Answer to Question #336189 in Geometry for clara

Question

In the corner of a cellar is a pyramidal hoop of coal. The base of the hoop is an isosceles right triangle whose hypotenuse is 30ft. and its height is 8ft. What is the volume of the pyramidal hoop of coal in cu.ft.?

Answer 1

isosceles right triangle:

$$hypotenuse=30ft, leg_1=leg_2=\dfrac{30}{\sqrt{2}}ft$$

$Area=A=\dfrac{1}{2}(\dfrac{30}{\sqrt{2}}ft)^2=225{ft}^2$ The volume of the pyramidal hoop of coal is

$$V=\dfrac{1}{3}(225{ft}^2)(8ft)=600{ft}^3$$

600 cubic ft.