2024 AMC 8 Problems/Problem 11

1 Problem
2 Solution 1
3 Solution 2
4 Solution 3
5 Video Solution by Central Valley Math Circle
6 Video Solution by Math-X (First understand the problem!!!)
7 Video Solution (A Clever Explanation You’ll Get Instantly)
8 Video Solution (easy to digest) by Power Solve
9 Video Solution by NiuniuMaths (Easy to understand!)
10 Video Solution 3 by SpreadTheMathLove
11 Video Solution by CosineMethod [🔥Fast and Easy🔥]
12 Video Solution by Interstigation
13 Video Solution by Daily Dose of Math (Certified, Simple, and Logical)
14 Video Solution by Dr. David
15 Video Solution by WhyMath
16 See Also

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

By the Shoelace Theorem, $\triangle ABC$ has area $\frac{1}{2}|(y \cdot 11 + 7 \cdot 5 + 7 \cdot 3) - (3 \cdot 7 + 11 \cdot 7 + 5 \cdot y)| = \frac{1}{2}|(11y + 56) - (98 + 5y)| = \frac{1}{2}|6y - 42|.$ From the problem, this is equal to $12$ . We now solve for y.

$\frac{1}{2}|6y - 42| = 12$

$|6y-42| = 24$

$6y - 42 = 24$ OR $6y - 42 = -24$

$6y = 66$ OR $6y = 18$

$y = 11$ OR $y = 3$

However, since, as stated in the problem, $y > 7$ , our only valid solution is $\boxed{\textbf{(D)} 11}$ .

~ cxsmi

Solution 3

As in the figure, the triangle is determined by the vectors $\begin{bmatrix}-2 \\ y-7\end{bmatrix}$ and $\begin{bmatrix}6\\0\end{bmatrix}$ . Recall that the absolute value of the determinant of these vectors is the area of the parallelogram determined by those vectors; the triangle has half the area of that parallelogram. Then we must have that $\frac{1}{2}|\begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix}| = 12 \implies \begin{vmatrix}-2 & y-7\\6 & 0\end{vmatrix} = \pm 24$ . Expanding the determinants, we find that $-6(y-7) = 24$ or $-6(y-7) = -24$ . Solving each equation individually, we find that $y = 3$ or $y = 11$ . However, the problem states that $y > 7$ , so the only valid solution is $\boxed{\textbf{(D)} 11}$ .

~ cxsmi (again!)

Video Solution by Central Valley Math Circle

https://youtu.be/D0pFHbZ5788

~mr_mathman

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

https://youtu.be/BaE00H2SHQM?si=qhPbhu8o5hamBrtb&t=2315

~Math-X

Video Solution (A Clever Explanation You’ll Get Instantly)

https://youtu.be/5ZIFnqymdDQ?si=6FzUoSOA5moM-gDP&t=1191

~hsnacademy

Video Solution (easy to digest) by Power Solve

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

Video Solution by NiuniuMaths (Easy to understand!)

https://www.youtube.com/watch?v=V-xN8Njd_Lc

~NiuniuMaths

Video Solution 3 by SpreadTheMathLove

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

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=-64aBL-lEVg

Video Solution by Interstigation

https://youtu.be/ktzijuZtDas&t=1063

Video Solution by Daily Dose of Math (Certified, Simple, and Logical)

https://youtu.be/8GHuS5HEoWc

~Thesmartgreekmathdude

Video Solution by Dr. David

https://youtu.be/0O4Y3RHzcR4

Video Solution by WhyMath

https://youtu.be/_r1Zh4HGA7g

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AJHSME/AMC 8 Problems and Solutions

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