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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
Are You Up For the Challenge: 700 Level Questions
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04 Nov 2019, 04:50
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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
\((x+y)^2=289\)
\(x^2+y^2+2xy=289\)
\(x^2+y^2+130=289\)
\(x^2+y^2=159\)
\(\sqrt{x^2+y^2}=\sqrt{159}\)
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
General Discussion
04 Nov 2019, 12:32
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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
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
