Question

The length of a rectangle is 4yd longer than its width. If the perimeter of the rectangle is 36yd, find its area

Answer 1

Answer:

$\boxed{\sf Area \ of \ the \ rectangle = 91 \ yd^{2}}$

Given:

Length of the rectangle = 4 yd longer than its width

Perimeter of the rectangle = 36 yd

To Find:

Area of the rectangle

Step-by-step explanation:

Let the width of the rectangle be 'w' yd

So,

Length of the rectangle = (w + 4) yd

$\therefore \\ \sf \implies Perimeter \: of \: the \: rectangle = 2(Length + Width) \\ \\ \sf \implies 36 = 2((4 + w) + w) \\ \\ \sf \implies 36 = 2(4 + w + w) \\ \\ \sf \implies 36 = 2(4 + 2w) \\ \\ \sf 36 =2(2w+4) \: is \: equivalent \: to \: 2(2w + 4) = 36: \\ \sf \implies 2(2w + 4) = 36 \\ \\ \sf Divide \: both \: sides \: of \: 2 (2w + 4) = 36 \: by \: 2: \\ \sf \implies 2w + 4 = 18 \\ \\ \sf Subtract \: 4 \: from \: both \: sides: \\ \sf \implies 2w = 14 \\ \\ \sf Divide \: both \: sides \: of \: 2w = 14 \: by \: 2: \\ \sf \implies w = 7$

So,

Width of the rectangle = 7 yd

Length of the rectangle = (7 + 4) yd

= 13 yd

$\therefore \\ \sf Area \ of \ the \ rectangle = Length \times Width \\ \\ \sf = 7 \times 13 \\ \\ \sf = 91 \: {yd}^{2}$