Question

length of a rectangle is 5 cm more than twice of its breadth if area of the rectangle is 1950 square cm find the perimeter of the rectangle​

Answer 1

Answer:

same answer as mvasu...told!!!

good my friend...keep on..

l=5+2b

area of rectangle= l×b

(5+2b)b

5b+2b²=1950

on finding D

we get, b=30cm

then l,

5+60

=65cm

hope this will help you....

Answer 2

The perimeter of the rectangle = 190 cm.

Solution:

Let x be the breadth of the rectangle.

Length = 5 more than twice of breadth

Length = 2x + 5

Area = 1950 sq. cm

Area of the rectangle = length × breadth

⇒ length × breadth = 1950

⇒ (2x + 5) × x = 1950

$\Rightarrow 2x^2+5x=1950$

$\Rightarrow 2x^2+5x-1950=0$

a = 2, b = 5 and c = –1950

Using quadratic formula,

$x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$x=\frac{-5 \pm \sqrt{5^{2}-4(2)(-1950)}}{2(2)}$

$x=\frac{-5 \pm \sqrt{15625}}{4}$

$x=\frac{-5+125}{4}$

$x=\frac{-5+125}{4}, \frac{-5-125}{4}$

$x=30,-32.5$

Breadth cannot be in negative terms. so neglect x = –32.5.

Therefore breadth = 30.

Length = 2(30) + 5 = 65

The perimeter of the rectangle = 2( l + b)

                                                    = 2(65 + 30)

                                                    = 190

Hence the perimeter of the rectangle = 190 cm.

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