Question

The area of rectangular play ground is 420 m². If its length is incrased by 7 m
and breadth is decreased by 5 m, the area remains the same. Find the length
and breadth of the rectanglular play ground.
allaurine is the frequency distribution with unknown frequencies.​

Answer 1

Step-by-step explanation:

LB=420

(L+7)(B-5)=420

LB-5L+7B-35=420

7B-5L=35

L=7B-35/5

(7B-35/5)B=420

B=20cm.

L = 21cm.

Answer 2

Explanation-

Area of rectangular ground is 420 m².

Length of the rectangle is increased by 7 m and breadth is decreased by 5 m.

Let the length of the rectangle be L and breadth be B.

$\implies\:\sf{Length\:=\:L\:+\:7\:and\:Breadth\:=\:B\:-\:5}$

We know that,

Area of rectangle = length × breadth

Substitute the known values in above formula

$\Rightarrow\:\sf{420\:=\:(L+7)(B-5)}$

$\Rightarrow\:\sf{420\:=\:LB\:-\:5L\:+\:7B\:-\:35}$

$\Rightarrow\:\sf{420\:+\:35\:=\:LB\:-\:5L\:+\:7B}$

But, given that area of rectangle is 420 m².

$\implies\:\sf{420\:=\:LB}$ ---- [1]

So,

$\Rightarrow\:\sf{455\:=\:420\:-\:5L\:+\:7B}$

$\Rightarrow\:\sf{455\:-\:420\:=\:-\:5L\:+\:7B}$

$\Rightarrow\:\sf{35\:=\:-\:5L\:+\:7B}$

$\Rightarrow\:\sf{35\:+\:5L\:=\:7B}$

$\Rightarrow\:\sf{\frac{35\:+\:5L}{7}\:=\:B}$

Substitute value of B in equation [1]

$\Rightarrow\:\sf{420\:=\:\frac{35L\:+\:5L^2}{7}}$

$\Rightarrow\:\sf{420(7)\:=\:35L\:+\:5L^2}$

$\Rightarrow\:\sf{2940\:=\:35L\:+\:5L^2}$

$\Rightarrow\:\sf{5L^2\:+\:35L\:-\:2940\:=\:0}$

Take 5 common

$\Rightarrow\:\sf{5(L^2\:+\:7L\:-\:588)\:=\:0}$

$\Rightarrow\:\sf{L^2\:+\:7L\:-\:588\:=\:0}$

$\Rightarrow\:\sf{L^2\:+\:28L\:-\:21L\:-\:588\:=\:0}$

$\Rightarrow\:\sf{L(L+28)\:-21(L+28)\:=\:0}$

$\Rightarrow\:\sf{(L-21)\:(L+28)\:=\:0}$

$\Rightarrow\:\sf{L\:=\:21,\:-28(neglected)}$

•°• Length of rectangular park is 21 m.

Substitute value of L = 21 in equation [1]

$\Rightarrow\:\sf{420\:=\:21B}$

$\Rightarrow\:\sf{\frac{420}{21}\:=\:B}$

$\Rightarrow\:\sf{20\:=\:B}$

$\Rightarrow\:\sf{B\:=\:20}$

•°• Breadth of the rectangular park is 20 m.