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Difference between revisions of "2024 AMC 8 Problems/Problem 11"
 
(41 intermediate revisions by 16 users not shown)
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==Solution 1==
 
==Solution 1==
−
The triangle has base <math>6,</math> which means its height satisfies
+
Since the triangle has a base of <math>6</math>, we can plug in that value as the base. Then, we can solve the equation for the height. Doing so gives us,
 
<cmath>\dfrac{6h}{2}=3h=12.</cmath>
 
<cmath>\dfrac{6h}{2}=3h=12.</cmath>
−
This means that <math>h=4, </math> so the answer is <math>7+4=\boxed{(D) 11}</math>
+
This means that <math>h=4</math>, so that means that we have to add 4 to the <math>y</math>-coordinate. So the answer is <math>7+4=\boxed{(D) 11}</math>
  
 
==Solution 2==
 
==Solution 2==
−
<asy>
+
By the Shoelace Theorem, <math>\triangle ABC</math> has area <cmath>\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|.</cmath> From the problem, this is equal to <math>12</math>. We now solve for y.
−
size(10cm);
+
 
−
draw((5,7)--(11,7)--(3,11)--cycle);
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<math>\frac{1}{2}|6y - 42| = 12</math>
−
draw((3,11)--(3,7)--(5,7),red);
+
 
−
draw((3,7.5)--(3.5,7.5)--(3.5,7));
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<math>|6y-42| = 24</math>
−
label("$A(5,7)$", (5,7),S);
+
 
−
label("$B(11,7)$", (11,7),S);
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<math>6y - 42 = 24</math> OR <math>6y - 42 = -24</math>
−
label("$C(3,y)$", (3,11),W);
+
 
−
label("$D(3,7)$", (3,7),SW);
+
<math>6y = 66</math> OR <math>6y = 18</math>
−
</asy>
+
 
−
Label point <math>D(3,7)</math> as the point at which <math>CD\perp DA</math>. We now have <math>[\triangle ABC] = [\triangle BCD] - [\triangle ACD]</math>, 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 <math>\triangle ACD</math> are <math>y-7</math> and <math>5-3=2</math>. The two side lengths of <math>\triangle BCD</math> are <math>y-7</math> and <math>11-3 = 8.</math> Now,
+
<math>y = 11</math> OR <math>y = 3</math>
 +
 
 +
However, since, as stated in the problem, <math>y > 7</math>, our only valid solution is <math>\boxed{\textbf{(D)} 11}</math>.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi]
 +
 
 +
==Solution 3==
 +
As in the figure, the triangle is determined by the vectors <math>\begin{bmatrix}-2 \\ y-7\end{bmatrix}</math> and <math>\begin{bmatrix}6\\0\end{bmatrix}</math>. 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 <math>\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</math>. Expanding the determinants, we find that <math>-6(y-7) = 24</math> or <math>-6(y-7) = -24</math>. Solving each equation individually, we find that <math>y = 3</math> or <math>y = 11</math>. However, the problem states that <math>y > 7</math>, so the only valid solution is <math>\boxed{\textbf{(D)} 11}</math>.
 +
 
 +
~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] (again!)
 +
 
 +
==Solution 4==
 +
Draw a rectangle so that the hypotenuse is the diagonal of the rectangle. The base is 8 and the height is \( y - 7 \), so the area is
 +
<math>{8(y - 7) = 8y - 56}</math>
 +
 
 +
Inside the rectangle, there are 2 extra right triangles along with the original triangle. The larger right triangle has an area of half the rectangle, so it has an area of
 +
 
 +
<math>{\frac{8y - 56}{2} = 4y - 28}</math>
 +
 
 +
The smaller right triangle has a base of 2 and a height of \( y - 7 \), so its area is
 +
 
 +
<math>{\frac{2(y - 7)}{2} = y - 7}</math>
 +
 
 +
Subtracting the extra right triangles from the area of the rectangle, you get
 +
 
 +
<math>{(8y - 56) - (4y - 28) - (y - 7) = 3y - 21}</math>
 +
 
 +
Since the problem told us that the original triangle had an area of 12, you get the equation
 +
 
 +
<math>{3y - 21 = 12}</math>
 +
 
 +
<math>{3y = 33}</math>
 +
 
 +
<math>{y = 11}</math>
 +
 
 +
So the answer to the problem is \( \boxed{\textbf{D}} \).
 +
 
 +
==Video by MathTalks 😉==
 +
 
 +
https://youtu.be/qAwRUj2N46c?si=QDUY8ZUVFP29Eg4c
 +
 
 +
  ~rc1219
  
−
<cmath>[\triangle ABC] = 12  = \frac{1}{2}\cdot (y-7)\cdot 8 - \frac{1}{2}\cdot (y-7)\cdot 2  = 3(y-7)</cmath>
 
−
   
 
−
Dividing by <math>3</math> gives <math>y -7 = 4,</math> so <math>y = \boxed{\textbf{(D)\ 11}}.</math>
 
  
−
-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!!!)==
 
==Video Solution by Math-X (First understand the problem!!!)==
Line 52: Line 87:
  
 
~Math-X
 
~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!)==
 
==Video Solution by NiuniuMaths (Easy to understand!)==
 
https://www.youtube.com/watch?v=V-xN8Njd_Lc
 
https://www.youtube.com/watch?v=V-xN8Njd_Lc
Line 63: Line 115:
  
 
https://www.youtube.com/watch?v=-64aBL-lEVg
 
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==
 
==See Also==
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{{MAA Notice}}
 
{{MAA Notice}}
  
−
==See Also==
+
[[Category:Introductory Geometry Problems]]
−
{{AMC8 box|year=2024|num-b=10|num-a=12}}
 
−
{{MAA Notice}}
 

Latest revision as of 18:44, 29 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

Draw a rectangle so that the hypotenuse is the diagonal of the rectangle. The base is 8 and the height is \( y - 7 \), so the area is ${8(y - 7) = 8y - 56}$

Inside the rectangle, there are 2 extra right triangles along with the original triangle. The larger right triangle has an area of half the rectangle, so it has an area of

${\frac{8y - 56}{2} = 4y - 28}$

The smaller right triangle has a base of 2 and a height of \( y - 7 \), so its area is

${\frac{2(y - 7)}{2} = y - 7}$

Subtracting the extra right triangles from the area of the rectangle, you get

${(8y - 56) - (4y - 28) - (y - 7) = 3y - 21}$

Since the problem told us that the original triangle had an area of 12, you get the equation

${3y - 21 = 12}$

${3y = 33}$

${y = 11}$

So the answer to the problem is \( \boxed{\textbf{D}} \).

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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