siddhans wrote:

http://img210.imageshack.us/img210/3589/gmatprepr.jpg

[Reveal] Spoiler:

What am i doing wrong here ? What am i doing wrong here ?

PQ^2 = QS^2 + PS^2

PQ^2 = PS^2 + 48 ------ (1)

QR^2 = SR^2 + 48 ------(2)

Adding 1 and 2

PQ^2 + QR^2 = PS^2 + SR ^2 + 96 ----- 3

But, PS^2 + SR ^2 = PR ^2 ------- 4

Substitute 4 in 3....

PQ^2 + QR^2 = PR^2 + 96 -----5

PQ = PR = QR since equilateral triangle

Therefore PQ^2 = 96

PQ = sqrt (96)

But this doesnt yield the correct answer ....

Red bold is not true... PS+SR = PR... so if you square (PS+SR)^2 = PR2... not what you have written

simplest way to solve this problem is sin60 = \(\sqrt{3}\)/2 = QS/PQ = 4\(\sqrt{3}\)/PQ

==> PQ = 8

so perimeter = 24

choice C.