Question

A right square pyramid with base edges of length $8\sqrt{2}$ units each and slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. What is the volume, in cubic units, of the new pyramid that is cut off by this plane?

Answer 1

Answer:

32 $unit^{3}$

Step-by-step explanation:

Given:

• The slant length 10 units
• A right square pyramid with base edges of length 8$\sqrt{2}$

Now we use  Pythagoras to get the slant height in the middle of each triangle:

$\sqrt{10^{2 - (4\sqrt{2} ^{2} }) }$ = $\sqrt{100 - 32}$ = $\sqrt{68}$  units

One again, you can use Pythagoras again to get the perpendicular height of the entire pyramid.

$\sqrt{68-(4\sqrt{2} ^{2} )}$ = $\sqrt{68 - 32}$ = 6 units.

Because slant edges of length 10 units each is cut by a plane that is parallel to its base and 3 units above its base. So we have the other dementions of the small right square pyramid:

• The height 3 units
• A right square pyramid with base edges of length 4$\sqrt{2}$

So the volume of it is:

V  = 1/3 *3* 4$\sqrt{2}$

= 32 $unit^{3}$