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If a rectangle with length sqrt(a) units and width sqrt(b)

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If a rectangle with length sqrt(a) units and width sqrt(b)  [#permalink]

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New post Updated on: 16 Dec 2013, 01:41
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If a rectangle with length \(\sqrt{a}\) units and width \(\sqrt{b}\) units is inscribed in a circle of radius 5 units, what is the value of a + b?

A. 10
B. 20
C. 25
D. 50
E. 100

Still taking prac exam right now, but this problem was driving me nuts, don't know OA yet, will post when I get done with exam.

The way I see it you can't solve this problem, there's a million different rectangles...I tried researching this on the internet, and everything I've read says there's an infinite number of solutions.

Originally posted by AccipiterQ on 15 Dec 2013, 17:25.
Last edited by Bunuel on 16 Dec 2013, 01:41, edited 1 time in total.
Edited the question and added the OA.
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Re: If a rectangle with length sqrt(a) units and width sqrt(b)  [#permalink]

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New post 16 Dec 2013, 02:00
6
1
AccipiterQ wrote:
If a rectangle with length \(\sqrt{a}\) units and width \(\sqrt{b}\) units is inscribed in a circle of radius 5 units, what is the value of a + b?

A. 10
B. 20
C. 25
D. 50
E. 100

Still taking prac exam right now, but this problem was driving me nuts, don't know OA yet, will post when I get done with exam.

The way I see it you can't solve this problem, there's a million different rectangles...I tried researching this on the internet, and everything I've read says there's an infinite number of solutions.


A right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Since a rectangle is made of two right triangles then a rectangle inscribed in a circle must have its diagonal as the diameter of the circle.

Thus \((\sqrt{a})^2+(\sqrt{b})^2=10^2\) --> \(a+b=100\).

Answer: E.

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Hope this helps.
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Re: If a rectangle with length sqrt(a) units and width sqrt(b)  [#permalink]

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New post 15 Dec 2013, 23:06
1
AccipiterQ wrote:
If a rectangle with length \(\sqrt{a}\)units and width \(\sqrt{b}\) units is inscribed in a circle of radius 5 units, what is the value of a + b?

A)10
B)20
C)25
D)50
E)100

Still taking prac exam right now, but this problem was driving me nuts, don't know OA yet, will post when I get done with exam.

The way I see it you can't solve this problem, there's a million different rectangles...I tried researching this on the internet, and everything I've read says there's an infinite number of solutions.


For a rectangle to be drawn in a circle, the diagonal of the rectangle= diameter of Circle=10
and in a rectangle of side c and d respectively we have c^2+d^2= l^2 where l is the diagonal...

In this case we side as \sqrt{a} and\sqrt{b} so a+b= 100....

Do share the OA
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Re: If a rectangle with length sqrt(a) units and width sqrt(b)  [#permalink]

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New post 16 Dec 2013, 14:36
Here is the thing though, why can't you have rectangles of different side dimensions inscribed in the circle as in my attached image?

Just for kicks, I drew the attached image out with a pen, paper, a circular template and a ruler. I understand that the radius is 5 (so diameter is 10) but if we drew a rectangle with almost no height it would be just slightly longer while being a lot more narrow. The ratio of length to width wouldn't change enough to make them the same area.

EDIT: Thinking about it, it makes a bit more sense. They're testing to see whether you completely understand the concept of a right triangle. In a right triangle (which this is, it's two of them) a^2 + b^2 MUST EQUAL c^2. Drawing it out throws you off. If you inscribe a rectangle in a circle and you change the dimensions of that rectangle while still keeping it inscribed, the change in width will offset the change in height and vice versa. I guess my hand drawn attempt at understanding this was incorrect. In other words:

a^2 + b^2 = c^2
6^2 + 8^2 = 100
36 + 64 = 100

(a-b)^2 + (b+a)^2 = 100 as well.

Simple, yet very tricky.
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Re: If a rectangle with length sqrt(a) units and width sqrt(b)  [#permalink]

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New post 31 Aug 2018, 09:54
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Re: If a rectangle with length sqrt(a) units and width sqrt(b) &nbs [#permalink] 31 Aug 2018, 09:54
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