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

Answer:

$V=196in^3$

Step-by-step explanation:

The volume of a pyramid is:

$V=\frac{A_{b}h}{3}$

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:

$h^2=100in^2-16in^2\\h^2=84in^2\\h=\sqrt{84in^2}\\ h=9.165in$

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

$V=\frac{A_{b}h}{3}$

$V=\frac{(64in^2)(9.165in)}{3} \\V=195.5in^3$

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