Dimension Problem

Question

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

Answer 1

Let the width = x

This means that the height is 2x+5

 

A = w * h

 

63 = x(2x+5)

63 = 2x^2 + 5x

Set equation equal to zero in order to solve the quadratic

0 = 2x^2 + 5x - 63

Factor to (2x-9)(x+7) = 0

Set each factor to 0 and solve(If you don't understand why, (something)(something else) can only equal 0 if one of them is 0)

2x-9 = 0; x+7 = 0

so x = 4.5 or -7 but a length cannot be negative so x = 4.5(which is the width) so plug into 2x+5 to get the height:

2(4.5)+5 = width = 14

14*4.5 = 63, we are correct

Answer 2

I'm going with trial and error on this one. It looks like we are working with all whole numbers, and the area of the rectangle is a number without many factors.

 

63 is divisible by 3, 7, or 9... The factor tree ends in: 

 

7 x 3 x 3

 

SO: The dimensions can be 7 x 9 .... OR .... 21 x 3.

 

Let's check.

 

width 7, x 2 = 14, +5 = 19.... The dimensions are not 7 x 19.

 

instead, maybe:

 

width 3, x 2 = 6, +5 = 11.... The dimensions are not 3 x 11.

 

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I believe that the problem was either copied or read incorrectly, and that it should read "the length of the rectangle is 15 m more..." If we just change the 5 to 15, the problem makes sense and the answer is width 3, length 21.

 

Good luck!