Navvye/Olympiad-Problems-CSV huggingface dataset:
Problem: Jamal wants to save 30 files onto disks, each with 1.44 MB space. 3 of the files take up 0.8 MB, 12 of the files take up 0.7 MB, and the rest take up 0.4 MB. It is not possible to split a file onto 2 different disks. What is the smallest number of disks needed to store all 30 files?
$\textbf{(A)}\ 12 \qquad \textbf{(B)}\ 13 \qquad \textbf{(C)}\ 14 \qquad \textbf{(D)}\ 15 \qquad \textbf{(E)} 16$

Solution: Similarly to Solution 1, we see that there must be $3$ disks to account for the $0.8$ MB file. Secondly, since there are $[30-(3+12)]-3 = 12$ files(for both 0.4 MB and 0.7 MB) left, it is easy to see that the optimal way to place the files would be $3$ files per disks for the 0.4 MB files and hence would require 4 disks.
We are left with $12$ files( $0.7$ MB), where the optimal number of files per disks is $2$ , so the optimal number of disks for this type of file would be $6$ disks. Therefore, the answer is $3+4+6=\boxed{13}$

Answer:  13
Problem: Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$

Solution: Let the time he needs to get there in be $t$ and the distance he travels be $d$ . From the given equations, we know that $d=\left(t+\frac{1}{20}\right)40$ and $d=\left(t-\frac{1}{20}\right)60$ . Setting the two equal, we have $40t+2=60t-3$ and we find $t=\frac{1}{4}$ of an hour. Substituting t back in, we find $d=12$ . From $d=rt$ , we find that $r$ , our answer, is $\boxed{48}$

Answer: B 48
Problem: Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$

Solution: Since either time he arrives at is $3$ minutes from the desired time, the answer is merely the harmonic mean of 40 and 60. 
Substituting $t=\frac ds$ and dividing both sides by $d$ , we get $\frac 2s = \frac 1{40} + \frac 1{60}$ hence $s=\boxed{48}$

Answer: B 48
Problem: Mr. Earl E. Bird gets up every day at 8:00 AM to go to work. If he drives at an average speed of 40 miles per hour, he will be late by 3 minutes. If he drives at an average speed of 60 miles per hour, he will be early by 3 minutes. How many miles per hour does Mr. Bird need to drive to get to work exactly on time?
$\textbf{(A)}\ 45 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 50 \qquad \textbf{(D)}\ 55 \qquad \textbf{(E)}\ 58$

Solution: Let x be equal to the total amount of distance he needs to cover. Let y be equal to the amount of time he would travel correctly.
Setting up a system of equations, $\frac x{40} -3 = y$ and $\frac x{60} +3 = y$
Solving, we get x = 720 and y = 15.
We divide x by y to get the average speed, $\frac {720}{15} = 48$ . Therefore, the answer is $\boxed{48}$

Answer: B 48
Problem: Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$

Solution: This is a Pythagorean triple (a $3-4-5$ actually) with legs $15$ and $20$ . The area is then $\frac{(15)(20)}{2}=150$ . Now, consider an altitude drawn to any side. Since the area remains constant, the altitude and side to which it is drawn are inversely proportional. To get the smallest altitude, it must be drawn to the hypotenuse. Let the length be $x$ ; we have $\frac{(x)(25)}{2}=\frac{(15)(20)}{2}$ , so $25x=300$ and $x$ is 12. Our answer is then $\boxed{12}$

Answer: B 12
Problem: Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$

Solution: By Heron's formula , the area is $150$ , hence the shortest altitude's length is $2\cdot\frac{150}{25}=\boxed{12}$

Answer: B 12
Problem: Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$

Solution: The sides are in the ratio of $3-4-5$ , and are therefore a Pythagorean Triple and a part of a right triangle.
We know two altitudes, the legs $15$ and $20$ . All that is left is to check the altitude from the right angle vertex to the hypotenuse. There is a rule that states that the three triangles formed when the altitude is dropped (the big triangle and two smaller ones it gets split into) are similar! We take the triangle with hypotenuse $20$ . The ratio of sides between that triangle and the bigger one is $4:5$ . Since the side of length $15$ in the big triangle corresponds with our altitude, we take $15\cdot\frac{4}{5}$ to get our answer ${12}$ or $\boxed{12}$

