Question

The edges of the bases of the frustum of a regular square pyramid have lengths 5 and 9, and the slant height of the frustum is 6.

What is the volume of the frustrum


I will give 100 points

Answer 1

Given↷

• a ⇢5
• b ⇢9
• l ⇢6

To find ↷

• The volume of the frustum

Solution ↷

We know that,

$V = \frac{1}{3} (a{}^{2} + ab + b{}^{2})h$

Also,

$l {}^{2} = h {}^{2} + (b - a) {}^{2}$

$6{}^{2} = h {}^{2} + (9 - 5) {}^{2}$

$36 = h {}^{2} +16$

$h {}^{2} = 36 - 16$

$h = \sqrt{20}$

$h = 4.47$

Now , putting the value in the formula of volume

we get,

$V = \frac{1}{3} (5{}^{2}+ 5 \times 9 + 9{}^{2})4.47$

$V = \frac{1}{3} (25 + 5 \times 9 + 81)4.47$

$V = \frac{1}{3} (25 + 45 + 81)4.47$

$V = \frac{1}{3} \times 151 \times 4.47$

$V = 224.99$

Hence, the volume of the given frustum is 224.99 cubic unit