Question

For her presentation on the Wonders of the World, Mary baked a square pyramid-shaped cake as pictured below. The slant height of the cake is 10 inches. Each edge of the square-shaped base is 8 inches. Find the volume of Mary's cake. Round your answer to the nearest cubic inch.

Answer 1

`V=196in^3`

Explanation:

The volume of a pyramid is:

`V=(A_(b)h)/(3)`

where
`A_(b)` is the area of the base and
`h` is the height (the perpendicular measurement between base and highest point, not the slant height)

Since the base is a square, the area is given by:

`A_(b)=l^2`

where
`l` is the length of the side:
`l=8in`, thus:

`A_(b)=(8in)^2\\A_(b)=64in^2`

Now we need to find the height, for this we use the right triangle that forms with half of a square side (8in/2 = 4in), the slant height (10in), and the height.

In this right triangle, the slant height is the hypotenuse, the leg 1 is the unknown height, and leg 2 is half of the square side.

Using pythagoras:

`hypotenuse^2=leg1^2+leg2^2`

substituting our values, and indicating that leg 1 is height h:

`(10in)^2=h^2+(4in)^2`

`100in^2=h^2+16in^2`

and solving for the height:

`h^2=100in^2-16in^2\\h^2=84in^2\\h=√(84in^2)\\ h=9.165in`

and finally we calculate the volume using this height and the area of the base:

`V=(A_(b)h)/(3)`

`V=((64in^2)(9.165in))/(3) \\V=195.5in^3`

rounding to the nearest cubic inch:
`V=196in^3`