A company produces a spherical object of radius 24 centimeters. A hole of radius 5 centimeters is drilled through the center of the object. Find the outer surface area of the object.
A) $73 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
B) $5 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
C) $24 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
D) $20 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
E) $96 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$

Question

A company produces a spherical object of radius 24 centimeters. A hole of radius 5 centimeters is drilled through the center of the object. Find the outer surface area of the object. 
A) $73 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
B) $5 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
C) $24 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
D) $20 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$
E) $96 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$

Quizplus · Accepted Answer

The surface area of the sphere is 
$$4 \pi r^2 = 4 \pi (24\text{ cm})^2 = 2304 \pi \text{ cm}^2.$$
The surface area of the hole is 
$$\pi r^2 = \pi (5\text{ cm})^2 = 25 \pi \text{ cm}^2.$$
Therefore, the outer surface area of the object is 
$$2304 \pi - 25 \pi = 2279 \pi \text{ cm}^2.$$
Using the approximation $\sqrt{551} \approx 23.47$, the answer can be simplified to 
$$2279 \pi \approx 96 \pi \cdot 23.47 = 2250 \pi \sqrt{551}.$$
So the answer is E, $96\pi\sqrt{551}\mathrm{~cm}^3$.

A Company Produces a Spherical Object of Radius 24 Centimeters 73π551 cm373 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }73π551​ cm3

A Company Produces a Spherical Object of Radius 24 Centimeters $73 \pi \sqrt { 551 } \mathrm {~cm} ^ { 3 }$