Problem Solving

The area of a circular sign whose radius is 6 feet, is 3 times the area of a rectangular sign whose length is twice its width. What is the length, in fett, of the rectangular sign?

$$\left(A\right)\sqrt{6\pi}$$
$$\left(B\right)\ 3\sqrt{\pi}$$
$$\left(C\right)\ 2\sqrt{6\pi}$$
$$\left(D\right)\ 6\sqrt{\pi}$$
$$\left(E\right)\ 12\sqrt{\pi}$$

The OA is C.

I'm really confused with this PS question. Please, can any expert assist me with it? Thanks in advanced.

Hi Luandato,


Okay, so let's start with what we know.

Radius, r = 6 ft.

Therefore, we can get the area of the circle:

$$A= \pi\times r^2$$
$$A =\pi\times 6^2$$
$$A = 36\pi$$

Now, we know that the area of the circle is 3 times the area of the rectangle, so

$$A (Rectangle) = \frac{36\pi}{3} = 12\pi$$

The equation for the area of ANY rectangle is as follows:

$$A (rectangle) = length\ \times\ width$$

And we know that the length is twice the width:

$$length\ =\ 2\times width$$

So we know that the area is:


$$A (rectangle) = \left(2\times width\right)\ \times\ \left(width\right)\ =\ 2\times width^2$$

Combining this equation with the above equation leads to:
$$12\pi = \ 2\times width^2$$

So the width is:
$$width\ =\ \sqrt{\frac{12\times \pi }{2}} =\sqrt{6\pi}$$

And since we know that the length is twice the width, we have


$$length\ =\ 2\ \times width\ =\ 2\sqrt{6\pi}$$

Answer choice C.

Best,
-OWN