Answer: B 12
Problem: Given a triangle with side lengths 15, 20, and 25, find the triangle's shortest altitude.
$\textbf{(A)}\ 6 \qquad \textbf{(B)}\ 12 \qquad \textbf{(C)}\ 12.5 \qquad \textbf{(D)}\ 13 \qquad \textbf{(E)}\ 15$

Solution: The triangle forms a $3-4-5$ right triangle, so we can first draw a triangle with legs $3$ and $4$ and hypotenuse $5$ . The two altitudes $3$ and $4$ are known, and the area is $6$ ; therefore $5x/2 = 6$ , and $x=12/5$ , which is clearly less than $3$ or $4$ . Now scale the triangle back up by multiplying every dimension by $5$ . The smallest altitude is then $12/5 * 5$ ${12}$ or $\boxed{12}$

Answer: B 12
Problem: Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$

Solution: Consider a general quadratic with the coefficient of $x^2$ being $1$ and the roots being $r$ and $s$ . It can be factored as $(x-r)(x-s)$ which is just $x^2-(r+s)x+rs$ . Thus, the sum of the roots is the negative of the coefficient of $x$ and the product is the constant term. (In general, this leads to Vieta's Formulas ).
We now have that the sum of the two roots is $63$ while the product is $k$ . Since both roots are primes, one must be $2$ , otherwise, the sum would be even. That means the other root is $61$ and the product must be $122$ . Hence, our answer is $\boxed{1}$

Answer: B 1
Problem: Both roots of the quadratic equation $x^2 - 63x + k = 0$ are prime numbers. The number of possible values of $k$ is
$\text{(A)}\ 0 \qquad \text{(B)}\ 1 \qquad \text{(C)}\ 2 \qquad \text{(D)}\ 4 \qquad \text{(E) more than 4}$

Solution: By Vieta's you have \[a+b=63\] \[ab=k\] Since we know odd + even = odd, we must have either $a$ or $b$ equal to $2$ and the other equal to $63-2=61.$ Both of these are prime so they satisfy the restraints. Thus there is $\boxed{1}$ solution.

Answer: B 1
Problem: Using the digits 1, 2, 3, 4, 5, 6, 7, and 9, form 4 two-digit prime numbers, using each digit only once. What is the sum of the 4 prime numbers?
$\text{(A)}\ 150 \qquad \text{(B)}\ 160 \qquad \text{(C)}\ 170 \qquad \text{(D)}\ 180 \qquad \text{(E)}\ 190$

Solution: Since a multiple-digit prime number is not divisible by either 2 or 5, it must end with 1, 3, 7, or 9 in the units place. The remaining digits given must therefore appear in the tens place. Hence our answer is $20 + 40 + 50 + 60 + 1 + 3 + 7 + 9 = 190\Rightarrow\boxed{190}$

Answer: E 190
Problem: Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$

Solution: We will simulate the process in steps.
In the beginning, we have:
In the first step we pour $4/2=2$ ounces of coffee from cup $1$ to cup $2$ , getting:
In the second step we pour $2/2=1$ ounce of coffee and $4/2=2$ ounces of cream from cup $2$ to cup $1$ , getting:
Hence at the end we have $3+2=5$ ounces of liquid in cup $1$ , and out of these $2$ ounces is cream. Thus the answer is $\boxed{25}$

Answer: D 25
Problem: Sarah places four ounces of coffee into an eight-ounce cup and four ounces of cream into a second cup of the same size. She then pours half the coffee from the first cup to the second and, after stirring thoroughly, pours half the liquid in the second cup back to the first. What fraction of the liquid in the first cup is now cream?
$\mathrm{(A) \ } \frac{1}{4}\qquad \mathrm{(B) \ } \frac13\qquad \mathrm{(C) \ } \frac38\qquad \mathrm{(D) \ } \frac25\qquad \mathrm{(E) \ } \frac12$

Solution: Let's consider this in steps. 
We have 4 ounces of coffee in the first cup. 
We hace 4 ounces of cream in the second cup. 
We take half of the coffee in the first cup(2 ounces), and add it to the second cup, yielding 6 ounces in total in the second cup(in a 1:2 ratio between coffee and cream, respecitvely).
We then take half of the second cup and pour it into the first cup. $6/2=3$ , so there is now 5 ounces in the first cup, 2 coffee and 3 the mixture.
Remember that the mixture is in a 1:2 ratio between coffee and cream. So, coffee has one ounce and cream has 2 ounces. 
In total, there is 2 ounces in the 5 ounce first cup. Putting 2 over 5, we get the answer. Therefore, the answer is $\boxed{25}$

