# How to find the length of the sides?

• Nov 1st 2009, 06:00 AM
JQ2009
How to find the length of the sides?
A plot of land is a square and its sides are x metres. It gets expanded so that the plot of 40 metres longer and 50 metres wider and so that it is 200 metres short of being double the area of its original size. What is x?

I have got as far as working out:

Original Perimeter is 4x
Original Area is 2x
New perimeter is 4x + 180
New Area = 4r - 200 = 2r - 100

I am unsure where to take it from here though. Help would be appreciated.
• Nov 1st 2009, 06:48 AM
ukorov
Quote:

Originally Posted by JQ2009
A plot of land is a square and its sides are x metres. It gets expanded so that the plot of 40 metres longer and 50 metres wider and so that it is 200 metres short of being double the area of its original size. What is x?

I have got as far as working out:

Original Perimeter is 4x
Original Area is 2x
New perimeter is 4x + 180
New Area = 4r - 200 = 2r - 100

I am unsure where to take it from here though. Help would be appreciated.

wrong. original area = (x)(x) = x^2
new area = (x + 50)(x + 40) = x^2 + 90x + 2000

"so that IT is 200 metres short of being double the area of its original size."
what is IT?
is it new area = original area x 2 - 200 square metres????
• Nov 1st 2009, 07:24 AM
JQ2009
It is the new area. Which is 200 metres away from being double the original area.
• Nov 1st 2009, 07:44 AM
ukorov
Quote:

Originally Posted by JQ2009
It is the new area. Which is 200 metres away from being double the original area.

new area = original area x 2 - 200 square metres <---- so is this correct?
• Nov 1st 2009, 07:48 AM
JQ2009
Quote:

Originally Posted by ukorov
new area = original area x 2 - 200 square metres <---- so is this correct?

yes.
• Nov 1st 2009, 07:51 AM
e^(i*pi)
Quote:

Originally Posted by JQ2009
yes.

\$\displaystyle (x+50)(x+40) = 2x^2-200\$

\$\displaystyle x^2+90x+2000 = 2x^2 - 200\$

\$\displaystyle x^2 - 90x - 2200 = 0\$

\$\displaystyle (x-110)(x+20) = 0\$

edit: I'm an idiot xD. Didn't see double area
• Nov 1st 2009, 07:54 AM
ukorov
x^2 + 90x + 2000 = 2(x^2) - 200
-x^2 + 90x + 2200 = 0
x^2 - 90x - 2200 = 0

Hence,
x = [-(-90) +/- (8100 + 4 x 2200)^0.5] / 2
= (90 +/- 130) / 2
= 110 or -20

There cannot be negative value for dimension, so x = 110 metres