these two things so

this squared means this thing multiplied

by itself and this squared means this

thing multiplied by itself so this 9x

plus 6 times 9x plus 6 you can multiply

it out or foil it it ends up giving us

81x squared plus 108x plus 36. and you

can do the same thing with this down

here here gives us 4x squared plus 24x

plus 36. 81x squared plus 4 is 85x

squared 108x plus 24x is 132x and 36

plus 36 is 72. let's subtract 169x

squared from both sides and this minus

this is negative 84x squared we can

bring the rest of this stuff down and

this is all going to be equal to zero

each of these coefficients is divisible

by 12 so let's divide it by 12. now

let's do negative 12 so we can make the

leading coefficient positive negative 84

divided by negative 12 is 7x squared

positive 132 divided by negative 12 is

negative 11x positive 72 divided by

negative 12 is negative 6. and 0 divided

by negative 12 is 0. this is a quadratic

so let's use the quadratic formula both

of these A's are equal to seven both of

these B's are equal to negative 11 and

the C is equal to negative 6. after we

simplify all this we end up getting x

equals 2 or x equals negative 3 7. and

looking up here 13 times a negative

number would give us a negative

hypotenuse and we can't have a negative

hypotenuse so this x negative 3 7 is not

a valid solution to our problem that

leaves us with x equals two and that

looks important so let's put a box

around it so going back to our original

figure let's copy down the smaller

triangle now for both our larger

triangle and the smaller triangle let's

plug in 2 for all the X's 9 times 2 is

18 2 times 2 is 4 9 times 2 is 18 plus 6

is 24. 2 times 2 is 4 plus 6 is 10 and

13 times 2 is 26. now we have specific

values for our triangles I think we're

ready to find the area of the red region

the area of the red region will be equal

to this big triangle minus the small

triangle so let's figure out the area of

each of those the area of either of

these triangles is going to be base

times height divided by 2. for this big

triangle the base is equal to 24 and the

height is equal to 10 and we're still

going to divide by two and for the

smaller triangle the base is 18 the

height is 4 and then we're going to

divide by two and now let's simplify

these fractions 21 4 times 10 is 240 and

18 times 4 is 72. 240 divided by 2 is

120 and 72 divided by 2 is 36. so our

big triangle is 120 so let's change this

to 120 and our small triangle is 36

let's change this to 36. and 120 minus

36 is 84. so the area of our red region

is 84 units squared even though some of

the measurements were in terms of X we

were still able to get an exact value

for the area how exciting

Area Challenge

I hope you guys like it!


have you guys ever heard of the birthday

Paradox imagine a group of 23 random

people what is the probability that at

least two people have the same birthday

and these are random people so no Twins

or anything like that so it's going to

be easiest to calculate the probability

that nobody has the same birthday and

then just do 100 minus whatever that

gives us first let's start with one pair

of people so the first person has a

birthday whatever that is then what is

the probability that the second person

does not have the same birthday well

there are 364 possible days that don't

match out of 365. so their probability

is 364 over 365. now let's bring in a

third person what is the probability

that that third person does not have the

same birthday as either of those other

two well that leaves 363 out of 365 days

so we can just multiply these and the

fourth person to not match any of these

three would be 362 over 365. and this

pattern continues for the rest of the

people in the group after we multiply

all of these probabilities we get

approximately 49.27 so the probability

is 49.27 that no one shares a birthday

so if we do a hundred percent minus that

we get a 50.73 probability that at least

two people have the same birthday so if

it feels weird that 23 people out of 365

days can do this it might be better to

look at this the number of pairs if we

look at the first person in the group he

has 22 pairs you can match with then the

next person has 21 new pairs they could

match with and the next person has 20

new pairs of people they can match with

and this pattern is going to continue so

there are going to be 253 pairs we're

just trying to get at least one of these

253 pairs to match and if you think of

it that way the 50.73 seems much more

realistic every one of these 253 pairs

is a possible chance for a birthday

match and this is the birthday paradox

Birthday Paradox Explained

Collab with Geeklyhub!


hey guys this looks like a fun one it

gives us that this is 9 this is 16 and

this is 12 and they're all at right

angles with each other and it wants us

to find the radius of this 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 for the first step

let's extend this over here and we're

completing this rectangle this bottom

will be equal to the top which is nine

and this side will be equal to this side

which is 16 and we've created this

little piece right here let's call that

X next let's draw two of our radi one

from here to here and one from here to

here and these are both radi so we'll

call them R first I want to focus on

this triangle right here this is a right

triangle so we can use the Pythagorean

theorem it'll be x + 16 2 + 9^ 2 is

equal to R 2 and then let's do the same

thing for this triangle down here it'll

be 9 + 12 2 + x^2 is = R 2 and we can

clean this up a little bit this 9 + 12

inside of the parentheses can be 21 and

now we have two equations with two

variables so let's solve for these

variables this is equal to R2 and that's

equal to R2 so we can set this equal to

that on the left hand side 212 is equal

to 441 and we can bring down the x^2 on

the right hand side this x + 16 2 is

equal to x + 16 * x + 16 and after we

multiply that out it ends up being x^2 +

32x + 256 and then this 9^ 2 is equal to

81 next let's subtract x^2 from both

sides of the equation on the left hand

side we have the 44 1 and then on the

right hand side we can bring down the

32x and 256 + 81 is equal to 337 let's

smush everything together and we want to

get this X all by itself so let's

subtract 337 from both sides of the

equation on the left hand side 441 - 337

is equal to 104 and on the right hand

side these will cancel each other out so

we're left with 32x next we can divide

both sides by 32 on the left hand side

both of these have eight in common so

it'll simplify to 13 over4 and on the

right hand side these 32s will cancel

each other out leaving us with X and now

we've solved for X it's equal to 13

over4 and let's update this x to be 13

over4 and we can also update this x to

be 13 over4 and now we're ready to solve

for R 212 is equal to 441 and this 13

over 4^ 2 that's the same thing as 132

over 4^2 which is equal to 169 over 16

and then we can copy down the R 2 in

order to add these two let's make this

441 a fraction by doing 441 over 1 to

give them a common denominator we can

multiply top and bottom of this by 16

and both of these 16s will just hop

right over to the fraction 16 * 441 is

equal to 756 and 16 * 1 is equal to 16

and now that these have the same

denominator we can write them as a

single fraction and 756 + 169 is equal

to

7,225 next we can square root both sides

of the equation on the left hand side

this square root can be split to top and

bottom the square root of

7,225 is equal to 85 and on bottom the

square root of 16 is equal to 4 and on

the right hand side the square root and

the square will cancel each other out

and then 85 / 4 is equal to

21.25 and now we found the radius to our

quarter Circle it's equal to

21.25 CM this is the answer to our

question let's put a box around it how

exciting

Find the Radius

This was a fun question, I hope you guys love it.

