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The shaded region in the figure above represents a frame shaped as a r

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The shaded region in the figure above represents a frame shaped as a r [#permalink]

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New post 23 Sep 2010, 21:29
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The shaded region in the figure above represents a frame shaped as a right triangle with sides 10 inches and 13 inches. The frame encloses a picture, shaped also as a right triangle, which has the same area as the frame itself. If the sum of the 2 legs of the picture is 17 inches, what is the length of the hypotenuse of the triangular picture?

(A) \(\sqrt{29}\)
(B) \(\sqrt{159}\)
(C) 13
(D) \(\sqrt{269}\)
(E) 17
[Reveal] Spoiler: OA

Last edited by Bunuel on 30 Aug 2014, 06:33, edited 1 time in total.
Renamed the topic and edited the question.

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Re: The shaded region in the figure above represents a frame shaped as a r [#permalink]

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New post 23 Sep 2010, 21:56
Area of frame + picture = 65

Since both have the same area, 65/2 = 32.5 - each are is 32.5.

The picture is a similar triangle to larger triangle. Therefore, the angles are the same and the sides are proportional.

Height must be less than 10. Base must be less than 13.

Area of picture = 1/2 b h = 32.5. bh = 65.

What two numbers add to 17 and multiply to 65, and are in proportion to h = 10, b = 13, and hypotenuse \(sqrt(269)\)?

I'm stuck on how to proceed. I can at least eliminate C, D, E.

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New post 23 Sep 2010, 22:08
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jpr200012 wrote:
Area of frame + picture = 65

Since both have the same area, 65/2 = 32.5 - each are is 32.5.

The picture is a similar triangle to larger triangle. Therefore, the angles are the same and the sides are proportional.

Height must be less than 10. Base must be less than 13.

Area of picture = 1/2 b h = 32.5. bh = 65.

What two numbers add to 17 and multiply to 65, and are in proportion to h = 10, b = 13, and hypotenuse \(sqrt(269)\)?

I'm stuck on how to proceed. I can at least eliminate C, D, E.


You almost nailed it ...! Let me complete the rest for you.

Lets assume the sides of the picture as \(X\) and \(Y\) and Hypotenuse as \(Z\)

Then we have \(X + Y = 17 -- > 1\)
and \(X*Y = 65 -- >2\)

We need \(\sqrt{X^2 + Y ^2} = Z\)

Square both the sides of equation \(1\)

\((X + Y)^2 = 17^2 = 289\)

\(X^2 + Y^2 + 2*X*Y = 289\)

\(X^2 + Y^2 + 2 *65 = 289\)

\(X^2 + Y^2 = 159 -- > 3\)

Take the square root of equation \(3\)

\(\sqrt{X^2 + Y ^2} = \sqrt{159}= Z\)

Hence B.

Hope it helps.

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New post 23 Sep 2010, 22:09
I went ahead and continued by approximating.

I guess another way of stating it is, what are the solutions to:

\(x^2 + 17x + 65 = 0\)

6, 10 are close to the solutions. Therefore 10/13 for the proportion of the sides.

Since \(sqrt(269)\) is near 16, \(10/13 * 16 = 160/13\) = near 12.

Since (A) is near 5, and (B) is near 12 - answer B.

Last edited by jpr200012 on 23 Sep 2010, 22:13, edited 1 time in total.

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Re: The shaded region in the figure above represents a frame shaped as a r [#permalink]

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New post 23 Sep 2010, 22:50
good attempts guys!

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Re: The shaded region in the figure above represents a frame shaped as a r [#permalink]

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New post 30 Aug 2014, 02:13
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New post 28 Aug 2016, 13:36
Hello from the GMAT Club BumpBot!

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Re: The shaded region in the figure above represents a frame shaped as a r [#permalink]

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New post 03 Sep 2017, 06:11
Ans is Clearly B
can be seen from image

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Re: The shaded region in the figure above represents a frame shaped as a r [#permalink]

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New post 04 Sep 2017, 03:25
Let height and base of right angled picture that is inside big right triangle = a and b
Frame = shaded area i.e. big triangle -picture area.

Now given sum of height and base of picture =17 => a+b=17
And we have to find hypotenuse of picture = \sqrt{a^2+b^2}


Now given area of picture= area of frame
=> Area of big triangle = 2 * area of picture
=> 1/2 * 10 * 13 = 2 * (1/2 *a *b)
=>ab =65

Now we have to find a^2 +b^2..

Given a+b =17 , squaring both sides
a^2+b^2 +2ab = 289
a^2+b^2+2*65 =289 ( ab =65 by above equation)
a^2+b^2 = 159

\sqrt{a^2+b^2} = \sqrt{159} = Hypotenuse of picture.


Answer: B

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Re: The shaded region in the figure above represents a frame shaped as a r   [#permalink] 04 Sep 2017, 03:25
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