Answer: D 25
Problem: A 3x3x3 cube is made of $27$ normal dice. Each die's opposite sides sum to $7$ . What is the smallest possible sum of all of the values visible on the $6$ faces of the large cube?
$\text{(A)}\ 60 \qquad \text{(B)}\ 72 \qquad \text{(C)}\ 84 \qquad \text{(D)}\ 90 \qquad \text{(E)} 96$

Solution: In a 3x3x3 cube, there are $8$ cubes with three faces showing, $12$ with two faces showing and $6$ with one face showing. The smallest sum with three faces showing is $1+2+3=6$ , with two faces showing is $1+2=3$ , and with one face showing is $1$ . Hence, the smallest possible sum is $8(6)+12(3)+6(1)=48+36+6=90$ . Our answer is thus $\boxed{90}$

Answer: D 90
Problem: The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$

Solution: As the unique mode is $8$ , there are at least two $8$ s.
As the range is $8$ and one of the numbers is $8$ , the largest one can be at most $16$
If the largest one is $16$ , then the smallest one is $8$ , and thus the mean is strictly larger than $8$ , which is a contradiction.
If we have 2 8's we can add find the numbers 4, 6, 7, 8, 8, 9, 10, 12.
This is a possible solution but has not reached the maximum.
If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14.
We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{14}$ . ~By QWERTYUIOPASDFGHJKLZXCVBNM

Answer: D 14
Problem: The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
$\text{(A) }11 \qquad \text{(B) }12 \qquad \text{(C) }13 \qquad \text{(D) }14 \qquad \text{(E) }15$

Solution: We could express this collection as integers $a\textsubscript{1}$ through $a\textsubscript{8}$ , with $a\textsubscript{1}$ being the smallest and $a\textsubscript{8}$ being the largest.
Since the mean is $8$ , we know that $a\textsubscript{4}$ and $a\textsubscript{5}$ must also be $8$ . If they were not, the other numbers, which are lesser and greater than $a\textsubscript{4}$ and $a\textsubscript{5}$ respectively, would not be able to satisfy the condition that $8$ is the mode.
There are $8$ terms and the mean is $8$ . This tells us that the sum of all the numbers is $64$
We want to maximize the value of $a\textsubscript{8}$ , so we set $a\textsubscript{6}$ and $a\textsubscript{7}$ to $8$ as well.
Knowing that we want to minimize numbers and that the range is $8$ , we set $a\textsubscript{1}$ $a\textsubscript{2}$ , and $a\textsubscript{3}$ equal to $a\textsubscript{8} - 8$
$a\textsubscript{1}$ $a\textsubscript{2}$ $a\textsubscript{3}$ $a\textsubscript{4}$ $a\textsubscript{5}$ $a\textsubscript{6}$ $a\textsubscript{7}$ $a\textsubscript{8}$ $=$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $a\textsubscript{8} - 8$ $8$ $8$ $8$ $8$ $a\textsubscript{8}$
Letting the sum of all the numbers be $64$ , we find that $32 + 4a_8 - 24 = 64$ , which simplifies to $4a_8 = 56$ . Solving, we get $\boxed{14}$ . ~By SK80, mod_x for minor edits

Answer: D 14
Problem: A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution: The pattern is quite simple to see after listing a couple of terms.
\[\begin{tabular}{|r|r|r|} \hline \#&\text{Removed}&\text{Left}\\ \hline 1&10&90\\ 2&9&81\\ 3&9&72\\ 4&8&64\\ 5&8&56\\ 6&7&49\\ 7&7&42\\ 8&6&36\\ 9&6&30\\ 10&5&25\\ 11&5&20\\ 12&4&16\\ 13&4&12\\ 14&3&9\\ 15&3&6\\ 16&2&4\\ 17&2&2\\ 18&1&1\\     \hline \end{tabular}\]
Thus, the answer is $\boxed{18}$

