As you know, the formula for the area of a circle is a power function. If the diameter of a circle is 37 μm, then for this circle to increase its area 32-times, what diameter, to the nearest whole μm, does this larger circle need to have?

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- Mar 10th 2010, 04:28 PMlittlesohiDiameter and Area!
As you know, the formula for the area of a circle is a power function. If the diameter of a circle is 37 μm, then for this circle to increase its area 32-times, what diameter, to the nearest whole μm, does this larger circle need to have?

- Mar 10th 2010, 06:03 PMSoroban
Hello, littlesohi!p

Quote:

If the diameter of a circle is 37 mm,

then for this circle to increase its area 32 times, what diameter,

to the nearest whole mm, does this larger circle need to have?

Let the original diameter be $\displaystyle d_1$

. . The area of the original circle is: .$\displaystyle A_1 \:=\:\pi\left(\frac{d_1}{2}\right)^2 \:=\:\frac{\pi}{4}(d_1)^2$

Let the diameter of the larger circle be $\displaystyle d_2$

. . The area of the larger circle is: .$\displaystyle A_2 \:=\:\pi\left(\frac{d_2}{2}\right)^2 \:=\:\frac{\pi}{4}(d_2)^2$

Since $\displaystyle A_2 = 32\!\cdot\!A_1$ we have: .$\displaystyle \frac{\pi}{4}(d_2)^2 \;=\;32\cdot\frac{\pi}{4}(d_1)^2 \quad\Rightarrow\quad (d_2)^2 \:=\:32(d_1)^2$

. . Hence: .$\displaystyle d_2\;=\;4\sqrt{2}\,d_1$

If $\displaystyle d_1 = 37$, then: .$\displaystyle d_2 \:=\:4\sqrt{2}\,(37) \;=\;209.3036072$

Therefore: .$\displaystyle d_2 \;\approx\;209\text{ mm} $