Question

s. Length of a rectangle is 3cm mo
than twice of its breadth If area
of the rectangle is 1950 sq.cm.
then find the perimeter of rectangle​

Answer 1

Question :

The length of a rectangle is 5 cm more than twice of its breadth. If the area of the rectangle is 1950 sq.cm then, find the perimeter of the rectangle.

Solution :

$\underline {\bold{Given:}}$

• Area of the rectangle is 1950 sq.cm.
• The length of the rectangle is 5 cm more than twice of its breadth.

$\underline {\bold{To\:Find:}}$

• The perimeter of the rectangle.

$\rule{193}{1}$

$\boxed{\purple{Area=length \times breadth}}$

Let the breadth of the rectangle be x cm.

Then, the length of the rectangle = (2x+5) cm

$\implies (2x + 5) \times x = 1950 \\ \implies 2 x^2 + 5x =1950\\ \implies 2 x^2 + 5x - 1950 = 1950 - 1950 \\ \implies 2 x^2 + 5x - 1950 = 0 \\ \implies 2x^2 + (65 - 60)x - 1950 = 0 \\ \implies 2 x^2 +65x - 60x - 1950 = 0 \\ \implies x(2x + 65) - 30(2x + 65) = 0 \\ \implies (2x + 65)(x - 30) = 0 \\ \implies 2x + 65 = 0 \: or \: x - 30 = 0 \\ \implies 2x = 0 - 65 \: or \: x = 0 + 30 \\ \implies 2x = - 65 \: or \: x = 30 \\ \implies x = \frac{ - 65}{2} \: or \: x = 30 \\ \implies x = - 32.5 \: or \: x = 30$

We can't take breadth in negative .

$\therefore{The\: breadth\: of\: the\: rectangle \:is\:30 \:cm.}$

$\therefore{The\: length \: of\: the\: rectangle \:is\:(2\times 30+5) \:cm=65\:cm.}$

$\rule{193}{2}$

$\fbox {\purple{Perimeter=2(length + breadth)}}$

$\implies Perimeter=2(length + breadth) \\ \implies Perimeter = 2(65 \: cm + 30 \: cm) \\ \implies Perimeter = 2 \times 95 \: cm \\ \implies Perimeter= 190 \: cm$

$\boxed {\green {\therefore {The\: perimeter\:of\:the\:rectangle\:is\:190\:cm.}}}$