Question

28. the length of a rectangle is increased by 5 metres and breadth is decreased by
3metres, the area would decrease by 5 sq.metres. If the length is increased by
3 metres and breadth is increased by 2 metres, the area would increase by
50 sq. metres. What are the length and breadth of the rectangle?​

Answer 1

Figure :

$\setlength{\unitlength}{0.78 cm}\begin{picture}(12,4)\thicklines\put(5.6,9.1){ A }\put(5.5,5.8){ B }\put(11.1,5.8){ C }\put(11.05,9.1){ D }\put(4.5,7.5){ 8\:cm }\put(8.1,5.3){ 10 \:cm }\put(11.5,7.5){ 8 \:cm }\put(8.1,9.5){ 10\:cm }\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(6,6){\line(5,3){5}}\end{picture}$

Answer:

Length of rectangle is 10 m.

Breadth of rectangle is 8 m.

Formula Required :

$\blacksquare \tt area \: of \: rectangle = l \times b.$

Solution :

Let the Length be x m

and it breadth be y m

A/Q,

Case 1.

$\implies\tt(x + 5)(y - 3) = xy - 5. \\ \implies \tt \cancel{xy }- 3x + 5y - 15 = \cancel{xy }- 5. \\ \implies \tt - 3x + 5y = 10. - - - (1).$

Case 2.

$\tt \implies(x + 3)(y + 2) = xy + 50. \\ \implies \tt \cancel{xy} + 2x + 3y + 6 = \cancel {xy} + 50. \\ \implies \tt2x + 3y = 44. - - - (2)$

In equation 1 * 2 and 2 * 3, we get.

$\tt \cancel{- 6x} + 10y = 20. \\ \: \: \: \tt \cancel{6x} + 9y = 132 \\ - - - - - - - - - - \\ \implies\tt19y = 152. \\ \implies\tt y = 8.$

Now,

Putting the value of y = 8 in equation (1)

$\implies\tt - 3 + 5y = 10. \\ \tt \implies- 3x + 5 \times 8 = 10. \\ \implies \tt x = 10.$

Therefore,the value of length be 10m and breadth be 8m.

Answer 2

$\huge\red{\underline{\underline{\mathcal{\purple{Given}}}}}$

The length of a rectangle is increased by 5m and breadth is decreased by

3m, the area would decrease by 5m.² If the length is increased by

3m and breadth is increased by 2 m, the area would increase by

increase by50m².

$\huge\red{\underline{\underline{\mathcal{\purple{To\:find}}}}}$

What are the length and breadth of the rectangle?

$\huge\red{\underline{\underline{\mathcal{\purple{Solution}}}}}$

Let the length of rectangle be x then it's breadth be y

According to the question

$\large{\boxed{\bf{\blue{In\:case\::1}}}}$

Length of rectangle is increased by 5m and breadth is decreased by 3m then the area decreased by 5m²

$\implies\sf length\times{breadth}=area-5m^2$

$\implies\sf (x+5)(y-3)=(xy-5)$

$\implies\sf x(y-3)+5(y-3)=xy-5$

$\implies\sf xy-3x+5y-15=xy-5$

$\implies\sf -3x+5y=15-5$

$\implies\sf -3x+5y=10$ ----(i)

$\large{\boxed{\bf{\blue{In\:case\::2}}}}$

Length of rectangle is increased by 3 and breadth is increased by 2 then the area increased by 50m²

$\implies\sf length\times{breadth}=area+50m^2$

$\implies\sf (x+3)(y+2)=(xy+50)$

$\implies\sf x(y+2)+3(y+2)=xy+50$

$\implies\sf xy+2x+3y+6=xy+50$

$\implies\sf 2x+3y=50-6$

$\implies\sf 2x+3y=44$ ----(ii)

Solve both the equation by elimination method

• -3x + 5y = 10 ----(i)
• 2x + 3y = 44 ----(ii)

Multiply (i) by 2 and multiply (ii) by 3

• -6x + 10y = 20
• 6x + 9y = 132

Adding both the equation

$\implies\sf (-6x+10y)+(6x+9y) = 132+20$

$\implies\sf -6x+10y+6x+9y = 152$

$\implies\sf 19y = 152$

$\implies\sf \Large\cancel\frac{152}{19}=8m$

Substitute the value of y in equation (i)

$\implies\sf -3x+5y=10$

$\implies\sf -3x+5\times{8}=10$

$\implies\sf -3x+40=10$

$\implies\sf -3x=10-40$

$\implies\sf \Large\cancel\frac{30}{3}=10m$

$\large{\boxed{\bf{\blue{Required\:length=x=10m}}}}$

$\large{\boxed{\bf{\blue{Required\:breadth=y=8m}}}}$

$\boxed{\begin{minipage}{7cm}\textbf{Some important formulae} \\ \\ Area of rectangle = length\times{breadth} \\ \\ Perimeter\:of\:rectangle = 2(length+breadth) \\ \\ Area\:of\:square = side\times{side} \\ \\ Perimeter\:of\:square = 4\times{side} \end{minipage}}$