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

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New post 04 Nov 2019, 04:00
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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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Spoiler: ::
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image052.jpg [ 4.88 KiB | Viewed 683 times ]
Spoiler: OA

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

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New post 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:
Image
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


Spoiler: ::
Attachment:
image052.jpg
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The shaded region in the figure above represents a frame shaped as a  [#permalink]

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

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New post 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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New post 12 Jan 2020, 06:36
Bunuel wrote:
Image
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


Spoiler: ::
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

Answer: B
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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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