The amount of material needed to cover a ball varies directly with the square of the radius. It takes about 221.63 cm2 of material to cover a ball with a radius of 4.2 cm. What is the radius of a ball that needs 352.92 cm2 of material to cover it?

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Answer 1 · 2016-09-29T18:58:51

$r = 5.3 c m$ $r=5.3 cm$

Since the amount of material necessary to cover the ball varies directly with the square of the radius, we know

${4.2}^{2} \propto 221.63$ $4.2^2 prop 221.63$
Or: ${4.2}^{2} = k \cdot 221.63$ $4.2^2 = k*221.63$ where k is some constant

Now we want to find $r_{2}$ $r_2$ such that ${(r_{2})}_{2}^{2} = k \cdot 352.92$ $(r_2)^2 = k*352.92$

We could solve for k and then plug in, but a simpler method would be to divide our equations and solve for $r_{2}$ $r_2$ directly:

$\frac{{4.2}^{2}}{{(r_{2})}_{2}^{2}} = \frac{k \cdot 221.63}{k \cdot 352.92}$ $4.2^2 / (r_2)^2 = (k*221.63)/(k*352.92)$

Notice how the k's cancel out, and we can solve for $r_{2}$ $r_2$ directly.

${4.2}^{2} \cdot 352.92 = {(r_{2})}_{2}^{2} \cdot 221.63$ $4.2^2*352.92 = (r_2)^2*221.63$

$\frac{{4.2}^{2} \cdot 352.92}{221.63} = {(r_{2})}_{2}^{2}$ $(4.2^2*352.92) / 221.63 = (r_2)^2$

$r_{2} = \sqrt{\frac{{4.2}^{2} \cdot 352.92}{221.63}}$ $r_2 = sqrt((4.2^2*352.92)/221.63)$

$r_{2} = 5.3 c m$ $r_2 = 5.3 cm$