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Difference between revisions of "2024 AMC 8 Problems/Problem 11"
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== Solution 4 ==
 
== Solution 4 ==
−
We construct a rectangle around the triangle such that the diagonal of the rectangle is the hypotenuse of the triangle. The base of the rectangle is \(8\), and the height is \(y - 7\), so the area of the entire rectangle is: <cmath> 8(y - 7) = 8y - 56 </cmath> The diagonal splits the rectangle into two right triangles. Since the original triangle lies **below** the diagonal, the area of the triangle below the diagonal is: <cmath> \frac{8y - 56}{2} = 4y - 28 </cmath> However, the triangle on the left (with base \(2\) and height \(y - 7\)) is not part of the original triangle. Its area is: <cmath> \frac{1}{2} \cdot 2 \cdot (y - 7) = y - 7 </cmath> So, the actual triangle's area is: <cmath> (4y - 28) - (y - 7) = 3y - 21 </cmath> We're told the triangle's area is 12, so we solve: <cmath> 3y - 21 = 12 \Rightarrow 3y = 33 \Rightarrow y = \boxed{11} </cmath>
+
We construct a rectangle around the triangle such that the diagonal of the rectangle is the hypotenuse of the triangle. The base of the rectangle is \(8\), and the height is \(y - 7\), so the area of the entire rectangle is: <cmath> 8(y - 7) = 8y - 56 </cmath> The diagonal splits the rectangle into two right triangles. Since the original triangle lies **below** the diagonal, the area of the triangle below the diagonal is: <cmath> \frac{8y - 56}{2} = 4y - 28 </cmath> However, the triangle on the left (with base \(2\) and height \(y - 7\)) is not part of the original triangle. Its area is: <cmath> \frac{1}{2} \cdot 2 \cdot (y - 7) = y - 7 </cmath> So, the actual triangle's area is: <cmath> (4y - 28) - (y - 7) = 3y - 21 </cmath> We're told the triangle's area is 12, so we solve: <cmath> 3y - 21 = 12 \Rightarrow 3y = 33 \Rightarrow y = \boxed{11} </cmath>. So the solution to this problem is <math>\boxed{\textbf{(D)} 11}</math>
  
 
==Video by MathTalks 😉==
 
==Video by MathTalks 😉==

Revision as of 15:47, 28 June 2025

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

Since the triangle has a base of $6$, we can plug in that value as the base. Then, we can solve the equation for the height. Doing so gives us, \[\dfrac{6h}{2}=3h=12.\] This means that $h=4$, so that means that we have to add 4 to the $y$-coordinate. 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!)

Solution 4

We construct a rectangle around the triangle such that the diagonal of the rectangle is the hypotenuse of the triangle. The base of the rectangle is \(8\), and the height is \(y - 7\), so the area of the entire rectangle is: \[8(y - 7) = 8y - 56\] The diagonal splits the rectangle into two right triangles. Since the original triangle lies **below** the diagonal, the area of the triangle below the diagonal is: \[\frac{8y - 56}{2} = 4y - 28\] However, the triangle on the left (with base \(2\) and height \(y - 7\)) is not part of the original triangle. Its area is: \[\frac{1}{2} \cdot 2 \cdot (y - 7) = y - 7\] So, the actual triangle's area is: \[(4y - 28) - (y - 7) = 3y - 21\] We're told the triangle's area is 12, so we solve: \[3y - 21 = 12 \Rightarrow 3y = 33 \Rightarrow y = \boxed{11}\]. So the solution to this problem is $\boxed{\textbf{(D)} 11}$

Video by MathTalks 😉

https://youtu.be/qAwRUj2N46c?si=QDUY8ZUVFP29Eg4c 

 ~rc1219



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

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

~Math-X

Video Solution by Central Valley Math Circle(Goes Through Full Thought Process)

https://youtu.be/D0pFHbZ5788

~mr_mathman

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

See Also

2024 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

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