Question

length of a rectangle 5 cm more than twice of its width if area of Rectangle 1950 sq cm then find the perimeter of rectangle​

Answer 1

length of a rectangle 5 cm more than twice of its width and area of Rectangle 1950 sq cm then the perimeter of rectangle​ is given by,

Area of rectangle = l × w

where, l = length of rectangle

w = width of rectangle

From given we have,

w = x (say)

l = 5 + 2x

A = l × w

1950 = (5 + 2x) × x

1950 = 5x + 2x²

2x² + 5x - 1950 = 0

solving the above quadratic equation, we get,

x = 30,  - 65/2

w = x = 30 cm

∴   w = 30 cm

l = 5 + 2x = 5 + 2 (30) = 65

∴  l  = 65 cm

Now consider,

The perimeter of the rectangle,

P = 2 ( l + w)

= 2 ( 65 + 30)

= 2 (95)

∴  P = 190 cm

Therefore the perimeter of rectangle is 190 cm.