PS - Geometry Q From GmatPrep

Question

Not sure how to solve it .

My approach was
Since Area of square garden = A(ie S^2) ,hence S = Sqrt(A)
Perimeter is P and A=2p+9

So to find value of P no

P = 4S = 4*Sqrt(A) = 4*Sqrt(2p+9)

but couldnt figure out how it resulted in P=36

could someone pls explain ?

could someone pls explain ?

yezz · Answer

question first if u plz :lol:

GmatInstinct · Answer

oh my bad .. i thought i already had the attachment to it 
Here it is

yezz · Answer

Let side of square = x

A = X^2
P = 4X

A=2P+9

THEN

X^2 = 8X+9

THUS

X^2-8X-9 = 0

(X+1)(X-9) = 0 X IS LENGTH CANT BE NEGATIVE THUS X = 9

P = 4X = 4*9 = 36

GmatInstinct · Answer

U Rock !!
 
I guess although i was rite with the equations involved but my logic was bit convoluted ,ie , instead of considering a S as side and taking it from there .I was trying to back solve it . 

i shall keep that in mind for future

thanks a ton

GMATT73 · Answer

If you just want to work with the variables given, you can also try this approach.

Two equations and two variables here.

1. sqrtA^2 = A (area) : (side length is sqrtA)
2. p/4*4 = P (perimeter) : (side lenght is p/4)

set the two equal to each other ---> sqrtA = p/4 ---> A = p^2/16

Given that A = 2P +9

Substitute

P^2/16 = 2P + 9

P^2-32P -144 =0

The only factor that works is (36 and -4)

PS - Geometry Q From GmatPrep

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