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

Answer: 4 units by 5 unitsStep-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.