SOLUTION: A rectangle is 10cm longer than it is wide. A line segment cute the area enclosed into two pieces, one of which is a square. The area of the rectangle os 118cm^2 more than the area

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Question 66019: A rectangle is 10cm longer than it is wide. A line segment cute the area enclosed into two pieces, one of which is a square. The area of the rectangle os 118cm^2 more than the area of the square. What is the width of the rectangle?
Answer by praseenakos@yahoo.com(507) (Show Source):
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QUESTION:
A rectangle is 10cm longer than it is wide. A line segment cute the area enclosed into two pieces, one of which is a square. The area of the rectangle os 118cm^2 more than the area of the square. What is the width of the rectangle?
ANSWER:

Given that length of a rectangle is 10cm longer than it is width,

So let's take width = x cm ( because length is given in terms of width.)

Then we have length = (x+10) cm

Then area of rectangle is = length * width

==> A = x ( x + 10 )

==> = x^2 + 10x

Next it is given that area of rectangle is 118 more that of square.

Here we can take square with side "x"

So area of square = x*x = x^2
that is,
area of rectangle = area of square + 118

==> x^2 + 10 x = x^2 + 118

Subtract x^2 from both sides,

==> x^2 + 10 x - x^2 = x^2 + 118 -x^2

==> 10x = 118

Divide both sides by 10

==> 10x/10 = 118/10

==> x = 11.8

That is width of the rectangle is 11.8 cm

Hope you understood.

Regards.
praseena.