Question 187·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A right rectangular pyramid has a rectangular base with length $L$ and width $W$ . The apex of the pyramid is directly above the center of the base.

The pyramid has volume 720 cubic inches, and $L=W+8$ . The slant height of each triangular face whose base edge has length $L$ is 13 inches. (The slant height of a face is the distance from the apex to the midpoint of the face’s base edge, measured along the face.)

The values of $L$ , $W$ , and the vertical height $H$ of the pyramid are all integers.

Which choice is the surface area of the pyramid, in square inches?

A right rectangular pyramid can have two different face slant heights. A face built on one side of the base uses half the other side as the flat leg of a right triangle. Use the given slant height and the integer condition to narrow the dimensions, then check the volume in Desmos. For surface area, add the base and both pairs of triangular faces.

Hints

Hint 1
The slant height on a face with bottom edge $L$ forms a right triangle with the vertical height $H$ . Its flat leg runs from the base center to that edge’s midpoint. Which base dimension does that leg cut in half?
Hint 2
The lengths $H$ and $W$ are integers. Use the Pythagorean theorem to see why $W/2$ must also be an integer. Then look for whole-number legs in a right triangle with hypotenuse $13$ .
Hint 3
A pyramid’s volume uses the vertical height, not a face’s slant height. Test the possible dimensions against the given volume. You’ll then need the slant height of the other pair of faces to find surface area.

Step-by-step

Find the dimensions, then count the faces
Step 1
Build the right triangle for the given slant height
The apex is above the base center, so $H$ is perpendicular to the base. For a face whose bottom edge is $L$ , the flat distance from the center to that edge’s midpoint is $W/2$ . Pair a face with half the other base side, not half its own edge. The Pythagorean theorem gives: $H^2+\left(\frac W2\right)^2=13^2$ .
Step 2
Use the integer condition to narrow the dimensions
Multiply the right-triangle equation by $4$ : $4H^2+W^2=676$ . Since $H$ and $W$ are integers, $W^2$ is a multiple of $4$ , so $W$ is even. Both legs, $H$ and $W/2$ , are integers. The only positive integer legs with hypotenuse $13$ are $5$ and $12$ , so $(H,W)=(12,10)$ or $(5,24)$ .
Step 3
Translate the volume into an equation
A pyramid’s volume is one-third its base area times its vertical height. The base area is $LW$ , and $L=W+8$ , so the given volume means: $720=\frac13(W+8)WH$ . Don’t put the slant height $13$ into this formula.
Step 4
Check which dimensions give the stated volume
Type $\frac{(24+8)(24)(5)}3$ and $\frac{(10+8)(10)(12)}3$ into Desmos to test the two possibilities. Desmos shows $1280$ and $720$ , respectively. Only the second pair matches the given volume, so $W=10$ and $H=12$ .
Step 5
Find the length of the base
Use $L=W+8$ with $W=10$ : $L=10+8=18$ . The length edge is $18$ inches.
Step 6
Find the other pair’s slant height
Each face with bottom edge $W$ uses half of $L$ as its flat leg: $L/2=9$ . Its other leg is $H=12$ . Those legs make a $9$ - $12$ - $15$ right triangle, so this pair’s slant height is $15$ inches, not $13$ .
Step 7
Count the base and all four triangular faces
Each triangular face has area one-half its bottom edge times its slant height. There are two faces of each kind, plus the rectangular base: $SA=LW+2\left(\frac12L\cdot13\right)+2\left(\frac12W\cdot15\right)$ . Using $13$ for all four faces would miss the different height of the $W$ -edge faces.
Step 8
Calculate the total surface area
Type $18(10)+18(13)+10(15)$ into Desmos. It shows $564$ , so the surface area is $564$ square inches. Choice B.

Question 187·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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