Super-Hard SAT Math Question 173 | Free SAT Prep | Aniko

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Question 173·200 Super-Hard SAT Math Questions·Geometry and Trigonometry

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A right square pyramid has a square base whose diagonal measures $30\sqrt{2}$ centimeters. The volume of the pyramid is 6,000 cubic centimeters.

Which choice is the area, in square centimeters, of one of the triangular faces of the pyramid?

Chain the geometry in the order dimensions are revealed: diagonal $\rightarrow$ side $s$ $\rightarrow$ base area $B$ $\rightarrow$ height $h$ from volume. Then switch from 3D to the 2D face: the triangular face area needs the slant height, found with a right triangle using legs $h$ and $s/2$ .

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Turn the diagonal into a side length

For a square, the diagonal and side length are related by $d=s\sqrt{2}$ . Use $d=30\sqrt{2}$ to find $s$ .

Use the pyramid volume formula

A pyramid’s volume is $V=\tfrac13Bh$ . Once you know the base area $B$ , you can solve for the height $h$ .

Be careful: face height is not the same as pyramid height

The triangular face uses the slant height. Find it using a right triangle with legs $h$ and $s/2$ .

Desmos Guide

Compute the base side length from the diagonal

Enter d=30*sqrt(2).

Then enter s=d/sqrt(2) and note the value of s.

Compute the height from the volume

Enter V=6000 and B=s^2.

Then enter h=3*V/B and note the value of h.

Compute the slant height

Enter l=sqrt(h^2+(s/2)^2).

Compute the triangular face area and match to a choice

Enter A=0.5*s*l.

Use the value shown for A to select the answer choice that matches it exactly.

Step-by-step Explanation

Find the side length of the square base

For a square with side length $s$ and diagonal $d$ , $d=s\sqrt{2}$ .

30\sqrt{2}=s\sqrt{2}\ \Rightarrow\ s=30.

Use volume to find the pyramid’s height

The base area is $B=s^2=30^2=900$ .

For a pyramid, $V=\tfrac13Bh$ , so

6000=\frac13(900)h \ \Rightarrow\ 6000=300h \ \Rightarrow\ h=20.

Find the slant height of a triangular face

The height of a triangular face (the slant height) forms a right triangle with legs:

the vertical height $h=20$
half the base edge $\tfrac{s}{2}=15$

\ell=\sqrt{20^2+15^2}=\sqrt{400+225}=\sqrt{625}=25.

Compute the area of one triangular face

Each triangular face has base $s=30$ and height (slant height) $\ell=25$ , so

\text{Area}=\frac12\cdot 30\cdot 25=375.

Therefore, the correct choice is 375.

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Step-by-step Explanation

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Step-by-step Explanation