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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) ==
Line 12: Line 41:


== Solution 2 ==
== Solution 2 ==
Note that the height is <imath>12</imath> from the previous method \\
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\\
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-\frac{2}{3}x)(13-\frac{1}{2}x) = \boxed{528}</imath>
 
 
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==
==See Also==
{{AMC12 box|year=2025|ab=A|before=[[2025 AMC 12A Problems/Problem 19]]|after=[[2025 AMC 12A Problems/Problem 21]]}}
{{AMC12 box|year=2025|ab=A|num-b=19|num-a=21}}
* [[AMC 12]]
* [[AMC 12]]
* [[AMC 12 Problems and Solutions]]
* [[AMC 12 Problems and Solutions]]

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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