2024 AMC 8 Problems/Problem 11

Problem

The coordinates of $\triangle ABC$ are $A(5,7)$ , $B(11,7)$ , and $C(3,y)$ , with $y>7$ . The area of $\triangle ABC$ is 12. What is the value of $y$ ?

$[asy] draw((3,11)--(11,7)--(5,7)--(3,11)); dot((5,7)); label("$A(5,7)$",(5,7),S); dot((11,7)); label("$B(11,7)$",(11,7),S); dot((3,11)); label("$C(3,y)$",(3,11),NW); [/asy]$

$\textbf{(A) }8\qquad\textbf{(B) }9\qquad\textbf{(C) }10\qquad\textbf{(D) }11\qquad \textbf{(E) }12$

Solution 1

The triangle has base $6,$ which means its height satisfies $\dfrac{6h}{2}=3h=12.$ This means that $h=4,$ so the answer is $7+4=\boxed{(D) 11}$

Solution 2

$[asy] size(10cm); draw((5,7)--(11,7)--(3,11)--cycle); draw((3,11)--(3,7)--(5,7),red); draw((3,7.5)--(3.5,7.5)--(3.5,7)); label("$A(5,7)$", (5,7),S); label("$B(11,7)$", (11,7),S); label("$C(3,y)$", (3,11),W); label("$D(3,7)$", (3,7),SW); [/asy]$ Label point $D(3,7)$ as the point at which $CD\perp DA$ . We now have $[\triangle ABC] = [\triangle BCD] - [\triangle ACD]$ , where the brackets denote areas. On the righthand side, both of these triangles are right, so we can just compute the two sides of each triangle. The two side lengths of $\triangle ACD$ are $y-7$ and $5-3=2$ . The two side lengths of $\triangle BCD$ are $y-7$ and $11-3 = 8.$ Now,

$[\triangle ABC] = 12 = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2 = 3(y-7)$

Dividing by $3$ gives $y -7 = 4,$ so $y = \boxed{\textbf{(D)\ 11}}.$

-Benedict T (countmath1)

Video Solution (easy to digest) by Power Solve

https://www.youtube.com/watch?v=2UIVXOB4f0o

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/LBcftVLvynE

~Math-X

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=RRTxlduaDs8

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AJHSME/AMC 8 Problems and Solutions

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