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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.)

Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the [#permalink]

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22 Jul 2015, 03:02

Bunuel 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.)

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.)

Let, AB = 3 and AD = 1 i.e. BD = \(\sqrt{3^2 + 1^2}\) = \(\sqrt{10}\) = Diameter of Circle

Area of Rectangle = AB x BD = 3 x 1 = 3

Area of Circle = (π/4)*Diameter^2 = (π/4)*10 = 5π/2

Area of Rectangle / Area of Circle = 3 / (5π/2) = 6/5π

Answer: Option E
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Re: In the figure above, ABCD is a rectangle inscribed in a circle. If the [#permalink]

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22 Jul 2015, 11:40

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Bunuel 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.)

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.)

The highlighted part is a mistake which has led to several mistakes in the three steps in between.
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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.)

Let AD = x. AB = 3x. Area of the rectangle is (AD) × (AB) = x × (3x) = 3x².

Using the Pythagorean theorem, take AB and AD to get the diameter AC of the circle. x² + (3x)² = AC² x² + 9x² = AC² 10x² = AC² AC = x√10 and the radius is [x√10] / 2.

The area of a circle is πr², so the area is π [(x√10) / 2]² = (10πx²)/4 = (5πx²)/2.

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.)

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