Answer:
[tex]V=196in^3[/tex]
Step-by-step explanation:
The volume of a pyramid is:
[tex]V=\frac{A_{b}h}{3}[/tex]
where [tex]A_{b}[/tex] is the area of the base and [tex]h[/tex] 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:
[tex]A_{b}=l^2[/tex]
where [tex]l[/tex] is the length of the side: [tex]l=8in[/tex], thus:
[tex]A_{b}=(8in)^2\\A_{b}=64in^2[/tex]
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:
[tex]hypotenuse^2=leg1^2+leg2^2[/tex]
substituting our values, and indicating that leg 1 is height h:
[tex](10in)^2=h^2+(4in)^2[/tex]
[tex]100in^2=h^2+16in^2[/tex]
and solving for the height:
[tex]h^2=100in^2-16in^2\\h^2=84in^2\\h=\sqrt{84in^2}\\ h=9.165in[/tex]
and finally we calculate the volume using this height and the area of the base:
[tex]V=\frac{A_{b}h}{3}[/tex]
[tex]V=\frac{(64in^2)(9.165in)}{3} \\V=195.5in^3[/tex]
rounding to the nearest cubic inch: [tex]V=196in^3[/tex]
