How to find the surface area of a sphere - ISEE Upper Level (grades 9-12) Quantitative Reasoning

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Question

In terms of $\pi$ , give the surface area, in square inches, of a spherical water tank with a diameter of 20 feet.

Answer

feet = $20 \times 12 = 240$ inches, the diameter of the tank; half of this, or 120 inches, is the radius. Set , substitute in the surface area formula, and solve for :

$A =4\pi r^{2}$

$A =4\pi \cdot 120^{2}$

$A =4\pi \cdot 14,400$

$A = 57,600 \pi$

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Question

Which is the greater quantity?

(a) The surface area of a sphere with radius 1

(b) 12

Answer

The surface area of a sphere can be found using the formula

$A = 4\pi r^{2}$ .

The surface area of the given sphere can be found by substituting :

$A = 4\pi \cdot 1^{2} = 4\pi$

$\pi > 3$ so $4 \times \pi >4 \times 3$ , or $4\pi > 12$

This makes (a) greater.

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Question

Sphere A has volume $972 \pi \textrm{ in}^{3}$ . Sphere B has surface area $400 \pi \textrm{ in}^{2}$ . Which is the greater quantity?

(a) The radius of Sphere A

(b) The radius of Sphere B

Answer

(a) Substitute $V = 972 \pi$ in the formula for the volume of a sphere:

$\frac{4}{3}\pi r^{3}= V$

$\frac{4}{3}\pi r^{3}= 972 \pi$

$\frac{4}{3}\pi r^{3} \div \frac{4}{3}\pi = 972 \pi \div \frac{4}{3}\pi$

$\frac{4}{3}\pi r^{3} \cdot \frac{3}{4\pi} = 972 \pi\cdot \frac{3}{4\pi}$

$r^{3} = 729$

$r =\sqrt[3]{ 729} = 9$ inches

(b) Substitute $A = 400 \pi$ in the formula for the surface area of a sphere:

$4\pi r^{2}= A$

$4\pi r^{2}= 400 \pi$

$4\pi r^{2} \div 4\pi = 400 \pi \div 4\pi$

$r^{2} = 100$

$r = \sqrt{100} = 10$ inches

(b) is greater.

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Question

is a positive number. Which is the greater quantity?

(A) The surface area of a sphere with radius

(B) The surface area of a cube with edges of length

Answer

The surface area of a sphere is $4 \pi$ times the square of its radius, which here is ; the surface area of the sphere in (A) is $4 \pi t^{2}$ .

The area of one face of a cube is the square of the length of an edge, which here is , so the area of one face of the cube in (B) is $(2t) ^{2} = 4t^{2}$ . The cube has six faces so the total surface area is $6 \cdot 4t^{2} = 24t^{2}$ .

$\pi < 4$ , so $4 \pi t^{2} < 16 t^{2} < 24 t^{2}$ , giving the sphere less surface area. (B) is greater.

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