A rectangle’s length is 5 cm less than twice its width. If the lengt

1.	A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.
Answer» Let width of the original rectangle = x cm Length of the original rectangle = (2x – 5)cm Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm New width of the rectangle = (x + 2) cm New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm Given; new perimeter = 74 cm 6x – 16 = 74 => 6x = 74 + 16 => 6x = 90 =>x = 15 Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm Width of the original rectangle = x = 15 cm

A rectangle’s length is 5 cm less than twice its width. If the length is decreased by 5 cm and width is increased by 2 cm; the perimeter of the resulting rectangle will be 74 cm. Find the length and the width of the origi¬nal rectangle.

Answer»

Let width of the original rectangle = x cm

Length of the original rectangle = (2x – 5)cm

Now, new length of the rectangle = 2x – 5 – 5 = (2x – 10) cm

New width of the rectangle = (x + 2) cm

New perimeter = 2[Length+Width] = 2[2x – 10 + x + 2] = 2[3x – 8] = (6x – 16) cm

Given; new perimeter = 74 cm

6x – 16 = 74

=> 6x = 74 + 16

=> 6x = 90

=>x = 15

Length of the original rectangle = 2x – 5 = 2 x 15 – 5 = 30 – 5 = 25 cm

Width of the original rectangle = x = 15 cm