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# Picture Frame

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Manager
Status: Keep fighting!
Joined: 31 Jul 2010
Posts: 238
WE 1: 2+ years - Programming
WE 2: 3+ years - Product developement,
WE 3: 2+ years - Program management
Followers: 4

Kudos [?]: 146 [2] , given: 104

Picture Frame [#permalink]  23 Sep 2010, 20:29
2
KUDOS
00:00

Difficulty:

35% (medium)

Question Stats:

58% (04:05) correct 43% (02:01) wrong based on 40 sessions
Attachment:

image052.jpg [ 4.88 KiB | Viewed 1215 times ]

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

-Give me kudos if you like this question.
[Reveal] Spoiler: OA
Manager
Joined: 30 May 2010
Posts: 191
Followers: 3

Kudos [?]: 41 [0], given: 32

Re: Picture Frame [#permalink]  23 Sep 2010, 20:56
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.
Senior Manager
Joined: 25 Jun 2009
Posts: 313
Followers: 2

Kudos [?]: 70 [1] , given: 6

Re: Picture Frame [#permalink]  23 Sep 2010, 21:08
1
KUDOS
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.
Manager
Joined: 30 May 2010
Posts: 191
Followers: 3

Kudos [?]: 41 [0], given: 32

Re: Picture Frame [#permalink]  23 Sep 2010, 21:09
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, 21:13, edited 1 time in total.
Manager
Status: Keep fighting!
Joined: 31 Jul 2010
Posts: 238
WE 1: 2+ years - Programming
WE 2: 3+ years - Product developement,
WE 3: 2+ years - Program management
Followers: 4

Kudos [?]: 146 [0], given: 104

Re: Picture Frame [#permalink]  23 Sep 2010, 21:50
good attempts guys!
Re: Picture Frame   [#permalink] 23 Sep 2010, 21:50
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