Question

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The area of a rectangle get reduced by 9 square metres, if it's length is reduced by 5 units and breath is increased by 3 units. If we increase the length by 3 units and the breath by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle ​

Answer 1

⇒Let the length of rectangle be 'x' units

and breadth of rectangle be 'y' units,

Hence Area × breadth

• Area = xy

Given that,

Area gets reduced by 9 square units,

if length is reduced by 5 units and breadth increased by 3 units

so, New Area = New Length × New Breadth

Old Area - 9 = ( length - 5 ) × ( Breadth + 3 )

→ xy - 9 = ( x - 5 ) ( y + 3 )

→ xy - 9 = x (y+3) - 5 (y+3)

→ xy - 9 = xy + 3x - 5y - 15

→ 0 = xy + 3x - 5y - 15 - xy + 9

→ 3x - 5y - 6 = 0

→ 3x - 5y = 6 ....( Eq. 1 )

Also,

Area increase by 67 square units.

If length is increased by 3 units and breadth increased by 2 units.

so,

New Area = New length × New Breadth

old Area + 67 = ( Length + 3) × ( Breadth + 2)

⇒xy + 67 = ( x + 3 ) ( y + 2 )

⇒xy + 67 = x ( y+2) + 3 (y+2)

⇒xy + 67 = xy + 2x + 3y + 6

⇒0 = xy + 2x + 3y + 6 - xy - 67

⇒2x + 3y - 61 = 0

⇒2x + 3y = 61 ....( eq. 2 )

Hence over the equations are,

• 3x - 5y = 6

• 2x + 3y = 61

from eq (1),

$3x - 5y - 6 = 0 \\ 3x = 6 + 5y \\ \\ x = \frac{6 + 5y}{3}$

Putting value of x in eq. 2

$2x + 3y = 61 \\ 2( \frac{6 + 5y}{3} ) + 3y = 61 \\ \\ multiplying \: both \: the \: sides \: by \: 3 \\ \\ 3 \times 2 \frac{( 6 + 5y) }{3} + 3 \times 3y = 3 \times 61 \\ \\ 2(6 + 5y) + 9y = 183 \\ 12 + 10y + 9y = 183 \\ 19y = 183 - 12 \\ 19y = 171 \\ \\ y = \frac{171}{9} \\ y = 9$

putting y = 9 in eq, 1

→3x - 5y = 6

→3x - 5×9 = 6

→3x - 45 = 6

→3x = 51

→x = 51/3

x = 17

So the length of rectangle is 17 units and breadth of rectangle is 9 units.

Answer 2

Answer:

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