Answer to Question #147379 in Geometry for solid mensuration

Question

A monument is a stone structure in the form of a frustum of a regular square pyramid whose height is 50m and whose base edges are 4.5m and 10m respectively. Through the center of the monument is a circular opening 3m in diameter at the top and 6m in diameter at the bottom. Refer to figure 1.

8. What is the slant height of the opening?
9. What is the lateral surface area of the opening?
10. Find the volume of stone in the monument.

Answer 1

Consider frustrum of a right circular cone

1. Let $r_1=$ the radius of the upper base, $r_2=$ the radius of the lower base, and $L=$ the slant height. Then from the right triangle by the Pythagorean Theorem

$$L^2=(r_2-r_1)^2+h^2$$

$$L=\sqrt{(r_2-r_1)^2+h^2}$$

Given

$r_1=\dfrac{3}{2}=1.5(m), r_2=\dfrac{6}{2}=3(m), h=50\ m$

$$L=\sqrt{(3-1.5)^2+50^2}=\sqrt{2502.25}\approx50.022(m)$$

The slant height of the opening is approximately 50.022 m.

9. The lateral area of the frustum of a right circular cone is

$$A_L=\dfrac{1}{2}(2\pi r_2+2\pi r_1)L$$

$$A_L=\dfrac{1}{2}(2\pi (3)+2\pi (1.5))\sqrt{2502.25}\approx707.176362(m^2)$$

the lateral surface area of the opening is approximately 707.176362 m².

10.

For any Frustum, the volume is

$$V=\dfrac{1}{3}(A_1+A_2+\sqrt{A_1A_2})h$$

The volume of the frustrum of the cone

$$V_c=\dfrac{1}{3}(\pi r_1^2+\pi r_2^2+\sqrt{\pi r_1^2\pi r_2^2})h$$

$$=\dfrac{\pi}{3}( r_1^2+ r_2^2+r_1r_2)h$$

$$V_c=\dfrac{\pi}{3}((1.5)^2+(3)^2+1.5(3))(50)$$

$$\approx824.668071567(m^3)$$

$$V_p=\dfrac{1}{3}((4.5)^2+(10)^2+4.5(10))(50)$$

$$\approx2754.666666667(m^3)$$

The volume of the stone is

$$V_{stone}=V_p-V_c$$

$$\approx2754.666666667-824.668071567$$

$$\approx1930\ (m^3)$$

The volume of the stone is 1930 m³.