Question

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$\underline{\bigstar\boldsymbol{\; Question:}}$
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The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.​

Answer 1

$\underline{\bigstar\boldsymbol{\; Question:}}$

　　　　　　　　　　　 

• The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

　　　　　　　　　　　 

$\underline{\bigstar\boldsymbol{\; Given:}}$

　　　　　　　　　　　 

• Area gets reduced by 9 square units. Length is reduced by 5 units and Breadth is increased by 3 units.

　　　　　　　　　　　 

• Area is increased by 67 square units and length is increased by 3 units and Breadth increased by 2.

　　　　　　　　　　　 

$\underline{\bigstar\boldsymbol{\; To\:find:}}$

　　　　　　　　　　　 

• Dimensions of the rectangle.

　　　　　　　　　　　 

$\underline{\bigstar\boldsymbol{\; Answer:}}$

　　　　　　　　　　　 

Let the length and the breadth of the rectangle be x and y units respectively.

Area = xy sq. units.

Now, if the length is reduced by 5 units and the Breadth is increases by 3 units, then area is reduced by 9 square units.

$\large\rm{xy-9=(x-5)(y+3)}$

$\large\rm{⇒\:x-9=xy+3x-5y-15}$

$\large\rm{⇒\:3x-5y-6=0}\:\:..(1)$

　　　　　　　　　　　 

Area is increased by 67 square units,

Length is increased by 3 units and Breadth increased by 2.

$\large\rm{xy+67=(x+3)(y+2)}$

$\large\rm{⇒\:xy+67=xy+2x+3x+6}$

$\large\rm{⇒\:2x+3y-61=0\:\:..(2)}$

　　　　　　　　　　　 

$\large \rm \gray { Hence,\:our\:equation\:is }$

　　　　　　　 ⛤ 3x - 5y = 6 ..(1)

　　　　　　　　⛤ 2x + 3y = 61 ..(2)

From ( 1 ) :

$\large \sf 3x - 5y - 6 = 0$

$\large \sf 3x = 6 + 5y$

$\large \sf x = \frac{6 + 5y}{3}$

Putting value of x in ( 2 ) :

$\large \sf 2x + 3y = 61$

$\large \sf 2( \frac{6 + 5y}{3} ) + 3y = 61$

Multiplying both sides by 3 :

$\large \sf 3 \times 2( \frac{(6 + 5y)}{3} + 3 \times 3y = 3 \times 61$

$\large \sf 2(6 + 5y) + 9y = 183$

$\large \sf 12 + 10y + 9y = 183$

$\large \sf 19y = 183 - 12$

$\large \sf 19y = 171$

$\large \sf y \: = \frac{171}{19}$

$\large \sf y \: = \: 9$

Putting y = 9 in one equation ..( 1 )

3x - 5y = 6

3x -5 (9) = 6

3x - 45 = 6

3x = 6 + 45

3x - 51

x = 51 / 3

x = 17

$\large{\underline{\boxed{\mathrm{x=17,\:y=9\:is\:the\:solution}}}}$

Hence ,

Length of rectangle = x = 17 units

Breadth of rectangle = y = 9 units

Answer 2

Question:-

• The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

To Find:-

• Find the dimensions of rectangle.

Solution:-

• Let the area be ' xy '

• The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units:-

$\sf \dashrightarrow \ xy - 9 = ( x - 5 ) ( y + 3 )$

$\sf \dashrightarrow \ xy - 9 = xy + 3x - 5y - 15$

$\sf \dashrightarrow \ 3x - 5y - 15 + 9 + xy - xy = 0$

$\sf \dashrightarrow \ 3x - 5y - 6 = 0 . . . . ( i )$

• If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units:-

$\sf \dashrightarrow \ xy + 67 = ( x + 3 )( y + 2 )$

$\sf \dashrightarrow \ xy + 67 = xy + 2x + 3y + 6$

$\sf \dashrightarrow \ 2x + 3y - 67 + xy - xy = 0$

$\sf \dashrightarrow \ 2x + 3y - 67 = 0 . . . . ( ii )$

Now ,

By cross multiplying both equation ( i ) and ( ii ) , we get:-

$\sf \dashrightarrow \ \dfrac { x } { 305 + 18 } = \dfrac { -y } { -183 + 12 } = \dfrac { 1 } { 9 + 10 }$

$\sf \dashrightarrow \ x = \cancel\dfrac { 323 } { 19 } and y = \cancel\dfrac { 171 } { 19 }$

$\sf \dashrightarrow \ x = 17 \: and \: y = 9$

Hence ,

• Length is 17 cm

• Breadth is 9 cm