Does anybody read these descriptions? I think it would be funny if everybody who read this paragraph commented their favorite color and then their favorite animal. Then so many people in the comments would be confused cause there is no prompt in the video. :-)


you guys this looks like a fun one it

says solve for x that's the square root

of x to the X to the x equals X and it

says X is greater than one so it does

look like x equals one would have been

one solution for this but they've

excluded that solution and they want us

to find another one if you want to try

this on your own pause it right now

because I'm going to solve it in three

two one so first we can multiply these

exponents at x times x is equal to x

squared and next to get rid of this

square root let's Square both sides of

the equation so this exponent and this

square root are going to cancel each

other out so now we have x to the x

squared equals x squared since X can't

equal one or zero that means the only

way this is true is if the exponents are

equal to each other so we can drop the

base Drop the Base

so we can drop the bass and set the

exponents equal to each other and then

from here we can square root both sides

and the square root of x squared is the

absolute value of x so we have the

absolute value of x is equal to square

root of 2 but since we know that X can't

be negative this is just going to be x

equals square root of 2 and that is the

answer to our question we can put a box

around it and let's check our work we're

going to plug in the square root of 2

for each of these x's and then we're

going to multiply these exponents square

root of 2 times square root of 2 is

equal to two and then this square root

and this Square cancel each other out

and then we end up with square root of 2

equals square root of 2 it checks out so

for this problem x equals the square

root of two how exciting

Fun Math Challenge

I thought this would be a fun problem. I also had a chance to play around with some animations and special effects. How exciting!


hey guys I love this problem it says the

width is twice the height if the

perimeter is 120 what is the area and

we're given that the width is x^2 - x -

16 and the height is x + 12 if you want

to try this on your own pause it right

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

these are a trap we don't need these so

we can just assign H for the height and

call the width 2 H because the width is

twice the height so we need the

perimeter to be 120 that means we need H

+ 2 H + H + 2 H to equal 120 and H H + 2

H + H + 2 H is equal to 6 h after we

divide both sides by 6 we get H equal to

20 so we can plug in 20 for these H's

and the two h's will become 40 40 + 20 +

40 + 20 gives us 120 and then the area

will be 40 * 20 which is 800 unit

squared and that's the answer to the

question how exciting

Sneaky Math Problem

I thought this was a fun one.


hey guys I made this video a couple days

ago I really like the problem because it

had a cool shortcut but so many people

asked me to solve it without the

shortcut if you want to try this on your

own pause it right now cuz I'm going to

solve for x and y in 3 2 1 so first

let's copy both of these things down so

we have X Y = 36 and x^2 + y^2 = 78

let's start with this one first let's

divide both sides by X and it becomes y

= 36 /x so in the lower equation in the

place of the Y let's plug in

36x now this exponent goes to both the

top and the bottom so 36 /x^2 becomes

362 over x^2 and 362 is equal to

1296 and I don't really like fractions

so let's get rid of this x^2 in the

denominator we can do that by

multiplying both sides of the equation

by X2 and we're allowed to multiply both

sides by X squ because we know that X

cannot be zero if x were zero XY would

have to be zero so on the left hand side

this x^ squ will distribute to both of

these terms we have x^2 * x^2 which is X

4th and then we've got x^2 * this thing

the x^2 on top and the x^2 on bottom

will cancel each other out so the second

term is going to be 1296 and that's all

going to be equal to 78 x^2 next I want

to set this all equal to zero let's

subtract 78 x^2 from both sides on the

left hand side we end up with x 4 - 78

x^2 + 1296 and on the right hand side

these cancel each other out leaving us

with zero now we can solve this by

factoring this x 4 will break down into

x^2 and x^2 and then we want to find the

two numbers p and Q such that p + Q is

equal to -78 and P * Q is equal to

1296 this could take a little bit of

effort it ends up being -24 and -54 and

I did use a calculator to find these

next we have this * that equals 0 that

means that either this equals z or that

equals z so now we just got to solve

these two equations we can add 24 to

both sides of this equation or add 54 to

both sides of this equation on the left

hand side these cancel each other out

and we're left with X2 is equal to 24

and on this equation these two cancel

each other out so we're left with x^2 is

equal to 54 now both of these are in

terms of X squ so we're going to want to

square root both sides of both equations

for both of these square root of x squs

they're going to be absolute value of x

and to solve for the absolute values of

X each of these will become plus or

minus and now we have our four Val vales

of X posi < tk24 < tk24 posun 54 orun 54

and we know that Y is equal to 36 overx

so for each of these values of X we can

find the corresponding value of y in the

place of this x we'll plug in < TK 24

and the same thing for the rest of these

X's so now we have our answers for X and

Y let's check our work in the place of X

we can plug in < TK 24 and in the place

of Y we can plug in 36 over < tk24 this

exponent will go to the top and the

bottom 36 6 4 is equal to a huge number

and both of these < TK 24 to 4ths are

equal to 576 and then this huge number

divided by 576 is equal to 2916 and then

576 + 2916 is equal to 3492 and that is

the correct answer for X 4th + y 4th so

it checks out and in fact all of these

will check out so 3492 is the answer to

our question and this was the difficult

way to do it let's put a box around him

how exciting and I also got a lot of

request to see the haircut here it is I

got hat hair right now I don't know why

there's the haircut how exciting

Without the Shortcut

This is part 2. I hope you like it!


ton of people get this problem wrong it

says what is the value of 5x and we have

a diagram we have a 90 degree angle

right here which means that this angle

right here is also going to be 90

degrees because this is a straight line

and since this angle right here is 90

degrees that means that if we were to

add up all of these inside angles x plus

2x plus 2x all of them have to add up to

90 degrees so let's go ahead and do that

down here so we have an x

we have a 2x and we have a 2x and all of

these angles added up has to equal 90

degrees okay so now we can just add up

like terms x 2x and 2x that's going to

be equal to 5x and we have an equals 90.

and just like that we don't have to do

any more work because our answer is

Missing Angles Geometry Problem | Tricky Math Question | JusticeTheTutor #maths #math #shorts


hey guys welcome to this isosceles

triangle video so we're going to solve

for

x in this isosceles triangle the whole

idea of an isosceles triangle is that

these base angles are equal

and the sides opposite of those base

angles are going to be equal

and since these two sides are equal we

can set their expressions equal to each

other

so we can say 3x plus 5 equals 5x plus

it says solve for x so our goal is get x

by itself

when we're done it should look something

like x equals something

a number but right now we have an x on

both sides of the equation we can't have

that so we're going to move 1 to the

other side

so i like to find the smaller of the 2

and the 3x is smaller than 5x

so i'm going to subtract that from both

sides the reason i do the smaller one is

because when i do 5x

minus 3x i get positive 2x and i prefer

positive numbers over negative numbers

and then this goes away so we're left

with just

5 on the left side and we have 2x plus

now we only have one side that contains

an x

our goal now is to get that x by itself

so first i'm going to subtract the 1.