Answer: C 18
Problem: A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution: Given $n^2$ tiles, a step removes $n$ tiles, leaving $n^2 - n$ tiles behind. Now, $(n-1)^2 = n^2 - n + (1-n) < n^2 - n < n^2$ , so in the next step $n-1$ tiles are removed. This gives $(n^2 - n) - (n-1) = n^2 - 2n + 1 = (n-1)^2$ , another perfect square.
Thus each two steps we cycle down a perfect square, and in $(10-1)\times 2 = 18$ steps, we are left with $1$ tile, hence our answer is $\boxed{18}$

Answer: C 18
Problem: A set of tiles numbered 1 through 100 is modified repeatedly by the following operation: remove all tiles numbered with a perfect square , and renumber the remaining tiles consecutively starting with 1. How many times must the operation be performed to reduce the number of tiles in the set to one?
$\text{(A)}\ 10 \qquad \text{(B)}\ 11 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 19 \qquad \text{(E)}\ 20$

Solution: We start of with $100 = 10 \cdot 10$ numbers. When we use the certain operation, call if $P(x)$ , have $100 - 10 = 90 = 10 \cdot 9$ .
Then we do $P(x)$ again, to subtract $9$ numbers and get $9 \cdot 9$ . In the end, we will want $1 = 1 \cdot 1$ . We can say we have to use $P(x)$ once to make $n \cdot n$ into $n \cdot (n-1)$ . Thus we must use it twice to get from $n \cdot n$ to $(n-1)(n-1)$ . For example, it takes us $2$ of $P(x)$ to get from $10 \cdot 10$ to $9 \cdot 9$ . Then $2$ of $P(x)$ to get from $9 \cdot 9$ to $8 \cdot 8$ . You should try this with $7$ and $6$ , and see it works. This means we can have $n$ be the number we start with, and $1$ be the number we want. Then we would use $P(x)$ $2(n - 1)$ times to get $1 \cdot 1$ . Substituting $n$ for $10$ we get $2(10-1) = 2 \cdot 9 = \boxed{18}$ .
- Wiselion

Answer:  18
Problem: Points $A,B,C$ and $D$ lie on a line, in that order, with $AB = CD$ and $BC = 12$ . Point $E$ is not on the line, and $BE = CE = 10$ . The perimeter of $\triangle AED$ is twice the perimeter of $\triangle BEC$ . Find $AB$
$\text{(A)}\ 15/2 \qquad \text{(B)}\ 8 \qquad \text{(C)}\ 17/2 \qquad \text{(D)}\ 9 \qquad \text{(E)}\ 19/2$

Solution: Let $M$ be the foot of the altitude from $E$ to $BC.$ Then $MB=MC=6$ because $\triangle BEC$ is isosceles. By the Pythagorean triple $(6,8,10)$ the altitude is $8.$ Since $(8,15,17)$ is the only primitive Pythagorean triple with leg $8,$ we test $AE=DE=17,AM=DM=15.$ Since $2(10+10+12)=(17+17+2\cdot 15)$ this works, giving us $AB=15-6=\boxed{9}.$

Answer: D 9
Problem: In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution: It shouldn't be hard to use trigonometry to bash this and find the height, but there is a much easier way. Extend $\overline{AD}$ and $\overline{BC}$ to meet at point $E$
Since $\overline{AB} || \overline{CD}$ we have $\triangle AEB \sim \triangle DEC$ , with the ratio of proportionality being $\frac {39}{52} = \frac {3}{4}$ . Thus \begin{align*} \frac {CE}{CE + 12} = \frac {3}{4} & \Longrightarrow CE = 36 \\ \frac {DE}{DE + 5} = \frac {3}{4} & \Longrightarrow DE = 15 \end{align*} So the sides of $\triangle CDE$ are $15,36,39$ , which we recognize to be a $5 - 12 - 13$ right triangle . Therefore (we could simplify some of the calculation using that the ratio of areas is equal to the ratio of the sides squared), \[[ABCD] = [ABE] - [CDE] = \frac {1}{2}\cdot 20 \cdot 48 - \frac {1}{2} \cdot 15 \cdot 36 = \boxed{210}\]

