Question

The length of a rectangle is 5 m more than twice the width, and the area of the rectangle is 63 m2 . find the dimensions of the rectangle.

Answer 1

A=63
L=2w+5
A=LW
63=(2W+5)*W
63=2W^2+5W
2W^2+5W-63=0
(2W-9)(W+7)=0
2W-9=0
W=9/2
W=4.5 (ANS. FOR THE WIDTH)
L=2*4.5+5
L=9+5
L=14 (ANS. FOR THE LENGTH)
PROOF:
63=14*4.5
63=63

Have a nice day :D

Answer 2

The dimensions of the rectangle are 14 meters by 4.5 meters.

How to find the dimensions of the rectangle?

For a rectangle of length L and width W, the area is given by the formula:

A = L*W

In this case, we know that:

L = 5m + 2*W

And the area is 63m², so we can write:

63m² = (5m + 2*W)*W

Now we need to solve this for W.

2*W^2 + 5m*W - 63m² = 0.

The solutions of this quadratic equation are:

$W = \frac{-5m \pm \sqrt{(5m)^2 - 4*2*(64m^2)} }{2*2} \\\\W = \frac{-5m \pm 23m }{4}$

We only care for the positive solution, which is:

W = (-5m + 23m)/4 = 18m/4 = 4.5m

Then the length is:

L = 5m + 2*4.5m = 14m

So the dimensions of the rectangle are 14 meters by 4.5 meters.

If you want to learn more about rectangles:

https://brainly.com/question/17297081

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