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GMAT Prep - Geometry

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by tutonaranjo » Mon Sep 03, 2007 6:07 pm
Image
In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S=
a) s . sqrt 2 / 2
b) s . sqrt 3 / 2
c) s . sqrt 2
d) s . sqrt 3
e) 2s

The . is multiplied
Answer is C. How did they get a root? Appreciatte an explanation!
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Re: GMAT Prep - Geometry

by jaspreet.sharma » Mon Sep 03, 2007 10:01 pm
tutonaranjo wrote:Image
In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S=
a) s . sqrt 2 / 2
b) s . sqrt 3 / 2
c) s . sqrt 2
d) s . sqrt 3
e) 2s

The . is multiplied
Answer is C. How did they get a root? Appreciatte an explanation!

for small triangle
Area = 1/2 (s*a) a is altitude

for bigger triangle
Area = 1/2 (S*A) A is altitude

ATQ

1/2 (S*A) = 2 * [1/2 (s*a)]
= s*a --------------------- 1st eqn

For similar triangles

a/A = s/S

This implies a = (A*s)/S

Putting values in 1st eqn

SA= 2 sa
= (2sAs)/S

S= s . sqrt 2

C is the answer :D
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by abkhan » Mon Sep 03, 2007 10:11 pm
let area of first triangle be A and B.
A=2B --5

let height of A be P and B be H

for first triangle
A= P.S/2 --1
tan x = P/S --2

for second triangle
B=H.s/2 --3
tan x= H/s --4

equating 2 = 4
-> P/S=H/s
-> P/H=S/s

equating 5
-> A=2B
-> replacing 1 and 3
-> P.S/2=2.H.s/2
-> P/H=2.s/S
-> S/s=2.s/S
-> S^2=2. s^2

taking root of both sides
-> S=s. 2^(1/2)
please correct me if i'm wrong
I am ..therefore I am..
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thanks

by tutonaranjo » Wed Sep 05, 2007 11:20 am
sweet
was missing

s/S = a/A

had no idea on how to set up those proportions
cheers!
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by [email protected] » Sat Aug 09, 2008 6:47 am
I had this same question come up in my practice exam and I dont really understand the explanation below, please can someone elaborate?

Thanks
again
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by smclean23 » Sun Sep 28, 2008 5:59 pm
Could someone give another spin on this? I'm confused by the explanations above.....
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by Stuart@KaplanGMAT » Sun Sep 28, 2008 8:09 pm
This question has been posted many times (if you search on "similar triangles" you'll get a big list).

Here's one:

https://www.beatthegmat.com/geometry-t11659.html#47672
Image

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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by Montreal06 » Sat Oct 04, 2008 2:45 am
we do not need trig to solve this!
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