Answer: C 210
Problem: In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution: Draw altitudes from points $C$ and $D$
Translate the triangle $ADD'$ so that $DD'$ coincides with $CC'$ . We get the following triangle:
The length of $A'B$ in this triangle is equal to the length of the original $AB$ , minus the length of $CD$ .
Thus $A'B = 52 - 39 = 13$
Therefore $A'BC$ is a well-known $(5,12,13)$ right triangle. Its area is $[A'BC]=\frac{A'C\cdot BC}2 = \frac{5\cdot 12}2 = 30$ , and therefore its altitude $CC'$ is $\frac{[A'BC]}{A'B} = \frac{60}{13}$
Now the area of the original trapezoid is $\frac{(AB+CD)\cdot CC'}2 = \frac{91 \cdot 60}{13 \cdot 2} = 7\cdot 30 = \boxed{210}$

Answer: C 210
Problem: In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution: Draw altitudes from points $C$ and $D$
Call the length of $AD'$ to be $y$ , the length of $BC'$ to be $z$ , and the height of the trapezoid to be $x$ .
By the Pythagorean Theorem, we have: \[z^2 + x^2 = 144\] \[y^2 + x^2 = 25\]
Subtracting these two equation yields: \[z^2-y^2=119 \implies (z+y)(z-y)=119\]
We also have: $z+y=52-39=13$
We can substitute the value of $z+y$ into the equation we just obtained, so we now have:
\[(13) (z-y)=119 \implies z-y=\frac{119}{13}\]
We can add the $z+y$ and the $z-y$ equation to find the value of $z$ , which simplifies down to be $\frac{144}{13}$ . Finally, we can plug in $z$ and use the Pythagorean theorem to find the height of the trapezoid.
\[\frac{12^4}{13^2} + x^2 = 12^2 \implies x^2 = \frac{(12^2)(13^2)}{13^2} -\frac{12^4}{13^2}  \implies x^2 = \frac{(12 \cdot 13)^2 - (144)^2}{13^2} \implies x^2 = \frac{(156+144)(156-144)}{13^2} \implies x = \sqrt{\frac{3600}{169}} = \frac{60}{13}\]
Now that we have the height of the trapezoid, we can multiply this by the median to find our answer.
The median of the trapezoid is $\frac{39+52}{2} = \frac{91}{2}$ , and multiplying this and the height of the trapezoid gets us:
\[\frac{60 \cdot 91}{13 \cdot 2} = \boxed{210}\]

Answer: C 210
Problem: In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution: We construct a line segment parallel to $\overline{AD}$ from point $C$ to line $\overline{AB},$ and label the intersection of this segment with line $\overline{AB}$ as point $E.$ Then quadrilateral $AECD$ is a parallelogram, so $CE=5, AE=39,$ and $EB=13.$ Triangle $EBC$ is therefore a right triangle, with area $\frac12 \cdot 5 \cdot 12 = 30.$
By continuing to split $\overline{AB}$ and $\overline{CD}$ into segments of length $13,$ we can connect these vertices in a "zig-zag," creating seven congruent right triangles, each with sides $5,12,$ and $13,$ and each with area $30.$ The total area is therefore $7 \cdot 30 = \boxed{210}.$

Answer: C 210
Problem: In trapezoid $ABCD$ with bases $AB$ and $CD$ , we have $AB = 52$ $BC = 12$ $CD = 39$ , and $DA = 5$ . The area of $ABCD$ is
[asy] pair A,B,C,D; A=(0,0); B=(52,0); C=(38,20); D=(5,20); dot(A); dot(B); dot(C); dot(D); draw(A--B--C--D--cycle); label("$A$",A,S); label("$B$",B,S); label("$C$",C,N); label("$D$",D,N); label("52",(A+B)/2,S); label("39",(C+D)/2,N); label("12",(B+C)/2,E); label("5",(D+A)/2,W); [/asy]
$\text{(A)}\ 182 \qquad \text{(B)}\ 195 \qquad \text{(C)}\ 210 \qquad \text{(D)}\ 234 \qquad \text{(E)}\ 260$

Solution: From Solution $2$ we know that the the altitude of the trapezoid is $\frac{60}{13}$ and the triangle's area is $30$ .
Note that once we remove the triangle we get a rectangle with length $39$ and height $\frac{60}{13}$ .
The numbers multiply nicely to get $180+30=\boxed{210}$ -harsha12345

Answer: C 210