Question

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

Answer 1

Answer:

Here is the solution for your question

Answer 2

The perimeter of the rectangle is 190 cm.

Step-by-step explanation:

Given : Length of a rectangle is 5 cm more than twice of its breadth.  If area of the rectangle is 1950 sq.cm.

To find : The perimeter  of rectangle ?

Solution :

Let the breadth of the rectangle be 'x'.

Length of a rectangle is 5 cm more than twice of its breadth.  

i.e. length = 2x+5

The area of the rectangle is $A=l\times b$

$1950=(2x+5)\times x$

$1950=2x^2+5x$

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

Applying quadratic formula, $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

a=2, b=5 and c=-1950

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

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

$x=\frac{-5\pm125}{4}$

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

$x=30,-32.5$

Reject x=-32.5

The width of the rectangle is 30 cm.

The length of the rectangle is 2(30)+5=65 cm

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

$P=2(65+30)$

$P=2(95)$

$P=190$

The perimeter of the rectangle is 190 cm.

#Learn more

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