Respuesta :
a. We know that the volume of a sphere is given by[tex] V=\frac{4}{3}\pi r^{3} [/tex].
Therefore, volume of the sphere with radius, r=5 feet will be:
[tex] V=\frac{4}{3}\pi (5)^{3} [/tex]=523.6 [tex] ft^3 [/tex].
We also know that the formula for the surface area of a sphere is given by:
[tex] S=4\pi r^{2} [/tex]
Therefore, surface area of the sphere of radius r=5 feet will be:
[tex] S=4\pi (5)^{2}=314.16 ft^2 [/tex]
b. We know that the volume of a pyramid is given by [tex] V=\frac{l\times w\times h}{3} [/tex], where h is the height which is 12 feet and w and l are the width and the length of the base which in this case are each equal to 8 feet. So, the volume of the pyramid will be:
[tex] V=\frac{8\times 8\times 12}{3}=256 ft^3 [/tex].
We also know that the formula for the surface area of the pyramid is:
[tex] S=lw+l\sqrt{(\frac{w}{2})^{2}+h^{2}}+w\sqrt{(\frac{l}{2})^{2}+h^{2}} [/tex]
After plugging in the given values we get:
[tex] S=8\times8+8\sqrt{(\frac{8}{2})^{2}+12^{2}}+8\sqrt{(\frac{8}{2})^{2}+12^{2}}\approx 266.4 [/tex]
c. For a Cone we know that the volume is given by [tex] V=\frac{1}{3}\pi r^{2}h [/tex]
We know that the height, h of the cone is 8 feet and the radius of the base of the cone, r is 5 feet. If we plug in these values we will get:
[tex] V=\frac{1}{3}\pi \times 5^{2}\times8\approx209.4 ft^{3} [/tex]
Likewise, the surface area of the cone is:
[tex] A=\pi r(r+\sqrt{h^{2}+r^{2}}) [/tex]
After plugging in the values we will get, the value of the surface area as
[tex] A=\pi \times 5(5+\sqrt{8^{2}+5^{2}})\approx226.7 ft^{2} [/tex]
d. The fourteenth hole has a rectangular solid. Its height, h is 16 feet. Width, w is 6 feet and length, l is 10 feet.
The formula for the volume of such rectangular solid is: [tex] V=lwh [/tex]
Plugging in the value gives [tex] V=10\times6\times 16=960 ft^{3} [/tex]
Likewise, the Surface Area formula is given by:
[tex] A=2(wl+hl+hw) [/tex]
Plugging in gives:[tex] A=2(6\times10+16\times10+16\times6)=632 ft^{3} [/tex]
