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2025 AMC 12A Problems/Problem 20: Difference between revisions

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The base of the pentahedron shown below is a <imath>13 \times 8</imath> rectangle, and its lateral faces are two isosceles triangles with base of length <imath>8</imath> and congruent sides of length <imath>13</imath>, and two isosceles trapezoids with bases of length <imath>7</imath> and <imath>13</imath> and nonparallel sides of length <imath>13</imath>.
The base of the pentahedron shown below is a <imath>13 \times 8</imath> rectangle, and its lateral faces are two isosceles triangles with base of length <imath>8</imath> and congruent sides of length <imath>13</imath>, and two isosceles trapezoids with bases of length <imath>7</imath> and <imath>13</imath> and nonparallel sides of length <imath>13</imath>.


[Diagram]
<asy>
import graph3;
size(200);
real l = 13;
real w = 8;
real offset = (l - 7)/2;  // 3
real midy = w/2;  // 4
real h = 12;
triple O1 = (0,0,0);
triple O2 = (l,0,0);
triple O3 = (l,w,0);
triple O4 = (0,w,0);
triple T1 = (offset, midy, h);
triple T2 = (l - offset, midy, h);
currentprojection=orthographic((-4,-6,3));
draw(O4--O1--O2, linewidth(1));
draw(O2--O3--O4, dashed + linewidth(1));
draw(O3--T2, dashed + linewidth(1));
draw(O1--T1, linewidth(1));
draw(O4--T1, linewidth(1));
draw(O2--T2, linewidth(1));
draw(T1--T2, linewidth(1));
label("13", (O1+O2)/2, 3*-Y);  // Bottom length
label("13", (O2+T2)/2, 1.5*X);
label("13", (O4+T1)/2, 2*-X);
label("8", (O1+O4)/2, 2*-X);  // Width
label("7", (T1+T2)/2, 1.5*Z);    // Top length
</asy>


What is the volume of the pentahedron?
What is the volume of the pentahedron?
<imath>\textbf{(A) } 416 \qquad \textbf{(B) } 520 \qquad \textbf{(C) }  528  \qquad  \textbf{(D) } 676 \qquad \textbf{(E) } 832</imath>


== Solution 1 (Split Into Three Parts) ==
== Solution 1 (Split Into Three Parts) ==
Notice that the triangular faces have a slant height of <imath>\sqrt{13^2-4^2}=\sqrt{153}</imath> and that the height is therefore <imath>\sqrt{153-(\frac{13-7}{2})^2} = 12</imath>. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of <imath>\frac{1}{3}(3)(8)(12) = 96</imath> and the prism has a volume of <imath>\frac{1}{2}(8)(12)(7) = 336</imath>. Thus the answer is <imath>336+96 \cdot 2 = \boxed{(C) \text{ } 528}</imath>
Notice that the triangular faces have a slant height of <imath>\sqrt{13^2-4^2}=\sqrt{153}</imath> and that the height is therefore <imath>\sqrt{153-(\frac{13-7}{2})^2} = 12</imath>. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of <imath>\frac{1}{3}(3)(8)(12) = 96</imath> and the prism has a volume of <imath>\frac{1}{2}(8)(12)(7) = 336</imath>. Thus the answer is <imath>336+96 \cdot 2 = \boxed{\text{(C) } 528}</imath>