5 minus 1 is 4 and this all cancels so

we just have 2x

on the right side and now this means

two times x so the way i'm going to undo

that is divide both sides by two

and we get x equals two is the answer

and we can check it

so we got this one right so that's how

these work when you're trying to solve

for x and you're dealing with isosceles

triangles you usually will just set

the two sides equal to each other so

here's another variation

once again it's an isosceles triangle

and so we know the base angles are equal

and the sides are equal well here

they're telling us the degrees of the

angles so we can just set these degrees

equal to each other we can say

x minus 10 equals 50. now this

parenthesis doesn't mean anything

really because it's already alone on

that side so we just get rid of it once

again this says

solve for x so our whole goal is get x

by itself right now there's a minus 10

the way we undo a minus 10 is adding 10

which leaves us

x on the left side and 50 plus 10 is 60.

so the answer to this one should be x

equals 60 and we can check it

and it says the answer is x equals 60.

if you want to try more of these i have

this page on andymath.com i'll include a

link down below in the description

there's also related pages for more

triangle related concepts and these all

have practice problems and answers as

well

please like this video and subscribe to

my channel and comment below if you have

any other math questions i'll try to

answer them for you

i'll talk to you guys soon bye

Isosceles Triangles Solve for x

https://andymath.com/isosceles-triangles/
Visit the above link for more practice problems and related pages!


what is e probably heard e is

approximately 2.718 and it is an

irrational number that will never repeat

but what is it I'll explain one way to

find it this writers the formula for

compounded interest let's set the

principal to a dollar set the interest

rate to 100 and set the time to one year

if we have one compounding per year it's

going to look like this this ends up

just being one plus one which is two now

if we did the same thing but with two

compoundings it would look like this

Which is higher than the two and then we

could do monthly compoundings that would

be n equals 12 and that would look

something like this which gives us

approximately 2.61304 so you can tell

we're getting closer and closer to that

value of e now if I keep going say do

daily compoundings that would look like

this right here and this number is

getting even closer to e if we did 10

000 compoundings this right here and we

could do 100 000 compoundings we're just

getting super close a million

compoundings gets us here and in the end

if we did it continuously that'd be like

n equal to Infinity the best way to

express that is with the limit as n

approaches infinity and that's going to

give us the number that is exactly e so

this right here is one definition of e I

hope this helps

Quick e Example #shorts

https://andymath.com/the-number-e/

#shorts


so we want to solve for x in this

triangle we're going to use the exterior

angle theorem

so if we call this a this b and this c

it is always true that a will equal b

plus c and these are the two angles that

are opposite

of this one so let's fill in for a b and

c a was 3x plus 2

b is x plus 25 and c is x plus 17.

i put the parentheses there so we could

see each of the angles x plus

x is 2x and 25 plus 17 is 42.

so we have an x on both sides of the

equation we want to subtract the smaller

one over to the other side

3x minus 2x is x bring down the 2 2x

minus 2x goes away

so we just have a 42 on this side so

subtract 2 from both sides

42 minus 2 is 40 for the right hand side

and that's the answer the question x

equals 40.

so we use the exterior angles theorem to

solve this

comment on this video for what you'd

like me to solve next and i'll talk to

you guys soon

bye

Exterior Angle Theorem #Shorts

𝐋𝐈𝐍𝐊𝐒 𝐓𝐎 𝐑𝐄𝐋𝐀𝐓𝐄𝐃 𝐓𝐎𝐏𝐈𝐂𝐒
𝐄𝐱𝐭𝐞𝐫𝐢𝐨𝐫 𝐀𝐧𝐠𝐥𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/exterior-angle-theorem/
𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐀𝐧𝐠𝐥𝐞 𝐒𝐮𝐦 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/triangle-angle-sum-theorem/

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

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

𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐞𝐚𝐧 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/pythagorean-theorem/

𝐏𝐲𝐭𝐡𝐚𝐠𝐨𝐫𝐞𝐚𝐧 𝐓𝐫𝐢𝐩𝐥𝐞𝐬: https://andymath.com/pythagorean-triples/

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/

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/

𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐋𝐨𝐧𝐠𝐞𝐫 𝐒𝐢𝐝𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/Triangle-longer-side-theorem/

𝐓𝐫𝐢𝐚𝐧𝐠𝐥𝐞 𝐋𝐚𝐫𝐠𝐞𝐫 𝐀𝐧𝐠𝐥𝐞 𝐓𝐡𝐞𝐨𝐫𝐞𝐦: https://andymath.com/Triangle-larger-angle-theorem/

