GMAT Prep - Geometry
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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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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
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
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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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
Thanks
again
- Stuart@KaplanGMAT
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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
Here's one:
https://www.beatthegmat.com/geometry-t11659.html#47672
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