Question 20·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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A decorative monument consists of a right square prism with a right square pyramid placed directly on its top face. The prism has height $12$ inches, and the base of the pyramid exactly covers the top face of the prism. The altitude of each triangular lateral face of the pyramid is $15$ inches.

The total exterior surface area of the monument, including the bottom of the prism but excluding the surfaces where the prism and pyramid meet, is $1728$ square inches. What is the volume, in cubic inches, of the monument?

Your answer

A monument made of two joined solids is a composite solid. Find the shared square’s side by counting only exposed faces and solving the surface-area equation with a positive-only Desmos regression. Then add the two volumes. The key distinction is that a pyramid’s triangular-face altitude is slanted; its volume needs the perpendicular height.

Hints

Hint 1
An exposed surface is one you could paint. The joined square is hidden, so count the bottom square, four prism rectangles, and four pyramid triangles. How can you write their total using one square side length?
Hint 2
The surface-area equation has one unknown: the square’s side length. A Desmos regression can find it. Use a positive restriction because a length cannot be negative.
Hint 3
The triangular face’s altitude is the slant height, not the height used for pyramid volume. Draw a right triangle from the tip to the base’s center to an edge’s midpoint. Its flat leg is half the square’s side.

Step-by-step

Count exposed faces, then add volumes
Step 1
Write the exterior surface-area equation
Let $s$ be the side length of the shared square. The exposed bottom has area $s^2$ ; the four prism sides each have area $12s$ ; and the four pyramid faces each have area $\frac12s(15)$ . The prism’s top and pyramid’s base are hidden where they meet, so the exterior surface area gives
$1728=s^2+4(12s)+4\left(\frac12s(15)\right).$
Skip the hidden faces for surface area, but include both solids for volume.
Step 2
Find the shared square’s side length
Type $s_1^2+4(12s_1)+4\left(\frac12s_1(15)\right)\sim1728\{s_1>0\}$ . In a Desmos regression, $\sim$ means “find a value that makes these equal,” and the braces require a positive length. Under PARAMETERS, Desmos shows $s_1=18$ , so each side of the shared square is $18$ inches.
Step 3
Build the pyramid’s height triangle
The $15$ -inch face altitude is a slant height: it runs along a triangular face. For volume, you need the perpendicular height $h$ from the tip to the base’s center. The pyramid is right, so that height, half of an $18$ -inch base side, and the slant height make a right triangle. Its flat leg is $18/2=9$ , so the Pythagorean theorem gives
$h^2+9^2=15^2.$
Step 4
Find the perpendicular pyramid height
Type $h_1^2+9^2\sim15^2\{h_1>0\}$ . The positive restriction keeps a negative solution from being used as a physical length. Desmos shows $h_1=12$ under PARAMETERS, so the pyramid’s perpendicular height is $12$ inches.
Step 5
Add the prism and pyramid volumes
A prism’s volume is base area times height; a pyramid’s is one-third of base area times perpendicular height. Type $s_1^2(12)+\frac13s_1^2h_1$ . Desmos prints $5184$ , so the monument’s volume is $5184$ cubic inches. Grid in 5184.

Question 20·25 Super-Hard SAT Geometry and Trigonometry Questions·Geometry and Trigonometry

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