Question

The perimeter of a rectangle is 44 cm. The area of the rectangle is 117 cm2. Find the length of the shorter side of the rectangle.

Answer 1

Let the length and width of the rectangle be x and y respectively.
As the perimeter of a rectangle is 44, we have
2x + 2y = 44 ......(1)
As the area of the rectangle is 117, we have
xy = 117 ........(2)
Simplifying equation (1), we have
x + y = 22 .......(3)
Solving the following simultaneous equations for x and y.
x + y = 22


xy = 117
Substituting y = 22 - x into xy = 117,we have
x(22 - x) = 117
22x - x^2 = 117
x^2 - 22x + 117 = 0
(x - 9)(x - 13) = 0
So x = 9 or x = 13
Substituting x = 9 into y = 22 - x,we have y = 13.
Substituting x = 13 into y = 22 - x,we have y = 9.
So the dimensions of the rectangle are 9 and 13. Mark as branlist

Answer 2

Perimeter = 44

Area = 117

P = 2×l+b , l+b= 44÷2 = 22

Length = 11+x

b = 22-11-x

b=11-x

Area=117

lb=117

(11+x)(11-x)=117

11²-x²=117

121-117=x²

4=x²

x²=4

x=+-2

l=11+x

l=11+2 & 11-2

l=13 & l=9

b=11-x

b=11-2 & 11+2

= 9 & =13

Length is 13 or 9