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If the radius of a sphere is increased by 50%, find the percent increase in surface area.

Solution

✅ Correct Option: 1

Original radius = rr

Radius increased by 50% means new radius = r+0.5r=1.5rr + 0.5r = 1.5r


Surface area of a sphere = 4πr24\pi r^2

Original surface area:

A1=4πr2A_1 = 4\pi r^2


New surface area with radius 1.5r1.5r:

A2=4π(1.5r)2A_2 = 4\pi(1.5r)^2

A2=4π×2.25×r2A_2 = 4\pi \times 2.25 \times r^2

A2=9πr2A_2 = 9\pi r^2


Increase in surface area:

Increase=9πr2−4πr2\text{Increase} = 9\pi r^2 - 4\pi r^2

Increase=5πr2\text{Increase} = 5\pi r^2


Percent increase:

Percent Increase=IncreaseOriginal×100\text{Percent Increase} = \dfrac{\text{Increase}}{\text{Original}} \times 100

=5πr24πr2×100= \dfrac{5\pi r^2}{4\pi r^2} \times 100

=54×100= \dfrac{5}{4} \times 100

=1.25×100= 1.25 \times 100

=125= 125

Therefore, the percent increase in surface area is 125%.

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