~ Shadowleafy
~ Shadowleafy
== Solution 2 ==
Note that the height is <imath>12</imath> from the previous method \n
Note that as you go up, the length and width both decrease linearly and reach <imath>0</imath> at the end
So the answer is <imath>\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}</imath>
== Solution (3D Pythagoras) ==
Let <imath>H</imath> be the height of the solid. By 3D Pythagoras (recursion of 2D Pythagoras),
<cmath>\left( \frac{8}{2} \right)^2+\left( \frac{13-7}{2} \right)^2+H^2=13^2=4^2+3^2+H^2=5^2+H^2</cmath>
This gives the <imath>5-12-13</imath> triple, so <imath>H=12</imath>. Continue as in other solutions.
~imosilver
== Note ==
It just reminds about one of old aime problem {https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11 1983 AIME Problem 11}.
~DRA777
==Video Solution 1 by OmegaLearn==
https://youtu.be/WwMDpnuZrJ4
==Video Solution by SpreadTheLove==
https://www.youtube.com/watch?v=BadjXLma84M
==See Also==
{{AMC12 box|year=2025|ab=A|num-b=19|num-a=21}}
* [[AMC 12]]
* [[AMC 12 Problems and Solutions]]
* [[Mathematics competitions]]
* [[Mathematics competition resources]]
{{MAA Notice}}

Latest revision as of 23:03, 11 November 2025

Problem

The base of the pentahedron shown below is a $13 \times 8$ rectangle, and its lateral faces are two isosceles triangles with base of length $8$ and congruent sides of length $13$, and two isosceles trapezoids with bases of length $7$ and $13$ and nonparallel sides of length $13$.

[asy] import graph3; size(200); real l = 13; real w = 8; real offset = (l - 7)/2; // 3 real midy = w/2; // 4 real h = 12; triple O1 = (0,0,0); triple O2 = (l,0,0); triple O3 = (l,w,0); triple O4 = (0,w,0); triple T1 = (offset, midy, h); triple T2 = (l - offset, midy, h); currentprojection=orthographic((-4,-6,3)); draw(O4--O1--O2, linewidth(1)); draw(O2--O3--O4, dashed + linewidth(1)); draw(O3--T2, dashed + linewidth(1)); draw(O1--T1, linewidth(1)); draw(O4--T1, linewidth(1)); draw(O2--T2, linewidth(1)); draw(T1--T2, linewidth(1)); label("13", (O1+O2)/2, 3*-Y); // Bottom length label("13", (O2+T2)/2, 1.5*X); label("13", (O4+T1)/2, 2*-X); label("8", (O1+O4)/2, 2*-X); // Width label("7", (T1+T2)/2, 1.5*Z); // Top length [/asy]

What is the volume of the pentahedron?

$\textbf{(A) } 416 \qquad \textbf{(B) } 520 \qquad \textbf{(C) } 528 \qquad \textbf{(D) } 676 \qquad \textbf{(E) } 832$

Solution 1 (Split Into Three Parts)

Notice that the triangular faces have a slant height of $\sqrt{13^2-4^2}=\sqrt{153}$ and that the height is therefore $\sqrt{153-(\frac{13-7}{2})^2} = 12$. Then we can split the pentahedron into a triangular prism and two pyramids. The pyramids each have a volume of $\frac{1}{3}(3)(8)(12) = 96$ and the prism has a volume of $\frac{1}{2}(8)(12)(7) = 336$. Thus the answer is $336+96 \cdot 2 = \boxed{\text{(C) } 528}$

~ Shadowleafy

Solution 2

Note that the height is $12$ from the previous method \n Note that as you go up, the length and width both decrease linearly and reach $0$ at the end


So the answer is $\int_0^{12} (8 - \tfrac{2}{3}x)(13 - \tfrac{1}{2}x) dx = \boxed{528}$

Solution (3D Pythagoras)

Let $H$ be the height of the solid. By 3D Pythagoras (recursion of 2D Pythagoras), \[\left( \frac{8}{2} \right)^2+\left( \frac{13-7}{2} \right)^2+H^2=13^2=4^2+3^2+H^2=5^2+H^2\] This gives the $5-12-13$ triple, so $H=12$. Continue as in other solutions.

~imosilver

Note

It just reminds about one of old aime problem {https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_11 1983 AIME Problem 11}. ~DRA777

Video Solution 1 by OmegaLearn

https://youtu.be/WwMDpnuZrJ4

Video Solution by SpreadTheLove

https://www.youtube.com/watch?v=BadjXLma84M


See Also

2025 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 19 		Followed by
Problem 21
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 AMC 12 Problems and Solutions

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