Question

Sketch a rectangle and label its dimensions that meets the following conditions perimeter 18 area 20

Answer 1

Answer:

4 units by 5 units

Step-by-step explanation:

To solve this problem, then we must develop two equations; one for the perimeter and another for the area of the rectangle.

The perimeter of a rectangle = 2(L +B)

The area of a rectangle = L x B

Let the length be represented by L while the breadth is denoted by B. Then, our equations are

         2L + 2B =18  ---  eq 1

and  

          LB = 20  ----- eq 2

From eq 2, then  L = 20/B

We shall substitute this in equation 1 such that

2(20/B) + 2B =18

= 40/B + 2B = 18

= 40/B + 2B/1 = (40 + 2B^2) /B =18

We do cross multiplication by multiplying both sides of the equation by B.

We get

 40 + 2B^2 = 18B

We then rearrange the equation as   2B^2- 18B + 40 = 0

Dividing both sides by 2, we get B^2- 9B =20 =0

We shall get the roots of the equation to be -4 and -5. Why?  

-4B-5B =-9  and -4x-5 =20

Thus, B^2- 9B =20 =0 can be reworked as  

(B-4) (B-5)= 0   From B-4=0, B= 4

                   From LB =20, If B =4, then L =20/4 = 5

                           From B-5=0; B=5

                  From LB =20, If B=5, L = 20/5 = 4

Thus, the dimensions of the rectangle are 5 units and 4 units.