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Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a

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New post Updated on: 27 Mar 2017, 12:00
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Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a radius of \(7.5\sqrt{2}\). What is the area of ABCD?

A. 150
B. 225
C. 450
D. 750
E. 900

Originally posted by tinku21rahu on 11 Jun 2014, 15:01.
Last edited by Bunuel on 27 Mar 2017, 12:00, edited 2 times in total.
Renamed the topic and edited the question.
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Re: Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a  [#permalink]

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New post 12 Jun 2014, 00:18
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tinku21rahu wrote:
Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a radius of 7.5\(\sqrt{2}\). What is the area of ABCD?
A) 150
B) 225
C) 450
D) 750
E) 900


let length be 'l' and breadth be 'b'

2(l+b)=60; l+b=30
also; 2r = diagonal of the rectangle ABCD, thus diagonal = 15\sqrt{2}

also we know that if l and b are length and breadth then length of diagonal = \sqrt{l^2 +b^2} =15\sqrt{2}
squaring both sides we have, l^2+b^2=450

area of the rectangle = lb

also, we know that \((l+b)^2\) = \(l^2\) + \(b^2\) +2lb;

substituting l+b=30; l^2+b^2 = 450 in the above equation we have;

900=450+2lb;
450=2lb;
or lb=225
hence area =225
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Re: Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a  [#permalink]

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New post 13 Jun 2014, 19:48
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Refer diagram below:

Perimeter = 60

If one side = x, then

other side would be = 30 - x

Diameter of circle = Diagonal of rectangle

\(= 2* 7.5\sqrt{2}\)

\(= 15\sqrt{2}\)

Setting up the equation

\((15\sqrt{2})^2 = x^2 + (30-x)^2\)

\(x^2 - 30x + 225 = 0\)

\((x - 15)^2 = 0\)

x = 15

Area = 15 * 15 = 225 (This rectangle is a square)

Answer = B

Bunuel, Kindly update the OA

One more thing,

I understand that every square is a rectangle & every rectangle NEED NOT have to be a square. However, in this question, they mentioned rectangle ABCD which turns up to be a square. Is the wording correct for this question?

Thanks :)
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Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a  [#permalink]

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New post 12 Nov 2017, 08:36
Guys, first post here...
Bunuel PareshGmat

Just checking if my workaround would be correct.
Since the radius = 7.5\sqrt{2}, it's correct to say that the 4 inner triangles are isosceles and that their angles are 45:45:90.

Following the rule for 45:45:90 triangles, the largest side would be 1:\sqrt{2}, leaving us with 7.5\sqrt{2} * \sqrt{2} = 7.5 * 2 = 15 (for all triangles).

With this information we can calculate the area of ABCD (which turns out to be a square) as 15 * 15 = 225.
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Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a  [#permalink]

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New post 14 Nov 2017, 01:02
tinku21rahu wrote:
Attachment:
Geo.png
Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a radius of \(7.5\sqrt{2}\). What is the area of ABCD?

A. 150
B. 225
C. 450
D. 750
E. 900


Since rectangle is inscribed in the circle. So. Diagonal of rectangle = diameter of circle = 7.5 \(\sqrt{2}\) * 2 = 15 \(\sqrt{2}\)
Area of rectangle = 1/2 * diagonal^2 = 1/2 * \({(15\sqrt{2})}^2\) = 225

Answer B
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Re: Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a  [#permalink]

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Re: Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a   [#permalink] 14 Jun 2019, 00:18

Rectangle ABCD with a perimeter of 60 is inscribed in a circle with a

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