The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2
OA is b.
Could someone please explain this answer?
thanks
Hypoteneuse of the triangle
This topic has expert replies
-
- Master | Next Rank: 500 Posts
- Posts: 107
- Joined: Wed Mar 04, 2009 4:39 am
- Location: Vancouver
- Thanked: 9 times
- GMAT Score:750
For any isosceles right triangle, the ratio of its 3 sides is 1:1:sqrt2
(sqrt = square root)
Let x is one side, the perimeter will be:
x+x+(sqrt2)x = 16+sqrt16
--> (2+sqrt2)x = 16+sqrt16
--> x = (16+sqrt16)/(2+sqrt2)
--> x = [(16+sqrt16)(2-sqrt2)]/[(2+sqrt2)(2-sqrt2)]
note: numerator and denominator each multiplies by (2-sqrt2), to get rid of sqrt
--> x = 8sqrt2
Hypotenuse = sqrt2 x 8sqrt2 = 16. (remember the ratio of all sides is 1:1:sqrt2)
(sqrt = square root)
Let x is one side, the perimeter will be:
x+x+(sqrt2)x = 16+sqrt16
--> (2+sqrt2)x = 16+sqrt16
--> x = (16+sqrt16)/(2+sqrt2)
--> x = [(16+sqrt16)(2-sqrt2)]/[(2+sqrt2)(2-sqrt2)]
note: numerator and denominator each multiplies by (2-sqrt2), to get rid of sqrt
--> x = 8sqrt2
Hypotenuse = sqrt2 x 8sqrt2 = 16. (remember the ratio of all sides is 1:1:sqrt2)
- sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
The perimeter of an isosceles right triangle, with its hypotenues as x, is always in the form x + x sqrt 2. So if perimeter is 16 + 16 sqrt 2, hypotenues is 16.Baldini wrote:The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2
OA is b.
Could someone please explain this answer?
thanks
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Hi,
Here is another way that is very similar.
If x is the hypotenuse, then the 2 sides will be equal to x/SQRT 2
then, x/(SQRT 2)+x/(SQRT 2)+x=16(1+SQRT 2)
then, x(2/(SQRT 2) +1)=16(1+SQRT 2)
and accordingly, x((SQRT 2 +1)=16 (1+SQRT 2), and so x=16
Here is another way that is very similar.
If x is the hypotenuse, then the 2 sides will be equal to x/SQRT 2
then, x/(SQRT 2)+x/(SQRT 2)+x=16(1+SQRT 2)
then, x(2/(SQRT 2) +1)=16(1+SQRT 2)
and accordingly, x((SQRT 2 +1)=16 (1+SQRT 2), and so x=16
Baldini wrote:thanks a lot - just wondering if a quicker way (shortcut) exists to find this answer?
Ossa Elhadary, PhD, CISA, PMP
Math Specialist
Test Prep New York
maximize your score, minimize your stress
www.testprepny.com
[email protected]
Math Specialist
Test Prep New York
maximize your score, minimize your stress
www.testprepny.com
[email protected]
- arora007
- Community Manager
- Posts: 1048
- Joined: Mon Aug 17, 2009 3:26 am
- Location: India
- Thanked: 51 times
- Followed by:27 members
- GMAT Score:670
45-90-45
has sides x, xroot2, x
2x+xroot2 = 16+16root2
taking common terms out from both sides....
xroot2( root2 + 1) = 16( 1 + root2)
now hypotenuse is xroot2 = 16
has sides x, xroot2, x
2x+xroot2 = 16+16root2
taking common terms out from both sides....
xroot2( root2 + 1) = 16( 1 + root2)
now hypotenuse is xroot2 = 16
https://www.skiponemeal.org/
https://twitter.com/skiponemeal
Few things are impossible to diligence & skill.Great works are performed not by strength,but by perseverance
pm me if you find junk/spam/abusive language, Lets keep our community clean!!
https://twitter.com/skiponemeal
Few things are impossible to diligence & skill.Great works are performed not by strength,but by perseverance
pm me if you find junk/spam/abusive language, Lets keep our community clean!!
very very quick, thanks a lotsanju09 wrote:The perimeter of an isosceles right triangle, with its hypotenues as x, is always in the form x + x sqrt 2. So if perimeter is 16 + 16 sqrt 2, hypotenues is 16.Baldini wrote:The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2
OA is b.
Could someone please explain this answer?
thanks
MBA admissions in leading universities of USA, Canada, UK, Australia, New Zealand, Singapore, Dubai.
Contact details: (+91)522-2200123 (+91)9369336687
Email: [email protected]
Located at Lucknow, India
Contact details: (+91)522-2200123 (+91)9369336687
Email: [email protected]
Located at Lucknow, India
- 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 - sRoot2.Baldini wrote:The perimeter of a certain isosceles right triangle is 16 + 16 sq root 2. What is the length of the hypoteneuse?
a. 8
b. 16
c. 4 sq rt 2
d. 8 sq rt 2
e. 16 sq rt 2
OA is b.
Could someone please explain this answer?
thanks
So if s=side and h=hypotenuse, h = sRoot2 and s = h/Root2.
Let's plug in the answers -- which represent the hypotenuse -- and determine the length of each leg and the perimeter of the whole triangle. We need the perimeter to be 16 + 16Root2.
We can see quickly that C, D and E won't work:
C:
h = 4Root2
s = 4
p = 4+4+4Root2 = 8 + 4Root2. No.
D:
h = 8Root2
s = 8
p = 8+8+8Root2 = 16 + 8Root2. No.
E:
h = 16Root2
s = 16
p = 16+16+16Root2 = 32 + 16Root2. No.
Let's try B:
h = 16
s = 16/Root2 = [(16*Root2)/(Root2*Root2)]= (16Root2)/2 = 8Root2
p = 8Root2 + 8Root2 + 16 = 16 + 16Root2. 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