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 somebody requested that I

solved this one my last two videos were

pretty intense so I thought we would do

a lighter one so we're trying to solve

for this question mark So this hundred

and this angle right here have to add to

180 so that's going to be 80 degrees and

same thing is going on down here this

110 and this has to be 180 so this will

be 70 degrees so now we have a triangle

here so 70 plus 40 plus blank has to

equal 180. that's going to be 70 degrees

so we can plug in 70 here and then by

vertical angles this is also 70. and now

we have another triangle with our

question mark it says 80 plus 70 plus

blank has to equal 180 and the blank is

going to be 30 and that means our

question mark is equal to 30. how

exciting

Swedish Math Olympiad #shorts

https://andymath.com/

#shorts


hey guys this looks like a fun one

someone sent me this in an email it

gives us two circles in a rectangle and

it wants us to solve for the height X

the base of the rectangle is 18 cm this

circle's diameter is 16 cm and this

circle's diameter is 10 cm if you want

to try on your own pause it right now

because I'm going to solve it in three

two 1 first let's rotate the two

diameters to be horizontal and let's

label the centers of both of the circles

this diameter of 10 is going to split

into two radiuses of 5 cm and this

diameter of 16 will split into two

radiuses of 8 cm and yes radiuses is a

plural of radius you could also say radi

and then from the center of the smaller

Circle let's drop a vertical line and

same thing for the center of the larger

Circle this piece is the same thing as

the radius of the smaller Circle so

it'll be equal to 5 cm and then this

portion right here is equal to the

radius of the larger Circle so it'll be

equal to 8 cm so now since the entire

base is 18 this is five and this is 8

that leaves 5 cm left over for this part

but this cannot be right because the

radius is 8 how is five greater than

eight so this thing is definitely not

drawn to scale if you're curious I

actually drew it in geogebra this is

what it would look like if it was drawn

correctly here's the same two circles

but this would be at a much steeper

angle up on top and then for the one

that is drawn to scale if you drop this

vertical line and this vertical line and

bring the measurements down we have five

five and8 and this makes makes more

sense the five fits within the eight so

this is definitely not to scale this is

what it should look like I looked into

this a little bit more this is something

from it looks like a textbook it doesn't

say what textbook so I can't find the

original Source but this might be the

original source and even in the textbook

the problem is definitely not drawn to

scale the textbook looks a lot more like

this incorrect one than it does the

correct one so I think it's funny even

the textbook is blatantly not to scale

but I think even though it's not to

scale we can still solve this so let's

keep going from the center here let's

draw this horizontal line and it'll be

perpendicular to the vertical line and

same thing from this Center let's draw a

horizontal line this portion of the

height is the same thing as the radius

of the larger Circle so this is going to

be equal to 8 cm and then this portion

of the height is the same thing as the

radius of the smaller Circle so it's

going to be equal to 5 cm so the

distance from here to here is still

unknown let's call it question mark cenm

or even better let's give it a variable

let's call it y CM so now we just got to

solve for y and it'll be pretty easy to

figure out X I see a right angle right

here so if we connect these two we'll

have a right triangle and this

hypotenuse is a distance of the five

radius plus the 8 radius so it'll be 13

cm and let's focus on this triangle the

height of this triangle will be y cm and

the base will be 5 cm and now we can do

Pythagorean theorem it says a^2 + b^2

will equal c^2 or in other words 5^ 2 +

y^2 will equal 132 52 is equal to 25 and

13 s is equal to 169 and we can bring

down everything else next we want to get

this y^2 alone on the left hand side so

let's subtract 25 from both sides on the

left hand side we have y^2 and on the

right hand side 169 minus 25 is 144 next

we can square root both sides of the

equation and that'll give us Y is equal

to 12 this is very important let's put a

box around it we're not going to need

this stuff anymore and let's bring this

up here and now we can rewrite this y CM

as 12 CM let's pull out these two

measurements and go back to our original

diagram our goal was to solve for x

which is the height of this rectangle

well the Height's also equal to 5 + 12 +

8 and then 5 + 12 + 8 is equal to 25

let's give it a label of centimet and

put a box around it this is the answer

to our question X is equal to 25 cm how

exciting

2 Circles 1 Rectangle

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


I saw this math Meme and so I thought I

would share it and I thought I would

explain it in case anybody doesn't get

it so it starts with a right triangle

with an angle X the side opposite of

that angle will call opposite the side

adjacent to it we'll call adjacent and

this side we'll call hypotenuse the S of

X is the ratio of the opposite over the

hypotenuse and the cosine of x is the

ratio of the adjacent over the

hypotenuse if we name the opposite side

a the adjacent side B and hypotenuse C

we can rewrite the S as a over C and the

cosine as B over C in the meme we have a

sin s x ring and a cosine squ X Ring

let's bring those two rings together

giving us sin^2 X plus cosine s x s of X

is a over C so sin squ of X would be a

over c^ 2 and then cosine is B over C

which would make this B over c^ 2 then

these squares can be distributed to all

the A's and the C's giving us a squ c s

plus b squ over c s now these have a

common denominator so we can just write

them as a single fraction now let's

rewrite this fraction let's bring the c^

S over and this is a right triangle so

that means that a s + b s is equal to

c^2 that's called the Pythagorean

theorem so now C2 over c s is equal to

one so when you have the sin squ ring

plus the cosine squ ring and they come

together they become one how wonderful

Sped up explanation

I sped it up to fit it in the 60 sec time limit. I hope you guys like it!


because I saw this thing the other day

this person texts an equation to

somebody and gets blocked for it that is

not cool it's 5 over 4X plus 3 equals

five over four X plus five over three I

wonder if we solve this thing if x will

have some secret message in it like this

expression here that when you erase the

top it says I love you let's bring this

piece down here I don't like fractions

so I want to get rid of all these

denominators in order to get rid of this

three I'm going to multiply everything

on both sides of the equation by three

and then to get rid of this 4X I'm going

to multiply everything on both sides of

the equation by 4X and then to get rid

of the 4X plus 3 I'm going to multiply

everything on both sides the equation by

4X plus 3. these threes will cancel this

4X and this 4X will cancel and this 4X

plus 3 will cancel with this 4X Plus 3.

on this left hand side 3 times 5 is 15

and 4x times 15 is 60x and on the right

hand side 5 times 3 is 15 and this 15

will distribute to both of these to give

us 60x plus 45 and then 5 times 4X is

20x and then 20x will distribute to 4X

plus 3 to give us 80x squared plus 60x

and now we can combine like terms by

adding these together so we have 80x

squared plus 120x plus 45. so now we

have a quadratic so let's set it equal

to zero let's subtract 60x from both

sides we have 0 on the left hand side

and then on the right hand side we have

80x squared plus 60x plus 45. we can

simplify this a bit all these

coefficients are divisible by 5. so

let's divide everything by five that

gives us 0 equals 16x squared plus 12x

plus 9. and in order to solve this let's

use the quadratic formula a is this

first coefficient B is the second

coefficient and C is the third

coefficient B here is equal to 12 so

we'll plug in 12 for both these B's a is

equal to 16 so we'll plug in 16 for both

the A's and C is equal to 9 so we'll

plug in 9 for the C 12 squared is equal

to 144 4 times 16 is 64 and 64 times 9

is 576. and 144 minus 576 is equal to

negative 432. and we can pull the

negative out of this to give us square

root of negative 1 times square root of

432 and this square root of negative 1

is going to be the imaginary number I

inside of the square root the 432 can be

broken out into 144 times 3 and we can

give each of these numbers their own

square root and square root of 144 is

equal to 12. so in the numerator we have

negative 12 plus or minus 12i root 3.

