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4vir4ik [10]
3 years ago
13

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 plan

e 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?

Mathematics
1 answer:
seropon [69]3 years ago
5 0

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}

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