Question

Ello....mate



if the rectangle, the length is increased and breadth is reduced each by 2 unit the area is reduced by 28 square unit if however the length is reduced by one unit and the breadth is increased by 2 unit the area increased by 33 square unit find the area of the rectangle.

solve this....________❣️

step by step ☺️❣️

don't spam ❎

​

Answer 1

(x+2)(y-2)=xy-28

xy-2x+2y-4=xy-28

-2x+2y-4

2x-2y+4=28

2x-2y=24

x-y=12

(x-1)(y+2)=xy-1

xy+2x-y-2=xy-1

2x-y-2=-1

2x-y=1

x=-11

substitute x=-11 in eq n(1) x-y=12

-11-y=12

-y=12+11

-y=23

x=-11 ;y=-23

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

Let, the length of the rectangle be 'x' unit

and breadth of the rectangle be 'y' unit

We know,

Area of a rectangle = length × breadth.

A/Q,

$\implies \: \sf{(x + 2)(y - 2) = xy - 28} \: \: \rightarrow(1)$

$\implies \: \sf{(x - 1)(y + 2) = xy + 33} \: \: \rightarrow(2)$

Now, solving equation (1)

$\implies \: \sf{(x + 2)(y - 2) = xy - 28} \:$

$\implies \: \sf{xy - 2x + 2y - 4 = xy - 28}$

$\implies \: \sf{- 2x + 2y = - 24}$

$\implies \: \sf{x - y = 12}$

$\implies \: \sf{y = x - 12} \: \: \: \rightarrow(3)$

Now solving (2), we get

$\implies \: \sf{(x - 1)(y + 2) = xy + 33} \:$

$\implies \: \sf {xy + 2x - y - 2 = xy + 33}$

$\implies \sf{2x - y = 35}$

Putting the value of 'y', we get

$\implies \: \sf {2x - (x - 12) = 35}$

$\implies \: \sf{2x - x + 12 = 35}$

$\implies \: \sf{x = 35 - 12}$

$\implies \: \sf{x = 23}$

Now, from (3)

$\implies \: \sf{y = x - 12} \:$

$\implies \: \sf{y = 23 - 12} \:$

$\implies \: \sf{y = 11} \:$

Hence we got,

Length = 23 units and

Breadth = 11 units

So,

$\sf{Area \: of \: the \: rectangle \: = l \: \times \: b }$

$= \: \sf{23 \times 11}$

$= \sf{253 \: {unit}^{2} }$

$\huge \boxed{ \sf{Area \: = \: 253 {unit}^{2} }}$