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

sweet
was missing

s/S = a/A

had no idea on how to set up those proportions
cheers!

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

Could someone give another spin on this? I'm confused by the explanations above.....

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

we do not need trig to solve this!