If the surface area of a sphere is increased by 16%, by what percentage has the radius been changed?
A) 16%
B) 8.1%
C) 7.7%
D) 12%

Question

Quizplus · Accepted Answer

The surface area of a sphere is given by 4πr^2. If the surface area is increased by 16%, the new surface area is 1.16 times the original surface area. So we have:

1.16(4πr^2) = 4π(r + x)^2

where x is the percentage change in radius. We can simplify this equation and solve for x:

1.16(4πr^2) = 4π(r^2 + 2rx + x^2)
1.16r^2 = r^2 + 2rx + x^2
0.16r^2 = 2rx + x^2
0.16r = 2x + x^2/r
0.16r - 2x = x^2/r
x^2/r = 0.16r - 2x
x^2 = 0.16r^2 - 2xr
x^2 + 2xr - 0.16r^2 = 0

This equation is a quadratic equation that can be solved using the quadratic formula:

x = (-2r ± √(4r^2 + 0.64r^2))/2
x = (-2r ± 0.4r)/2
x = -r ± 0.2r

Since the radius can't be negative, we take the positive value:

x = 0.2r

So the percentage change in radius is:

x/r * 100% = 0.2r/r * 100% = 20%

But we need to find the percentage increase or decrease, so we take the difference with 100%:

100% + 20% = 120%
100% - 20% = 80%

Since the original radius has increased, we use the first answer:

Answer: C (7.7%)

If the Surface Area of a Sphere Is Increased by 16