and in the denominator we have 2 times

16 which is 32. now all these

coefficients have a 4 in common this 12

can change into 4 times 3 this 12 can

change into 4 times 3 and the 32 can

change into 4 times 8 and then we can

cancel out all these fours and then we

don't want to mix this real portion and

this imaginary part inside of a fraction

so we'll split it up into two separate

fractions so the solution to this

equation is x equals negative 3 8 plus

or minus 3 I root 3 over 8. so let's put

a box around it I don't see a secret

message in here that's probably why they

got blocked

Math Meme Equation


hey guys this looks like a fun one we're

trying to find the area of this blue

region if you want to try it on your own

pause it right now cuz I'm going to

solve it in 3 2 one first I'm noticing

we can split this up in a couple

different ways and I want to focus on

this part right here let's get rid of

the question mark and continue this line

right here so let's call this portion X

and then since 28 is double the 14 this

blue portion would be 2X and now that

we've introduced this line we can also

split it up this way next let's focus on

this portion right here called this side

of the purple rectangle Y and then 105

is 3 * 35 so this would be 3 y so now

let's split this up we know that all of

this stuff is going to be 1/3 of all of

this stuff so we can set up a proportion

on top it would be this whole area which

would be x + 14 + 35 and on bottom it

would be this whole area which would be

2x + 28 + 21 + 105 now let's clean this

up on top here we can do 14 + 35 to give

us 49 and then on bottom the 28 + 21 is

also 49 and then 49 + 105 is 154 next we

can cross multiply we get 3 * the

quantity x + 49 is equal to 1 * thean 2x

+ 154 and then we can distribute this

three to both of these terms 3 * X is 3x

and 3 * 49 is 147 and then on the right

hand side this 1 isn't doing anything so

we have 2X + 154 now we have an x on

both sides of the equation let's get all

the the X's to one side we can do that

by subtracting 2x from both sides of the

equation after we do this we'll have an

x on this side so let's get rid of this

147 by subtracting 147 from both sides

these two 147's will cancel each other

out and these two 2 XS will cancel each

other out we end up with 3x - 2x is

equal to X and 154 - 147 is 7 x is equal

to 7 so now we can come back up here and

in the place of the X we can plug in

seven and in the place of the 2x we can

plug in 14 and 7 + 14 is equal to 21 so

the area of the blue region is equal to

21 so our question mark is equal to 21

square units this is the answer to our

question let's put a box around it I

think this problem was brilliant

speaking of brilliant let's talk about

brilliant brilliant has thousands of

lessons to choose from and all of them

are interactive you learn by doing I

really enjoyed this course on scientific

thinking it starts with simple machines

like gears and pulley and then it

eventually worked up to Einstein's

special theory of relativity when my

favorite parts were all the fun thought

experiment questions for example if we

have this distance time graph where the

y- AIS is time and years and the x-axis

is distance in light years so while

you're at the Space Rock concert a rumor

is spread that the staral concert in 2

years is not going to be any good and

the question is asking will the negative

rumor affect the attendance of the

staral concert in 2 years and the answer

is no because while the staral concert

is 2 years in the future the distance is

three light years away so the news of of

the rumor would not even reach anybody

who's going to attend this concert I'm

struggling to explain it in just one

sentence if you want to go through it

yourself brilliant currently has a free

30-day trial visit brilliant.org Andy or

check out the link in the description

you can also get 20% off an annual

premium plan if you wanted to go through

this one the name of the course is

scientific thinking how exciting

Brilliant Blue Area Question

To try everything Brilliant has to offer for free for a full 30 days, visit https://brilliant.org/AndyMath/ . You'll also get 20% off an annual premium subscription.


hey guys this looks like a fun one what

are the values of A and B such that a

plus b is equal to a times B which is

equal to a divided by B so let's do this

first thing I'm noticing is B cannot be

zero because of the a divided by B and

we can't have division by zero next

let's try a equal to zero let's look at

these two right here plug in zero for

both of these A's zero plus b is equal

to B and zero times B is equal to zero

so we have b equal to zero but we said

earlier B cannot equal zero so we have a

contradiction that means that a cannot

equal zero and we could put a boxer on

that let's get rid of this stuff and

bring this up here now let's look at a

times b equals a divided by B let's

rewrite it like this and multiply both

sides by B and on the right hand side

since these B's cancel we just have a b

times B is B squared so we have a times

B squared is equal to a if a divide both

sides by a we get B squared is equal to

one so that means there are two possible

values for B it could be positive one or

negative one for both scenarios let's

look at a plus b equals a times B for

both of these B's let's plug in one a

times one is eight so we end up with a

plus one equals a let's subtract a from

both sides on the left hand side we have

one and on the right hand side we have

zero so we end up with one equals zero

that is a contradiction that means that

b cannot equal one let's see what

happens when we plug in b equals

negative one we end up with a minus one

is equal to negative a now we can

subtract a from both sides we get

negative one equals negative two a

divide both sides by negative two we get

one half is equal to a and there's no

contradiction here so a equals one half

and b equals negative one this is our

solution how exciting

Math Riddle Challenge


hey guys this looks like a fun one we

have a semicircle and a quarter circle

inside of a square and it wants to know

what is the area of the square this is

part two in part one we found the length

of this segment right here if you want

to check out that video I'll put a link

up here and in the description and if

you want to try to find the area of this

Square pause it right now because I'm

going to solve it in 3 2 1 we have two

unknown values here I want to call those

X and Y since we know it's a square we

know that this 10 + x is equal to the 8

+ y and then we can subtract 8 from both

sides and we get Y is equal x + 2 this

looks important let's put a box around

it next let's focus on this radius right

here which we know is equal to 8 and

let's pivot it over here to this point

of tangency and we also know this radius

is equal to 5 and let's pivot it over

here to this point of tangency and then

let's complete this right triangle and

we can now do the Pythagorean theorem we

have the quantity x + 5^ 2 plus the

quantity y + 8 2 is equal 2 13 2 right

now we have two variables but in the

place of this y we can plug in x + 2 2 +

8 is equal to 10 so now we have x + 5^ 2

+ x + 10^ 2 this x + 5^ 2 means the same

thing as x + 5 * x + 5 and the x + 10 2

is the same thing as x + 10 * x + 10

after we multiply this out we get x^2 +

10 x + 25 and after we multiply this out

we get x^2 + 20x + 100 and then we can

add these together X2 + x is = 2x^2 10 x

+ 20x is = 30X and 25 + 100 is 125 and

that's all going to equal 13 s which is

169 nice since we have a quadratic we

want to set this whole thing equal to

zero so let's subtract 169 from both

sides the 2x^2 and the 30X can be copied

down 125 - 169 is equal to -44 and 169 -

169 is equal to 0 next let's divide

everything by two on the left hand side

2x2 / 2 is x^2 30X / 2 is 15x and -44 /

2 is -22 and then on the right hand side

0 / 2 is equal to 0 now there's a couple

different ways we can solve this but

let's do the quadratic formula it says x

equal to B + orus the < TK of B ^2 - 4 a

c whole thing over 2 a a is the 1 in

front of the x^2 B is the 15 in front of

the X and C is the constant -22 in the

place of both of these B's we can plug

in 15

in the place of both of these A's we can

plug in 1 and in the place of this C we

can plug in -22 152 is equal to 225 and

-4 * 1 * -22 is equal to POS 88 inside

the < TK 225 + 88 is equal to

313 and in the denominator 2 * 1 is

equal to 2 so this gives us a positive

value and a negative value and we know

that X is a positive length so we can

get rid of the negative value so X is

equal to -15 plus the < TK of 313 whole

thing ID 2 and we know that the side

length of the square is equal to x + 10

and in the place of the X we can plug in

all of this stuff in order to combine

these let's change this 10 into a 20

over2 we can combine this ne-15 over2

and this POS 20 over2 to give us POS 5

over2 and now we know the side length is

equal to 5 + the sarek of 313 / 2 and

that is approximately 11.3 and now we're

ready for our question the area of this

square is going to be this side length

squared we can multiply this times this

and that gives us an exact value of 169

+ 5un 313 everything over two and we

give that a label of square units and

the approximate value of this is 128.

