The figure above shows a circular flower bed, with its cente : GMAT Problem Solving (PS)
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# The figure above shows a circular flower bed, with its cente

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20 Dec 2012, 05:43
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The figure above shows a circular flower bed, with its center at O, surrounded by a circular path that is 3 feet wide. What is the area of the path, in square feet?

(A) $$25\pi$$
(B) $$38\pi$$
(C) $$55\pi$$
(D) $$57\pi$$
(E) $$64\pi$$
[Reveal] Spoiler: OA
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Re: The figure above shows a circular flower bed, with its cente [#permalink]

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20 Dec 2012, 05:48
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The figure above shows a circular flower bed, with its center at O, surrounded by a circular path that is 3 feet wide. What is the area of the path, in square feet?

(A) $$25\pi$$
(B) $$38\pi$$
(C) $$55\pi$$
(D) $$57\pi$$
(E) $$64\pi$$

The radius of the bigger circle is 8 + 3 = 11 feet, thus its area is $$\pi{r^2}=121\pi$$.

The area of the smaller circle is $$\pi{8^2}=64\pi$$.

The difference is $$121\pi-64\pi=57\pi$$.

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Re: The figure above shows a circular flower bed, with its cente [#permalink]

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19 Jan 2015, 16:32
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The figure above shows a circular flower bed, with its cente [#permalink]

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26 Apr 2016, 14:15
$$Radius of total garden$$ = 8 + 3 = 11 $$feet$$
$$Area of flower bed including path$$ = $$\pi$$$$11^2$$
$$Area of flower bed$$ = $$\pi$$ $$8^2$$
$$Area of only path$$ = 57$$\pi$$ sq.feet
$$correct answer$$ - D
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Re: The figure above shows a circular flower bed, with its cente [#permalink]

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27 Apr 2016, 07:44
Attachment:
Circle.png
The figure above shows a circular flower bed, with its center at O, surrounded by a circular path that is 3 feet wide. What is the area of the path, in square feet?

(A) $$25\pi$$
(B) $$38\pi$$
(C) $$55\pi$$
(D) $$57\pi$$
(E) $$64\pi$$

In solving this problem we first must recognize that the flower bed is the right triangle with sides of y yards, x yards, and z yards. We are given that the area of the bed (which is the right triangle) is 24 square yards. Since we know that area of a triangle is ½ Base x Height, we can say:

24 = ½(xy)

48 = xy

We also know that x = y + 2, so substituting in y + 2 for x in the area equation we have:

48 = (y+2)y

48 = y^2 + 2y

y^2 + 2y – 48 = 0

(y + 8)(y – 6) = 0

y = -8 or y = 6

Since we cannot have a negative length, y = 6.

We can use the value for y to calculate the value of x.

x = y + 2

x = 6 + 2

x = 8

We can see that 6 and 8 represent two legs of the right triangle, and now we need to determine the length of z, which is the hypotenuse. Knowing that the length of one leg is 6 and the other leg is 8, we know that we have a 6-8-10 right triangle. Thus, the length of z is 10 yards.

If you didn't recognize that 6, 8, and 10 are the sides and hypotenuse of a right triangle, you would have to use the Pythagorean to find the length of the hypotenuse: 6^2 + 8^2 = c^2 → 36 + 64 = c^2 → 100 = c^2. The positive square root of 100 is 10, and thus the value of z is 10.
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Re: The figure above shows a circular flower bed, with its cente [#permalink]

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05 Jul 2016, 05:16
Attachment:
Circle.png
The figure above shows a circular flower bed, with its center at O, surrounded by a circular path that is 3 feet wide. What is the area of the path, in square feet?

(A) $$25\pi$$
(B) $$38\pi$$
(C) $$55\pi$$
(D) $$57\pi$$
(E) $$64\pi$$

8+3=11
11^2=121

8^2= 64

121pi -64 pi= 57 pi
Re: The figure above shows a circular flower bed, with its cente   [#permalink] 05 Jul 2016, 05:16
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