Triangle T is a right triangle

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by Brent@GMATPrepNow » Sun Jan 11, 2009 9:15 am

Triangle T is a right triangle with a 12-inch hypotenuse and an area of 28 square inches. What is the perimeter, in inches, of triangle T?
(A) 20
(B) 28
(C) 12 + sqrt(210)
(D) 32
(E) 45

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Source: Beat The GMAT — Problem Solving | Sun Jan 11, 2009 9:15 am

DanaJ: Site Admin; Posts: 2567; Joined: Thu Jan 01, 2009 10:05 amFollowed by: 550 members; Thanked: 712 times; GMAT Score:770

by DanaJ » Sun Jan 11, 2009 10:41 am

Here it goes:
you have the area of 28. That means that (c1*c2)/2=28, so 2c1*c2=4*28=112.
Hyp is 12, so (c1)^2+(c2)^2 = 12^2 = 144.
Since (c1+c2)^2= (c1)^2+(c2)^2 +2c1*c2, (c1+c2)^2 = 256, which is 16^2. So c1+c2 = 16.
Perimeter = c1+c2+hyp= 16+12=28

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Jan 11, 2009 11:25 am

Awesome work!
Here's my solution as well:

Let x and y be the two legs of the triangle.
A 12-inch hypotenuse means that x^2 + y^2 = 12^2 (144)
An area of 28 means that 1/2xy=28 or xy=56 or (most importantly) 2xy=112
Note: We aren’t required to find the value x and the value of y. We need only the sum x+y to determine the entire perimeter.
Important part: Notice that when we expand (x+y)^2 = x^2 + 2xy + y^2, we get some recognizable equalities from above. Let’s perform some substitutions: (x+y)^2 = x^2 + 2xy + y^2
=(x^2 + y^2) + 2xy
= 144 + 112 = 256
If (x+y)^2 = 256, then x+y = 16 and so the perimeter is 16+12=28 (Answer = B)

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drewd313: Newbie | Next Rank: 10 Posts; Posts: 8; Joined: Fri Nov 14, 2008 8:31 pm

by drewd313 » Sun Jan 11, 2009 1:07 pm

What about the 45-45-90 triangle rule.

Divide 12 by sq. rt. 2 = roughly 8

2 (8) + 12 = roughly 28

(I actually got 8.5 for the legs and 29 for perimeter, but figured this was due to approximations and picked 28 since it was the closest to 29)

Is that an acceptable way to solve the problem or was it just a coincidence that it worked?

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canada_sms: Senior | Next Rank: 100 Posts; Posts: 84; Joined: Tue Nov 04, 2008 3:53 pmFollowed by: 9 members; Location: Boston, MA; Thanked: 50 times; GMAT Score:760

by canada_sms » Sun Jan 11, 2009 1:58 pm

Great question Brent. Would you mind telling us the source?

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Brent@GMATPrepNow: GMAT Instructor; Posts: 16207; Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members; Location: Vancouver, BC; Thanked: 5254 times; GMAT Score:770

by Brent@GMATPrepNow » Sun Jan 11, 2009 2:54 pm

canada_sms wrote:Great question Brent. Would you mind telling us the source?

Hi fellow Canadian,

I have been creating my own questions for my posts.

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arzanr: Senior | Next Rank: 100 Posts; Posts: 60; Joined: Wed Jan 07, 2009 1:56 pm; Thanked: 12 times; GMAT Score:750

by arzanr » Mon Jan 12, 2009 2:15 pm

Since this is a right triangle, the sides would be in the ratio of 3:4:5. The hypotenuse being the largest side solve for 7:5 with 5 as 12. This will give you 84/5. (84/5)+20 = 28 4/5 and the closest answer would be B.

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logitech: Legendary Member; Posts: 2134; Joined: Mon Oct 20, 2008 11:26 pmFollowed by: 25 members; Thanked: 237 times; GMAT Score:730

by logitech » Mon Jan 12, 2009 2:31 pm

arzanr wrote:Since this is a right triangle, the sides would be in the ratio of 3:4:5. T

WOW!! A right angle triangle can have INFINITE number of variations and 3,4,5 is JUST one of them.

Watch out!

5,12 and 13 for example.

LGTCH
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"DON'T LET ANYONE STEAL YOUR DREAM!"

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arzanr: Senior | Next Rank: 100 Posts; Posts: 60; Joined: Wed Jan 07, 2009 1:56 pm; Thanked: 12 times; GMAT Score:750

by arzanr » Mon Jan 12, 2009 2:47 pm

You are correct, not all right triangles would be 3:4:5, but a triangle with the ratio 3:4:5 would be a right triangle. Thanks for making that important clarification!

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