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In the figure above, ABCD
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11 Jun 2021, 06:26
11 Jun 2021, 06:26
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In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.)
A. 1:2
B. 3:2π
C. 2:5
D. 4:3π
E. 6:5π
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Joined: 10 Apr 2015
Posts: 6218
Given Kudos: 136
Re: In the figure above, ABCD
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11 Jun 2021, 06:30
11 Jun 2021, 06:30
1
Carcass wrote:
In the figure above, ABCD is a rectangle inscribed in a circle. If the length of AB is three times the length of AD, then what is the ratio of the area of the rectangle to the area of the circle? (Figure not drawn to scale.)
A. 1:2
B. 3:2π
C. 2:5
D. 4:3π
E. 6:5π
Let's say AD has length 1 and AB has length 3 (this also means that side CD also has length 3)
Since the blue triangle is a right triangle, we can use the Pythagorean Theorem to find the length of the 3rd side (the hypotenuse)
If we let x = the length of the hypotenuse, we can write: 1² + 3² = x²
Simplify: 1 + 9 = x²
So, x² = 10, which means x = √10
At this point, we can see that the diameter of the circle is √10, which means the radius is (√10)/2
What is the ratio of the area of the rectangle to the area of the circle?
Area of rectangle = (length)(width)
= (1)(3)
= 3
Area of circle = π(radius)²
= π(√10)/2)²
= π(10/4)
= π(5/2)
= 5π/2
So, the RATIO = 3/(5π/2)
= (3)(2/5π)
= 6/5π
= 6 : 5π
Answer: E
Cheers,
Brent
Re: In the figure above, ABCD
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13 Jan 2023, 11:33
13 Jan 2023, 11:33
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
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