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In a certain right triangle, the sum of the lengths of the

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Senior Manager
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In a certain right triangle, the sum of the lengths of the  [#permalink] New post 01 Nov 2005, 07:52
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In a certain right triangle, the sum of the lengths of the
two legs and the hypotenuse is 60 inches. If the hypotenuse
is 26 inches, which of the following is the length of one of
the legs?

A. 24 inches
B. 34 inches
C. 29 inches
D. 16 inches
E. 13 inches
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Re: PS: Length of Triangle [#permalink] New post 01 Nov 2005, 08:09
sudhagar wrote:
In a certain right triangle, the sum of the lengths of the
two legs and the hypotenuse is 60 inches. If the hypotenuse
is 26 inches, which of the following is the length of one of
the legs?

A. 24 inches
B. 34 inches
C. 29 inches
D. 16 inches
E. 13 inches


Let a, b be the two legs and c be the hypotenuse
we have: a+b+c = 60 ----> a+b= 60-26=34 ---> b= 34-a
c^2= a^2+b^2= 26^2= 676
Substitute b into the 2nd equation, we get:
a^2+ (34-a)^2= 676 ....solve this quadratic equation, we get a = 24
A it is.
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 [#permalink] New post 01 Nov 2005, 08:11
Answer : A

hyp = 26. So sum of lengths = 60-26 = 34.

Taking answer choices, 24^2 + 10^2 = 26^2.
Hence A.
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 [#permalink] New post 01 Nov 2005, 11:35
yep, is one of the typical right triangles

13-12-5

so

26-24-10
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 [#permalink] New post 01 Nov 2005, 11:44
A it is...

tried a few choices...wasnt getting the right answer, then, picked one of the famous triangles, 3-4-5 that didnt work, 5-12-13...works fine...
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 [#permalink] New post 01 Nov 2005, 13:12
Thats correct guys, OA is A

Here is the detailed explanation:
The Pythagorean theorem states that the sum of the squares of the lengths of the legs of a triangle is equal to the square of the length of the hypotenuse.

Let A and B be the lengths of the sides. Let C be the length of the hypotenuse. Thus, we can set up the following equation:

A2 + B2 = C2

We are told that the sum of the legs and hypotenuse is 60 inches. We are also told that the hypotenuse is 26 inches. Thus, we can set up the following equation:

A + B + 26 = 60
A + B = 34
A = 34 - B

Plugging the value of A and C into the first equation:

(34 - B)2 + B2 = 262
(34 - B)(34 - B) + B2 = 676
1156 - 68B + B2 + B2 = 676
2B2 - 68B + 1156 = 676
2B2 - 68B + 480 = 0
(2B - 48)(B - 10) = 0
B = 10 or B = 24
  [#permalink] 01 Nov 2005, 13:12
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