7291 and this is the answer to our

question this is the area of this Square

let's put a box around it how exciting

Part 2 Area of the Square


hey guys welcome to this functions video

let f of x equal

5x plus 3 and g of x equal 3x plus 5.

find f plus g of 3. so what does this

mean well you can think of that as f of

plus g of three so to do f of three we

take the f

function and we plug three in for the x

so it's going to be five times three

plus three plus g of three means we look

at the g

function and we plug in three four x so

this will be three

times three plus five now we can just

simplify each of these five times three

is fifteen

plus three and three times three is nine

plus the five

fifteen plus three is eighteen plus and

nine plus five is fourteen

eighteen plus fourteen equals 32 and

that's our answer to the question

f plus g of 3 is equal to 32. comment

below on what math problem you'd like me

to solve next i'll talk to you soon

bye

Operations on Functions (Addition) Fast Example #Shorts

𝐋𝐈𝐍𝐊𝐒 𝐓𝐎 𝐑𝐄𝐋𝐀𝐓𝐄𝐃 𝐓𝐎𝐏𝐈𝐂𝐒
𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐢𝐧𝐠 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬 𝐮𝐬𝐢𝐧𝐠 𝐆𝐫𝐚𝐩𝐡𝐬: https://andymath.com/evaluating-functions-using-graphs/
𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐢𝐧𝐠 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬: https://andymath.com/evaluating-functions/

𝐎𝐩𝐞𝐫𝐚𝐭𝐢𝐨𝐧𝐬 𝐨𝐟 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/operations-of-functions/

𝐂𝐨𝐦𝐩𝐨𝐬𝐢𝐭𝐞 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/composite-functions/

𝐈𝐬 𝐢𝐭 𝐚 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧?:
https://andymath.com/is-it-a-function/

𝐈𝐧𝐯𝐞𝐫𝐬𝐞 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬 𝐚𝐧𝐝 𝐑𝐞𝐥𝐚𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/inverse-functions-and-relations/

𝐃𝐞𝐬𝐜𝐫𝐢𝐛𝐢𝐧𝐠 𝐓𝐫𝐚𝐧𝐬𝐟𝐨𝐫𝐦𝐚𝐭𝐢𝐨𝐧𝐬 𝐨𝐟 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/transformations-of-functions/

𝐏𝐚𝐫𝐞𝐧𝐭 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/parent-functions/

𝐄𝐯𝐞𝐧 𝐚𝐧𝐝 𝐎𝐝𝐝 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/even-and-odd-functions/

𝐄𝐯𝐚𝐥𝐮𝐚𝐭𝐢𝐧𝐠 𝐏𝐢𝐞𝐜𝐞𝐰𝐢𝐬𝐞 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/evaluating-piecewise-functions/

𝐆𝐫𝐚𝐩𝐡𝐢𝐧𝐠 𝐏𝐢𝐞𝐜𝐞𝐰𝐢𝐬𝐞 𝐅𝐮𝐧𝐜𝐭𝐢𝐨𝐧𝐬:
https://andymath.com/piecewise-functions/

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


draw three mutually tangent unit circles

like this these three circles enclose an

area as shown here can you determine the

area of the enclosed shape pause the

video and think about it before I show

you an answer one way to answer this

problem is to realize that the centers

of the three circles can be connected to

form an equilateral triangle with a side

length of two this has an area ofare <

TK of three but then this equilateral

triangle can be decomposed into these

three wedge shaped pieces in the area we

want each of the wed shaped pieces has

an area ofk 6 so the area we want is the

< TK 3 - 3 * > 6 which simplifies to the

< of 3 - pi/ 2 let me know in the

comments if you solved it a different

way

What’s the area?

This is a short, animated visual proof finding the area bounded between three mutually tangent unit circles.

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hey guys so we want to solve the radius

of this circle first let's find the

center of the circle and this right here

is going to be the radius I first

thought it was maybe it equals 1 but it

doesn't cuz if we bring this down that's

not in the middle here cuz the edge of

the circle is not in the middle either

so we can't really do that next let's

look at this diagonal here we can

actually find out what this diagonal is

cuz this is a right angle so if we call

this side a this side B and this side C

we know that a s + B2 is equal to c^2 a

right here is equal to 2 the B is the

same as this right here so it's equal to

2 and then C will just leave as c^ S 2 s

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

is 8 we square root both sides on the

left hand side we can break < TK of8

into < of 4 * < TK of two and the < TK

of 4 is 2 and on the right hand side it

just becomes C so we can clean this

stuff up up here and let's put 2 < tk2

right here for this piece of the

diagonal let's get rid of all this stuff

unfortunately the diameter here is not 2

< tk2 because it doesn't go all the way

to the corner but we can say that the

rest of the diagonal right up here is

also 2k2 so we can calculate the

diagonal of the big square and it's

going going to be equal to the 2 < tk2

plus the 2 < tk2 which is just 4 < tk2

and we can put a box around that let's

clean this all up a little bit and this

right here is the radius we'll call that

R this is what we're trying to solve for

so we know this we know this we just got

to find this piece right here that's not

too bad because we can make a square

right here we know each of these are

equal to the radius as well so we can

call both the sides of this Square R

then we can do another diagonal we'll

move this R over to here and if we make

the diagonal equal to C then we know

that R 2 + R2 = c^2 well R2 + r 2 is 2 r

squ we can square root both sides the

left hand side can be broken out into <

TK of 2 * < TK of R 2 and the right hand

side would just bequ < TK of c squ and

then s of R squ is just R and squ of c^

squ is just C and we can smush these

together and we have > of 2 R is equal

to C up here was our diagonal C so we

can change that into < tk2 R clean

things up a little bit and now what do

we have big diagonal again this time

it's equal to this diagonal here the 2 <

tk2 plus the r plus the < tk2 r and we

can put a box around that but we know

that the big diagonal has to equal the

big diagonal so we're just going to set

the 4 < tk2 for this side of the big

diagonal equal to all of this and now we

only have one variable R and one

equation so once we solve for R we know

the answer we can subtract 2 < tk2 from

both sides the left hand side becomes 2

< tk2 the right hand side that's gone so

it's just this right here and we can

scoot these over and let's Factor an R

out of the right hand side so R * what

gives us R that's one so R * 1 is r and

then R * what gives us < tk2 R well

that's going to be < tk2 so if you

imagine Distributing this it would bring

us back to where we were up here so we

factored out the r then on the left hand

side leave that as 2 < tk2 the 1 plus <

tk2 is just a number so we can divide

both sides by that number and we end up

with r being equal to 2 < tk2 over 1 + <

tk2 and we can plug it into a calculator

and get a decimal approximation of 17157

and that keeps going it's irrational and

now we have the answer to our question

the radius is approximately

17157 and we put a box around that how

exciting

Fun Geometry Challenge

https://andymath.com/geometry-challenges/
For More Geometry Challenges, Check out the above link!


