Answer:
32 [tex]unit^{3}[/tex]
Step-by-step explanation:
Given:
- The slant length 10 units
- A right square pyramid with base edges of length 8[tex]\sqrt{2}[/tex]
Now we use Pythagoras to get the slant height in the middle of each triangle:
[tex]\sqrt{10^{2 - (4\sqrt{2} ^{2} }) }[/tex] = [tex]\sqrt{100 - 32}[/tex] = [tex]\sqrt{68}[/tex] units
One again, you can use Pythagoras again to get the perpendicular height of the entire pyramid.
[tex]\sqrt{68-(4\sqrt{2} ^{2} )}[/tex] = [tex]\sqrt{68 - 32}[/tex] = 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[tex]\sqrt{2}[/tex]
So the volume of it is:
V = 1/3 *3* 4[tex]\sqrt{2}[/tex]
= 32 [tex]unit^{3}[/tex]
