Question

The length of a rectangle is 5 cm more than its width and the area is
50
c
m
2
. Find the length, width and the perimeter

Answer 1

Step-by-step explanation:

hope it helps u.........

Answer 2

Fig:-

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ x\:cm }\put(8.1,5.3){ (x+5) \:cm }\put(11.5,7.5){ 45 \:cm }\put(8.1,9.5){ 60\:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\end{picture}$

Solution:-

Let, Width= x cm.

length = (x+5) cm.

we know that for the rectangle,

Area =(length × breadth)

$\sf\implies 50= (x+5)\times x$

$\sf\implies 50= x^2+5x$

$\sf\implies x^2+5x-50=0$

$\sf\implies x^2+10x-5x-50=0$

$\sf\implies x(x+10)-5(x+10)=0$

$\sf\implies x(x+10)(x-5)=0$

Either,

(x+10)=0

=> x=-10 [Not taken into under consideration]

as, dimension can't be zero or negative,

Or,

x-5=0

=> x=5

therefore , breadth= 5 cm

and length =(5+5) cm= 10 cm.

∴ perimeter = 2(10+5) cm= 30 cm.