a lot of people ask me to do calculus in

area so here it is we'll start with the

horizontal line first so let's say we

want to know the area underneath this

line we pick two end points a lower

bound and an upper bound and we want to

find the area underneath our line in

between our two end points this stopped

at the x axis and there's a reason for

that if we want to know what the height

is here it's just going to be our

function value at each of these values

of X the area of this is just a

rectangle which is base time height now

what if we wanted to find the area under

a curve we don't have any formulas from

geometry for this let me use a rectangle

to make an estimation so I set the

height of this one to be the height of

my left bound we can tell this isn't

very accurate it's an underestimate by

quite a bit I can cut it into two pieces

and start using two figures to estimate

it now I've gotten closer this is the

base of our first rectangle this is the

base of our second rectangle this is the

height of our first rectangle and this

will be the height of our second

rectangle our estimate for the area

would be the area of this first

rectangle B1 * H1 plus the area of the

second rectangle B2 * H2 and we can

write this in Sigma notation as b i h i

that way the I can change from 1 to two

we'll start at I equal 1 and we stop at

I equal 2 these are two ways to say the

same thing next let's do four rectangles

here are all the bases and here are all

the heights to add all these up it would

just have four of them or we can change

Sigma notation and make this a four we

could do it again with eight rectangles

which would make eight here or change

this to an eight we do it with 16 which

would add up 16 of these or change our

Sigma notation to a 16 this shows the

advantage of Sigma notation this takes

up way less space than this right here

so let's get rid of all this stuff let's

look closely at the bi if this right

here is base I the left side would then

be x i and this right hand side would be

x sub I + 1 it's one more than this I if

we want to know how big bi is it's the

difference between the two x's and we

call that Delta X let's change this bi

to Delta X so now let's talk about the

height this height hi since this stops

when it hits our function the height is

actually defined by the function value

right there so we would say F of XI so

we can change our hi into F of XI now

let's move this Delta x to the other

side let's change this 16 to n so we can

get any number of rectangles we want and

if we take the limit as n approaches

Infinity we will get closer and closer

to the exact area under that curve we've

come up with new notation for this that

visually mimics that smoothness this

right here gets Rewritten as the

integral symbol which is much more

smooth this f ofx i doesn't make sense

anymore because it's discrete jumps we

want it to be smooth so we're just going

to say f ofx and the Delta X we also

have a smooth way of expressing this as

that distance goes to zero we call it DX

and this is an integral areas are equal

when we do the integral as we chop it

into Infinite pieces we find the exact

area that's the end of this video if you

guys want to learn how to calculate

these integrals please follow me and I

will have more videos coming out you

guys are awesome I'll talk to you soon

bye

Quick Intro Calculus & Area

The original footage is a couple years old. I edited it this weekend to fit Youtube format. I hope you guys like it!


hey guys this looks like a fun one it

was sent to me by one of my patrons it's

got a circle inscribed in a rhombus that

means all four of these lines are

tangent lines they all touch the circle

in exactly one point and that's called

the tangent point and the rhombus has

diagonals 12 and 16 and it wants to know

what is the area of the 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 let's get rid of these colors

and let's get rid of the question mark

and let's construct this vertical

diagonal this has a length of 12 m and

then if we construct the other diagonal

it's going to cut this 12 in half and

that's one of the properties of the

diagonals of a rhombus they cut each

other in half and there's another

property of the diagonals of a rhombus

they also meet at right angles let's

mark this intersection Point that's

going to be the center of the rhombus

and the center of the circle next let's

look at this longer diagonal it has a

length of 16 M and it'll split in half

to be 8 and 8 next let's focus on this

right triangle let's label the

hypotenuse C and we can do the

Pythagorean theorem it's going to be 62

+ 8^ 2 is equal to c^ 2 6 2 is 36 + 8 2

is 64 and then we can bring down the c^

2 36 + 64 is equal to 100 and then we

can square root both sides of the

equation to give us C is equal to 10 so

we now know this hypotenuse is equal to

10 and since it's a rhombus we know all

these sides are equal to 10 next let's

connect this radius to here and we know

this will also be at right angles the

radius going to the tangent point will

always be at right angles to that

tangent line let's break this into two

parts let's call this p piece a and that

means this other piece would be 10 minus

a and now we can do Pythagorean theorem

two more times let's focus on this

triangle first it'll give us r^ 2 + a^ 2

is equal to 6^ 2 next let's do this

other triangle and it will give us r^ 2

plus the quantity 10 - a^ 2 is equal to

8^2 and then let's clean these up a

little bit on this top one 62 is equal

to 36 and then we can bring down

everything else for this bottom one

let's look at the quantity 10 - a^ 2

means the same thing as 10 - A * 10 - A

and after we multiply this out it'll

give us 100 - 28 + a^ 2 and then we can

bring down the R 2 and this 8^ 2 is

equal to 64 next we can subtract 100

from both sides of the equation the left

hand side we'll have r^ 2 - 20 a + a 2

and then on the right hand side 64 - 100

is equal to -36 now we've cleaned up our

two equations and let's rearrange them

like this let's drag the r squ over here

and I want to subtract this row minus

this row this r 2us r 2 are going to

cancel other out -2 a minus nothing is

-2 a and then positive a^2 minus

positive a 2 are also going to cancel

each other out and then on the right

hand side we have -36 - 36 which will

give us -72 now to get the a by itself

we can divide both sides of the equation

by -20 and this will give us a is equal

to 3.6 let's bring this up here now that

we have a I don't think we need this top

equation anymore let's move this up here

and now for this bottom equation in the

place of the a let's plug in 3.6 6

Plus's simplify the 3.6 squar to give us

12.96 and then we can bring down

everything else let's subtract 12.96

from both sides of the equation and

that'll give us r^2 is equal to

23.04 this looks important let's clean

this up a little bit and let's bring

this up here ultimately we want to find

the area of the circle the area of a

circle is < R 2 we now know what R 2 is

equal to so in the place of this R 2

let's plug in

23.04 and this is the answer to our

question the area is equal to

23.04 piun M squar let's put a box

around it and the approximate area is

72.3 M squ how exciting

Is there an easier way?

