The perimeter of a certain isosceles right triangle is 16 + 16(sqrroot 2). What is the length of the hypotenuse of the triangle?
(A) 8
(B) 16
(C) 4(sqrroot 2)
(D) 8(sqrroot 2)
(E) 16(sqrroot 2)
OA: B
Hypotenuse of Triangle?
This topic has expert replies
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
- 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
- [email protected]
- Elite Legendary Member
- Posts: 10392
- Joined: Sun Jun 23, 2013 6:38 pm
- Location: Palo Alto, CA
- Thanked: 2867 times
- Followed by:511 members
- GMAT Score:800
Hi bml1105,
Mitch's approach to TEST THE ANSWERS is perfect for this scenario. There's a Number Property about Isosceles triangles that will help you to avoid most of the wrong answers...
The ratio of sides in an Isosceles Triangle is: x, x, and x√2
If, for example, one of the "short sides" was 5, then we'd have 5, 5, 5√2 and the perimeter would be 10 + 5√2. Notice that the 10 and 5 are different numbers....That will ALWAYS happen IF the short side is an integer and the hypotenuse has a √2 attached to it....
We're given a perimeter of 16 + 16√2; here, both numbers are 16, so we are NOT dealing with the above scenario. The "short sides" CANNOT be integers and the hypotenuse CANNOT have a √2 attached to it. Eliminate C, D and E.
From there, we can TEST A and B to find the match.
GMAT assassins aren't born, they're made,
Rich
Mitch's approach to TEST THE ANSWERS is perfect for this scenario. There's a Number Property about Isosceles triangles that will help you to avoid most of the wrong answers...
The ratio of sides in an Isosceles Triangle is: x, x, and x√2
If, for example, one of the "short sides" was 5, then we'd have 5, 5, 5√2 and the perimeter would be 10 + 5√2. Notice that the 10 and 5 are different numbers....That will ALWAYS happen IF the short side is an integer and the hypotenuse has a √2 attached to it....
We're given a perimeter of 16 + 16√2; here, both numbers are 16, so we are NOT dealing with the above scenario. The "short sides" CANNOT be integers and the hypotenuse CANNOT have a √2 attached to it. Eliminate C, D and E.
From there, we can TEST A and B to find the match.
GMAT assassins aren't born, they're made,
Rich
- bml1105
- Master | Next Rank: 500 Posts
- Posts: 113
- Joined: Sun Nov 24, 2013 9:19 pm
- Thanked: 1 times
- Followed by:5 members
Hi Brent,Brent@GMATPrepNow wrote:
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
I'm a little confused. How can we factor out x√2 from 2x + x√2 and get x√2(√2 + 1) when the 2x doesn't have an x√2. If anything wouldn't it be 2x/x√2 (√2/√2) = (2√2)/2?
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
Good question.bml1105 wrote:
Hi Brent,
I'm a little confused. How can we factor out x√2 from 2x + x√2 and get x√2(√2 + 1) when the 2x doesn't have an x√2. If anything wouldn't it be 2x/x√2 (√2/√2) = (2√2)/2?
Notice that 2 = (√2)(√2)
So, let's start at the point where we have: 2x + x√2 = 16 + 16√2
We can rewrite this as (√2)(√2)x + x√2 = 16 + 16√2
So, on the left side, both terms have an x and a √2
So we can factor x√2 from the left side to get : x√2(√2 + 1) = 16 + 16√2
I hope that helps.
Cheers,
Brent
I'm so disappointed of myself. I tried to answer the problem but couldn't get the answer from the choices. Hmmmmn. Maybe I should start reviewing on how to get the hypotenuse of a triangle. Should have been an easy question since it's been taught all over from high school to college. Thanks for the explanations you've provided here. I am enlightened as well. Thanks for posting the question too. It was mind-boggling and tickled the brain cells in me. Best of luck to all test takers. Especially on the Math section. This section proved to be the most challenging for me.
-Garmahis
-Garmahis
CheatSheetOne
All GMAT Math Formulas you need to succeed!
All GMAT Math Formulas you need to succeed!
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
Don't be too hard on yourself, Garmahis. This is a tricky question. In fact, I'd put it in the 700 range (difficulty-wise).
Cheers,
Brent
Cheers,
Brent