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Leg of a triangle

Expert replies
by singhsa » Wed Sep 22, 2010 4:49 pm
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

OA - A
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Source: — Problem Solving |

by beatthegmatinsept » Wed Sep 22, 2010 5:05 pm
singhsa 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

OA - A
You already know the hypotenuse is 26, and that this is a right angle triangle.
Does 26 and right triangle ring a bell? Does 26 make you think of 13? Does 13 make you think of the right traingle that has 5, 12 and 13 as its 3 sides, with 13 being the hypotenuse?
That's the approach I took.. 5, 12, 13 will also give you 10, 24, 26 as a possible right triangle.
Pick A.
Last edited by beatthegmatinsept on Wed Sep 22, 2010 7:21 pm, edited 1 time in total.
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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by beatthegmatinsept » Wed Sep 22, 2010 5:06 pm
Oh and a quick check would be to add up the 3 sides and see if you get 60 as the perimeter.

Hope this helps :)
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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by garudcvg » Wed Sep 22, 2010 7:11 pm
Use Pythagoras first. This will give you the following equation

676 = b^2 + C^2 ---- Equation 1

b + C + a = 60 which gives you b + c = 34 ........ Equation 2

(b + c)^2 = b^2 +2bc + c^2
which leads us to

b^2 + c^2 = (b + c)^2 - 2bc .........Equation 3

Plug in b^2 + c^2 = 676 from Equation 1 and Plug in b+c = 34 from equation 2 in Equation 3. Doing this will give you the following:

676 = 34 X 34 -2bc
which means bc = 34 ......Equation 4

Based on Equation 4 and Equation 2 we can say that b and c are 24 and 10.

24 is your answer choice A and hence the correct answer is A
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by diebeatsthegmat » Wed Sep 22, 2010 7:13 pm
beatthegmatinsept wrote:
singhsa 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

OA - A
You already know the hypotenuse is 26, and that this is a right angle triangle.
Does 26 and right triangle ring a bell? Does 26 make you think of 13? Does 13 make you think of the right traingle that has 5, 12 and 13 as its 3 sides, with 13 being the hypotenuse?
That's the approach I took.. 5, 12, 13 will also give you 10, 24, 26 as a possible right triangle.
Pick B.
B cant be answer. the answer is A
a+b+26=60 so a+b=34 and a^2+b^2=26^2
solve this we will have the answer is 10 and 24
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by garudcvg » Wed Sep 22, 2010 7:14 pm
Correction


bc = 240

b + c = 34

which means b and c are 24 and 10 respectively.

I sincerely apologies for the typo.
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by beatthegmatinsept » Wed Sep 22, 2010 7:20 pm
diebeatsthegmat wrote:
beatthegmatinsept wrote:
singhsa 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

OA - A
You already know the hypotenuse is 26, and that this is a right angle triangle.
Does 26 and right triangle ring a bell? Does 26 make you think of 13? Does 13 make you think of the right traingle that has 5, 12 and 13 as its 3 sides, with 13 being the hypotenuse?
That's the approach I took.. 5, 12, 13 will also give you 10, 24, 26 as a possible right triangle.
Pick B.
B cant be answer. the answer is A
a+b+26=60 so a+b=34 and a^2+b^2=26^2
solve this we will have the answer is 10 and 24
My explanation says 24, which is what A is, I guess I just wrote B by mistake in the final line. Explanation stays the same. Good catch though. I'll edit B for A.
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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by garudcvg » Wed Sep 22, 2010 7:39 pm
I like that approach better. Mine is brutal number crunching. I guess i will memorize some standard Pythagorean triplets.
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by beatthegmatinsept » Wed Sep 22, 2010 8:05 pm
garudcvg wrote:I like that approach better. Mine is brutal number crunching. I guess i will memorize some standard Pythagorean triplets.
Yeah, it took me less than a minute to solve this one. On the actual test, saving 1 minute on a question using tricks like these will surely help :)
Being defeated is often only a temporary condition. Giving up is what makes it permanent.
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