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

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Math Expert
Joined: 02 Sep 2009
Posts: 60515
The shaded region in the figure above represents a frame shaped as a  [#permalink]

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04 Nov 2019, 04:00
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Difficulty:

75% (hard)

Question Stats:

57% (03:26) correct 43% (03:12) wrong based on 44 sessions

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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

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

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04 Nov 2019, 04:50
1
Area of Picture+Area of frame=$$\frac{1}{2}*10*13$$=65

We know that
Area of Picture=Area of frame

Area of picture= 65/2

Assume 2 legs of the picture are x and y

$$\frac{x*y}{2}= \frac{65}{2}$$
xy=65

Also x+y=17

$$(x+y)^2=289$$
$$x^2+y^2+2xy=289$$
$$x^2+y^2+130=289$$
$$x^2+y^2=159$$

$$\sqrt{x^2+y^2}=\sqrt{159}$$

Bunuel wrote:

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

Are You Up For the Challenge: 700 Level Questions

Attachment:
image052.jpg
Intern
Joined: 23 Nov 2017
Posts: 8
The shaded region in the figure above represents a frame shaped as a  [#permalink]

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04 Nov 2019, 12:32
Total area of triangle = 1/2*13*10=65
Area of Picture=Area of frame= 65/2
let smaller triangle sides (picture legs) as a & b
sum of smaller triangle sides (sum of picture legs) = a+b=17
$$1/2a*b=65/2$$(area of the picture)
ab=65

$$(a+b)2=289$$
$$a2+b2+2ab=289$$
$$a2+b2+130=289$$
$$a2+b2=159$$
$$\sqrt{a2+b2}=\sqrt{159}$$ (hypotenuse)
Ans:B
Intern
Joined: 08 Nov 2019
Posts: 3
GMAT 1: 580 Q47 V22
Re: The shaded region in the figure above represents a frame shaped as a  [#permalink]

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02 Dec 2019, 12:24
I dont get it.

Question says picture has the same area as the frame itself, but the solution takes the picture area to be 65/2. What am I missing ? question seems to be poorly worded
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Re: The shaded region in the figure above represents a frame shaped as a  [#permalink]

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12 Jan 2020, 06:36
Bunuel wrote:

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

Attachment:
image052.jpg

Let’s let x and y be the lengths of the sides of the picture and c be the length of the hypotenuse of the picture. We have the following two equations:

Area of frame = area of picture

(10 * 13)/2 - xy/2 = xy/2

130 - xy = xy

130 = 2xy

65 = xy

and

x + y = 17

Squaring the above equation, we have:

x^2 + y^2 + 2xy = 17^2

Substituting, we have:

x^2 + y^2 + 2(65) = 289

x^2 + y^2 +130 = 289

x^2 + y^2 = 159

Since x^2 + y^2 = c^2:

c^2 = 159

c = √159

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Re: The shaded region in the figure above represents a frame shaped as a   [#permalink] 12 Jan 2020, 06:36
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