Question

A circle has a radius of 9 inches. The Radius is multiplied by 2/3 to form a second circle. How is the ratio of the areas related to the ratio of the radii?

Answer 1

Answer:

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{r_{1} }{r_{2}}) ^{2}$

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii.

Step-by-step explanation:

Radius of first circle $(r_{1})$ = 9 inches

Area of first circle = $\pi r_{1} ^{2}$

Area of first circle = 9 × 9 × π = 81 π

Now, since the radius is multiplied by 2/3 for from a new circle.

∴ Radius of the second circle = $9 \times \frac{2}{3} = 6\ inches$

Area of second circle =  $\pi r_{2} ^{2}$

Area of second circle = 6 × 6 × π = 36 π

Now,

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81\pi }{36\pi }$

$\frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)} = \frac{81}{36} = (\frac{9}{6}) ^{2} = (\frac{r_{1} }{r_{2}}) ^{2}$

∵ $(r_{1})$ = 9 inches and $(r_{2})$ = 6 inches

The above expression shows that ratios of the areas of the circles are equal to the square of the ratio of their radii. i.e., $\frac {radius\ of\ first\ circle)^{2} }{(radius\ of\ second\ circle)^{2} } = \frac {(Area\ of\ first\ circle) }{(Area\ of\ second\ circle)}$