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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? Jan 29, 2019

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Hi Lightning,

A right square pyramid with base edges of length 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?

You have the edge slant height 10, and the base length  8sqrt2

Using pythagoras you can get the slant height in the middle of each triangle (not that the sides are isoceles triangles)

this will be

$$\sqrt{10^2-(4\sqrt2)^2}=\sqrt{100-32}=\sqrt{68}\;units$$

Now you can use pythagoras again to get the perpendicular height of the entire pyramid.

$$=\sqrt{68-(4\sqrt2)^2}=\sqrt{68-32}=\sqrt{36}=6 units.$$

sionce you are halving the height when you cut the top off, you can halve all the other dimensions for the little pyramid too.

$$volume\;\;of \;\;little\;\;pyramid\\ =\frac{1}{3}*(4\sqrt2)^2*3\\ =\frac{1}{3}*32*3\\ =32\;\;units^3$$

See if you can draw the pics that should go with tihs explanation :)

Jan 29, 2019