Hey Guys, here is the question
The perimeter of a certain isosceles triangle is, 16+ 16√2. What is the length of the hypotenuse of the triangle?
8; 16 ; 4√2; 8√2; 16√2
In order to resolve this question I wrote down the following;
y= hypothenus
x= one of the two equal sides.
y+2x=16+16√2 --> y=8+8√2-y/2
y2(exponent)=2x2(exponent)
Then I would try the function with the given answers for y, but the calculations were long and prompt to errors. Any other way to resolve this?
Many thanks
Lukas
hypotenus of an isoceles right triangle
This topic has expert replies
-
- Senior | Next Rank: 100 Posts
- Posts: 65
- Joined: Wed Mar 26, 2014 6:56 am
- Followed by:1 members
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
When you transcribed the question, you missed an important part. I've posted the original question below.
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
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
Last edited by Brent@GMATPrepNow on Thu Apr 10, 2014 3:43 pm, edited 1 time in total.
-
- Senior | Next Rank: 100 Posts
- Posts: 65
- Joined: Wed Mar 26, 2014 6:56 am
- Followed by:1 members
- 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 proportioned: 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 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
-
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Tue Apr 08, 2014 2:23 pm
-
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Tue Apr 08, 2014 2:23 pm
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
The question tells us that we have a right isosceles triangle.susahntgupta402 wrote:Mitch and Brent is the reason for assuming right isosceles triangle occurrence of 2^1/2?
Cheers,
Brent
-
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Tue Apr 08, 2014 2:23 pm
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
Oops, sorry.
Looks like the person who posted the question transcribed it incorrectly.
The original question states that it's an isosceles right triangle.
I've since noted this in my response above.
Cheers,
Brent
Looks like the person who posted the question transcribed it incorrectly.
The original question states that it's an isosceles right triangle.
I've since noted this in my response above.
Cheers,
Brent