S3omolaballcan
Answered

A hot water heater measures 20" in diameter. What is the area of a metal base for the water heater, assuming a four-inch overhang for the base beyond the edge of the heater? (Round to the nearest tenth.)

Verified Answer :

pingwin09

A four-inch part is connected around the hot water heater as to expand. So d = 20 + 4 + 4 = 28 inches.

The area of a circle is A = π(d/2)^2
A = π(28/2)^2 = 3615.7521... inches = 3620 inches (rounded to the nearest tenth)

Answer:

\(307.72 in^2\\\)

Step-by-step explanation:

The diameter of hot water heater \(= 20"\\\)

There is overhang of \(4"\\\)

Thus the new extended radius of the base of hot water heater is

\(= \frac{20"}{2} + 4"\\= 10" + 4"\\= 14"\\\)

The heater must be of circular shape, thus the area of the water heater

\(=\pi r^{2} \\= \pi (\frac{d}{2})^2\\= (3.14)(0.5)(14")^2\\= 307.72 in^2\\\)