The perimeter of a certain isosceles right triangle is 16+16root2. What is the length of the hypotenuse of the triangle?
A.8
B.16
C.4root2
D.8root2
E.16root2
GMATPrep Test 2 Q8
This topic has expert replies
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
The sides of an isosceles right triangle are in the following ratio: s : s : s√2.The perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse?
a. 8
b. 16
c. 4√2
d. 8√2
e. 16√2
So if s=side and h=hypotenuse, then h = s√2 and s = h/√2.
We can plug in the answers, which represent the length of the hypotenuse.
Answer choice C: h = 4√2
s = (4√2)/√2 = 4.
p = 4 + 4 + 4√2 = 8 + 4√2.
Eliminate C. The perimeter needs to be quite a bit larger.
Answer choice B: h = 16
s = 16/√2 = (16*√2)/(√2*√2) = (16√2)/2 = 8√2.
p = 8√2 + 8√2 + 16 = 16 + 16√2. Success!
The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
An IMPORTANT point to remember is that, in any isosceles right triangle, the sides have length x, x, and x√2 for some positive value of x.The Perimeter of a certain isosceles right triangle is 16 + 16√2. What is the length of the hypotenuse of the triangle?
A) 8
B) 16
C) 4√2
D) 8√2
E) 16√2
Note: x√2 is the length of the hypotenuse, so our goal is to find the value of x√2
From here, we can see that the perimeter will be x + x + x√2
In the question, the perimeter is 16 + 16√2, so we can create the following equation:
x + x + x√2 = 16 + 16√2,
Simplify: 2x + x√2 = 16 + 16√2
IMPORTANT: Factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
Now factor 16 from the right side to get: x√2(√2 + 1) = 16(1 + √2)
Divide both sides by (1 + √2) to get: x√2 = 16
Answer = B
Cheers,
Brent
-
- GMAT Instructor
- Posts: 2630
- Joined: Wed Sep 12, 2012 3:32 pm
- Location: East Bay all the way
- Thanked: 625 times
- Followed by:119 members
- GMAT Score:780
We'll start with x + x + x√2 = 16 + 16√2.
We know that ONE of these numbers (16 or 16√2) is the hypotenuse, and the other is the sum of the two legs. If 2x = 16, then x = 8, and x√2 = 8√2. That doesn't match our numbers, so it can't work.
Hence the OTHER number, 16√2, must be the sum of the two legs. This gives us legs of 8√2, and a hypotenuse of 8√2(√2), or 16. Success!
If we wanted to do this algebraically, we could. Let x = one of the legs, and x√2 = the hypotenuse.
2x + √2x = 16 + 16√2
x * (2 + √2) = 16 + 16√2
x = (16 + 16√2)/(2 + √2)
x = ((16 + 16√2) * (2 - √2)) / ((2+√2)(2-√2))
x = (16√2)/2
x = 8√2
So our legs are each 8√2, and our hypotenuse is 16.
We know that ONE of these numbers (16 or 16√2) is the hypotenuse, and the other is the sum of the two legs. If 2x = 16, then x = 8, and x√2 = 8√2. That doesn't match our numbers, so it can't work.
Hence the OTHER number, 16√2, must be the sum of the two legs. This gives us legs of 8√2, and a hypotenuse of 8√2(√2), or 16. Success!
If we wanted to do this algebraically, we could. Let x = one of the legs, and x√2 = the hypotenuse.
2x + √2x = 16 + 16√2
x * (2 + √2) = 16 + 16√2
x = (16 + 16√2)/(2 + √2)
x = ((16 + 16√2) * (2 - √2)) / ((2+√2)(2-√2))
x = (16√2)/2
x = 8√2
So our legs are each 8√2, and our hypotenuse is 16.