Question

the perimeter of a rectangular lawn is 54m it is reduced in size so that length is 3/5th and the breadth is 3/4th of the original dimensions the perimeter of the reduced rectangle is 36m what were the original dimensions of the lawn​

Answer 1

AnsWer :

Length = 15 m and Breadth = 12 m.

Solution :

Let Length of Dimension be ( L ) m

and Breadth of Dimension be ( B ) m

A/Q,

Case 1.

The Perimeter of a Rectangle lawn is 54m.

• 2( L + B ) = 54.

$\rule{90}1$

Case 2.

It is reduced in size so that length is 3/5th and the breadth is 3/4th of the original dimensions the perimeter of the reduced rectangle is 36m

• 2 [ 3/5 L + 3 /4 B ] = 36.

$\rule{90}1$

Our Equation become,

$\tt \dagger \: \: \: \: \: 2( L + B ) = 54. \: \: \: \: \: - (1)$

$\tt \dagger \: \: \: \: \: 2 [ \dfrac{3}{5} L + \dfrac{3}{4} B ] = 36.$

$\tt \mapsto \dfrac{12 L + 15 B}{10} = 36.$

$\tt\mapsto 12L+ 15B= 360. \: \: \: \: \: - (2)$

$\rule{90}1$

Now,

Taking Equation ( 1 )

=> 2 ( L + B ) = 54.

=> L + B = 27.

=> L = 27 - B ________( 3 )

Substitute the value of L in Equation ( 3 )

=> 12 L + 15B = 360.

=> 12( 27 - B ) + 15B = 360.

=> 324 -12B + 15B = 360.

=> 3B = 360 - 324.

=> 3B = 36.

=> B = 12 m.

Putting the value of B in Equation ( 3 )

=> L = 27 - B.

=> L = 27 - 12.

=> L = 15 m.

Therefore, the Length of rectangle is 15m and Breadth is 12m.

Answer 2

$\huge\purple{\underline{\huge{\rm{Solution:}}}}$

$\small{\underline{\red{\rm{Let:}}}}$

• $\rm{The\: orginal\: dimensions\:of\:rectangle\:are}$
• $\rm{The\: length\:of\: rectangle=x\:m}$
• $\rm{The\: breadth\:of\: rectangle=y\:m}$

$\small{\underline{\red{\rm{To\: Find:}}}}$

• $\rm{The\: orginal\: dimensions\:of\:rectangle}$

$\small{\underline{\red{\rm{Given,\:in\:1st\:case:}}}}$

• the perimeter of a rectangular lawn is 54m

$\rm\green{\implies Perimeter\:of\: lawn=2(length+breadth)}$

$\rm{\implies \cancel{2}(x+y)=\cancel{54}}$

$\rm{\implies x+y=27}$

$\rm\purple{\implies x+y=27---------(i)}$

$\small{\underline{\red{\rm{Given,\:in\:2nd\:case:}}}}$

• The dimensions of the rectangle is reduced in size so that length is 3/5th and the breadth is 3/4th of the original dimensions the perimeter of the reduced rectangle is 36m.

$\rm{\implies 2(\dfrac{3}{5}x+\dfrac{3}{4}y)=36}$

$\rm{\implies \dfrac{3}{5}x+\dfrac{3}{4}y=18}$

$\rm{\implies \dfrac{12x+15y}{20}=18}$

$\rm{\implies 12x+15y=360}$

• Dividing by 3 on both sides

$\rm{\implies \cancel{12}x+\cancel{15}y=\cancel{360}}$

$\rm{\implies 4x+5y=120}$

$\rm\purple{\implies 4x+5y=120---------(ii)}$

• Now solving equation I and ii here

$\rm{in\:eq\:1st\: multiplying\:by\:4\:than\: subtracting\:by\:eq\:2nd}$

$\rm{\implies 4x+4y=108}$

$\rm{\implies 4x+5y=120}$

• $\rm{by\: solving equation\:we\:get,}$

$\rm{\implies -y=-12}$

$\rm\red{\implies y=12}$

• Now putting the value of y=12 in eq 1

$\rm{\implies x+y=27}$

$\rm{\implies x+12=27}$

$\rm{\implies x=27-12}$

$\rm\red{\implies x=15}$

Hence,

$\rm\green{\implies The\: length\:of\: lawn=x=15\:m}$

$\rm\orange{\implies The\: breadth\:of\: lawn=y=12\:m}$