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.
The area of a circular sign whose radius is 6 feet, is 3...
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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
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
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Area of a circular sign whose radius is 6 feetLUANDATO wrote: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?
$$\left(A\right)\sqrt{6\pi}$$
$$\left(B\right)\ 3\sqrt{\pi}$$
$$\left(C\right)\ 2\sqrt{6\pi}$$
$$\left(D\right)\ 6\sqrt{\pi}$$
$$\left(E\right)\ 12\sqrt{\pi}$$
Area = π(radius)²
= π(6)²
= 36Ï€
Area of a rectangular sign whose length is twice its width.
Let x = width
So, 2x = length
Area of rectangle = (length)(width)
= (2x)(x)
= 2x²
Area of a circular sign is 3 times the area of a rectangular sign.
Area of a circular sign = 3(area of a rectangular sign)
36π = 3(2x²)
Simplify right side: 36π = 6x²
Divide both sides by 6 to get: 6π = x²
So, x = √(6π)
In other words, the width = √(6π) feet
Since the length is TWICE the width, the length = 2√(6π) feet
Answer: C
Cheers,
Brent
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The area of the circular sign is 36Ï€.BTGmoderatorLU wrote: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.
Thus, the area of the rectangle is 12Ï€. Since the length is twice the width, we have:
2w * w = 12Ï€
w^2 = 6Ï€
w = √(6π)
So the length is 2√(6π).
Answer: C
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