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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) \(64\pi\)
Attachment:
Circle.png [ 12.22 KiB | Viewed 45169 times ]
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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 \(121\pi-64\pi=57\pi\).
Answer: D.
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General Discussion
Posts: 386
\(Radius of total garden\) = 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
\(Area of flower bed including path\) = \(\pi\)\(11^2\)
\(Area of flower bed\) = \(\pi\) \(8^2\)
\(Area of only path\) = 57\(\pi\) sq.feet
\(correct answer\) - D
JeffTargetTestPrep
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Target Test Prep Representative
Joined: 04 Mar 2011
Last visit: 05 Jan 2024
Posts: 2,973
Status:Head GMAT Instructor
Affiliations: Target Test Prep
Expert reply
Posts: 2,973
Updated on: 19 Apr 2020, 11:21
Originally posted by JeffTargetTestPrep on 27 Apr 2016, 07:44.
Last edited by JeffTargetTestPrep on 19 Apr 2020, 11:21, edited 1 time in total.
Last edited by JeffTargetTestPrep on 19 Apr 2020, 11:21, edited 1 time in total.
Walkabout
The circular path (i.e., the shaded region) is called an annulus in geometry. Basically an annulus is a ring-shaped figure bounded by two concentric circles. To find the area of an annulus, we use the following formula:
π(R^2 – r^2)
where r is the radius of the inner circle and R is the radius of the outer circle.
We know:
R = 8 + 3 = 11 feet
r = 8 feet
Plugging into our equation, we have:
area = π(11^2 – 8^2)
area = π(121 - 64)
area = 57π
Answer: D
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Walkabout
8+3=11
11^2=121
8^2= 64
121pi -64 pi= 57 pi
Posts: 7
Area of the path = Area of larger circle (with radius 8+3=11) - Area of smaller circle (with radius 8)
=>Area of the path = π∗(11)^2 - π∗(8)^2
=>Area of the path = 121π - 64π = 57π
Answer: D
=>Area of the path = π∗(11)^2 - π∗(8)^2
=>Area of the path = 121π - 64π = 57π
Answer: D
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