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# The area of a rectangle inscribed in a circle is 12. Area of

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The area of a rectangle inscribed in a circle is 12. Area of [#permalink]  19 Dec 2010, 06:29
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The area of a rectangle inscribed in a circle is 12. Area of the circle is 25/4 pi. What is the perimeter of the rectangle?
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Re: Perimeter of the inscribed rectangle [#permalink]  19 Dec 2010, 07:00
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smitakokne wrote:
The area of a rectangle inscribed in a circle is 24. Area of the circle is 25/4 pi. What is the perimeter of the rectangle?

I do not have an answer to this question. How do we solve this?

As explained by Karishma below it should be: $$area_{rectangle}=ab=12$$ instead of 24.

Given: $$area_{circle}=\pi{r^2}=\frac{25}{4}*\pi$$ --> $$r=2.5$$ and $$area_{rectangle}=ab=12$$, where $$a$$ and $$b$$ are the lengths of the sides of the rectangle.
Question: $$perimeter_{rectangle}=2(a+b)=?$$

Now, a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.

A diagonal of a rectangle is the hypotenuse for its sides, so as rectangle is inscribed in a circle then its diagonal is the diameter of the circle. Which means that $$diagonal=diameter=2*radius=5$$. Also $$d^2=25=a^2+b^2$$.

Now, square $$perimeter_{rectangle}=2(a+b)$$ --> $$P^2=4(a^2+2ab+b^2)$$, as given that $$ab=12$$ and $$25=a^2+b^2$$ then $$P^2=4(a^2+2ab+b^2)=4(25+24)=4*49$$ --> $$P=\sqrt{4*49}=2*7=14$$.

Answer: $$14$$.

Hope its clear.
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Re: Perimeter of the inscribed rectangle [#permalink]  19 Dec 2010, 07:19
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smitakokne wrote:
The area of a rectangle inscribed in a circle is 24. Area of the circle is 25/4 pi. What is the perimeter of the rectangle?

I do not have an answer to this question. How do we solve this?

25/4 pi is around 20. The area of inscribed rectangle cannot be more than this so it cannot be 24. From the way this question is formed, I would expect the area of the rectangle to be 12. (I will explain this later). I will solve the question assuming this value of area.

Area of circle = 25/4 pi = pi*r^2
Radius of circle = 5/2 and diameter is 5.
The rectangle inscribed in the circle will have its center at the center of the circle since its opposite sides are equal and parallel. The diagram below will show you why it should be so.
Attachment:

Ques1.jpg [ 7.04 KiB | Viewed 3034 times ]

So diameter of the circle should be the diagonal of the rectangle i.e. 5.
So area of rectangle = ab = 12
and $$a^2 + b^2 = 5^2$$ (pythagorean theorem)
The most common pythagorean triplet is 3,4,5 and since the diagonal was 5, I was looking for the sides to be 3 and 4 possibly giving the area as 12. In GMAT, numbers fall in place very well)
Sides must be 3 and 4 giving the perimeter as 2(3+4) = 14
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Last edited by VeritasPrepKarishma on 19 Dec 2010, 07:24, edited 1 time in total.
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Re: Perimeter of the inscribed rectangle [#permalink]  19 Dec 2010, 07:21
Very clear, thank you. This does not seem like 600 level question though.
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Re: Perimeter of the inscribed rectangle [#permalink]  19 Dec 2010, 07:51
Guess i misread the area of rectangle. This makes sense.
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Re: Perimeter of the inscribed rectangle [#permalink]  13 Nov 2013, 15:39
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Re: The area of a rectangle inscribed in a circle is 12. Area of [#permalink]  11 Sep 2015, 15:58
Hello from the GMAT Club BumpBot!

Thanks to another GMAT 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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Re: The area of a rectangle inscribed in a circle is 12. Area of   [#permalink] 11 Sep 2015, 15:58
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