I hope you guys like this one! Let me know if there is an easier method.


you guys this looks like a fun one 99

factorial minus 98 factorial over 99

factorial plus 98 factorial would you

want to try this on your own pause it

because I'm going to solve it in three

two one a lot of calculators aren't

going to be able to handle these numbers

these are very large numbers but we

don't need a calculator we can solve it

by hand First Step let's rewrite this 99

factorial as 99 times 98 factorial and

let's subtract this and let's rewrite

this as 1 times 98 factorial and then

we're going to do the same thing on the

bottom except we're going to be adding

the two terms all of these terms have a

98 factorial so let's factor that out of

top and bottom if we do 98 factorial

times 99 it'll bring us back to this

right here and then we're going to

subtract 98 factorial times 1. and same

thing on bottom but it's going to be a

plus and then the 98 factorial on top

and bottom those can cancel each other

out so we're just left with 99 minus 1

which is 98 and 99 plus 1 which is 100

and 98 over 100 simplifies to 49 over

50. and let's put a box around it so

this simplified is 49 over 50. how

exciting

Factorial Challenge

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


hey guys let's subtract two numbers that

are in binary notation so we do one

minus one which gives us zero zero minus

zero which gives us zero one minus zero

which gives us one and then zero minus

one we can't do so we're gonna have to

borrow so we're gonna scratch out this

one and make it a zero and we're going

to be basically borrowing two for this

column for binary a one in this column

is equal to two in this column and then

two minus one gives us one and then we

have another zero minus one which we

can't do so we're going to borrow again

and making this a two two minus one

gives us one and then once again we

can't do zero minus one so we have to

borrow but we can't borrow from here we

have to borrow all the way from over

here so that gives us a two here so

we're going to make this two a one so we

can move one over here but a one in this

column is equal to a two in this column

and two minus one is equal to one and

then one minus one is zero and then zero

minus zero is zero so we don't need

these last two zeros so one zero one one

zero one zero one minus one one one one

zero zero one is equal to one one one

one zero zero and here's the same

problem in decimal notation 181 minus

one twenty one equals sixty how exciting

Binary Subtraction Example

Here is a quick example on how to do subtraction with numbers in binary notation.


hey guys this looks like a fun one it

says a semicircle and a quarter circle

inside of a square how long is the black

tangent line this is another puzzle from

Katrina a I'm trying to solve all of

these if you want to try this one on

your own pause it right now because I'm

going to solve it in three 2 1 first

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

radius of this quarter Circle they're

telling us right here that it's equal to

8 and let's shift it over here to this

point of tangency and that means it's

going to be at right angles and from

here let's duplicate this one we're

going to do something different on this

one and this one for the top one let's

look at the radius of five and let's

shift it down here and then let's look

at this side length and let's call that

s for side length and since it's a

square we know both these side lengths

are equal to S and then last let's

complete this triangle and we know this

is a right angle because it's a square

we don't know this angle down here so

let's call it Theta and that's going to

be the same angle as this down here so

this one's also Theta and while we're

down here let's draw this side and we

know that's equal to s because it's a

side of the square let's bring this down

and keep it parallel and then let's

complete this triangle and we know this

is also a right triangle because it's

parallel to the top next let's try to

find this angle inside here since this

is a right triangle and this acute angle

is Theta that leaves 90 minus Theta for

the other angle since this is a right

angle and this piece is 90 minus Theta

that means that this angle in this

triangle will also be equal to Theta and

now we have two congruent triangles

let's focus on those triangles for this

triangle up here on the hypotenuse the

five and the eight will add together to

give us 13 and now let's rotate this one

now we have these two triangles lined up

and they are congruent the reason we

know they're congruent is because of Asa

which stands for angle side angle that's

because we have these two angles

congruent we have these two sides

congruent and we have these two angles

congruent and now since we know these

two triangles are congruent that means

if this side is 13 this side is also 13

and this side length right here is the

same thing as this side length right

here so we've now solved the problem

it's equal to 13 units let's put a box

around it and there is a part two to

this problem I also found the area of

this square if you want to check out

that video I'll try to put the link up

here and also put it in the description

how exciting

Part 1 Length of the Tangent


this video is sponsored by brilliant hey

guys this looks like a fun one it wants

to know what is the Shaded area and it's

telling us that the largest semicircle

and the smallest semicircle have the

same Center first let's look at the

radius of the smaller semicircle and

let's call it little r and then let's

look at the radius of the larger

semicircle and call it Big R when set up

like this these form a right triangle

with this distance of two so that means

we can do Pythagorean theorem it'll be

2^2 + little R2 = Big R 2 we can

simplify this we can make 2^2 equal to 4

and let's get both of these RS onto the

same side of the equation we can

subtract little R 2 from both sides and

that'll give us 4 is equal to Big R 2

minus little R 2 and this looks

important let's put a box around it this

blue area is split up into two sections

there's the left area and the right area

and if we add these together we'll have

the total blue area so to calculate this

left side it's going to be the larger

semicircle minus the smaller semicircle

and the equation for both those

semicircles is pi/ 2R 2 and then on the

right hand side we can do this

semicircle minus this semicircle so now

let's find this radius it's going to be

capital r so we can plug that in right

here and then to find the radius of this

one the diameter of this one is the Big

R plus the little r so the radius is

going to be Big R plus little r / 2 so

we can put that right here and then for

this semicircle it's going to be the

same diameter Big R plus little r so the

radius would be Big R R plus little r /

2 and last we have this little guy which

has a radius of little r now we have our

left and our right so we just want to

add all of these together let's

rearrange things a little bit these two

match and one's positive and one's

negative so they're going to cancel each

other out and that'll just leave us with

Pi / 2 big r^ 2 minus P piun / 2 little

R 2 both of these have a common factor

of Pi / 2 so let's factor that out and

Pi / 2 * r^ 2 gives us the first term

and then we subtract little R 2 to get

the second term so we're left with < / 2

* big r^ 2 - little R 2 and if you look

at our box we already know what big r^ 2

minus little R 2 is it's equal to four

so in the place of this we can

substitute four we can bring the divided

two in here with the four and then 4 / 2

is equal to 2 and now we have the answer

to our question the area is equal to 2

pi square units and we can put a box

around it I think this is a brilliant

problem speaking of brilliant I want to

give you guys an update on brilliant

I've been going through this for over a

month now and I absolutely love it there

are thousands of interactive lessons in

math data analysis programming AI all

kinds of different things it's the best

way to learn it's not about memorizing

steps it's about understanding the

concepts I was originally excited about

the math courses yet the last two weeks

I've been exploring other topics right

now I want to focus on how AI works I

feel like AI is going to play a huge

role in content Creation in the future

so I kind of want to learn how it works

I just started this course how language

models work I've been able to learn how

they build vocabulary and choose the

next word and there's also a part that

shows the differences based on what you

use for training data for example the

difference between training on Taylor

Swift lyrics versus a cookbook I'm

pretty excited to see what comes next if

you want to try a brilliant for free for

30 days visit brilliant.org Andy or

click on the link in the description you

can also get 20% off an annual premium

plan how exciting

Concentric Circle Challenge