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tutonaranjo: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Wed Jun 20, 2007 8:45 am; Thanked: 1 times

GMAT Prep - Geometry

by tutonaranjo » Mon Sep 03, 2007 6:07 pm

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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Source: Beat The GMAT — Problem Solving | Mon Sep 03, 2007 6:07 pm

jaspreet.sharma: Junior | Next Rank: 30 Posts; Posts: 24; Joined: Wed Aug 22, 2007 4:49 pm; Thanked: 3 times

Re: GMAT Prep - Geometry

by jaspreet.sharma » Mon Sep 03, 2007 10:01 pm

tutonaranjo wrote:
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

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abkhan: Junior | Next Rank: 30 Posts; Posts: 27; Joined: Thu May 03, 2007 2:20 am; Location: India

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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tutonaranjo: Senior | Next Rank: 100 Posts; Posts: 42; Joined: Wed Jun 20, 2007 8:45 am; Thanked: 1 times

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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[email protected]: Senior | Next Rank: 100 Posts; Posts: 62; Joined: Thu Jul 03, 2008 4:52 am

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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smclean23: Master | Next Rank: 500 Posts; Posts: 172; Joined: Mon Jun 23, 2008 12:00 pm; Thanked: 3 times; Followed by:1 members

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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Stuart@KaplanGMAT: GMAT Instructor; Posts: 3225; Joined: Tue Jan 08, 2008 2:40 pm; Location: Toronto; Thanked: 1710 times; Followed by:614 members; GMAT Score:800

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