Math, asked by pranav12278, 1 year ago

the area of a rectangle get reduced by 9 units when it length is reduced by 5 unit and breadth is incresed by 3 units if we increase the length by 3 unis and breadth by 2 units the area increses to 67 unit square find the dimension of the rectangle​

Answers

Answered by Anonymous
29

Solution:

Let length and breadth of rectangle be x unit and y unit.

So, Area = xy

According to question,

(x - 5)(y + 3) = xy - 9

=> 3x - 5y - 6 = 0         ..........(1)

(x + 3)(y + 2) = xy + 67

=> 2x + 3y - 61 = 0      ...........(2)

We solve it by Cross multiplication method,

=> 3x - 5y - 6 = 0         ..........(1)

=> 2x + 3y - 61 = 0      ...........(2)

\sf{\implies \dfrac{x}{305-(-18)} = \dfrac{y}{-12-(-183)} =\dfrac{1}{9-(-10)}}

\sf{\implies \dfrac{x}{323}=\dfrac{y}{171}=\dfrac{1}{19}}

\sf{\implies \dfrac{x}{323}=\dfrac{1}{19}}

=> 19x = 323

=> x = 17

\sf{\implies \dfrac{y}{171}=\dfrac{1}{19}}

=> 19y = 171

=> y = 9

Hence the Length and Breadth of Rectangle is 17 and 9 units.

Answered by Anonymous
38

SOLUTION:-

We know that the area of rectangle is of the form x & y

Where,

x= length

y= breadth

A/q

(x-5)(y+3)= xy-9............(1)

(x+3)(y+2)= xy+67..........(2)

On solving the equations, we get:

Equation from (1):

xy +3x-5y-15= xy-9

=) 3x-5y= 6................(3)

Equation from (2):

=) xy+2x+3y+6= xy +67

=) 2x +3y= 61..............(4)

Now by elimination method:

On multiplying equation(3) by 2 & equation (4) by 3.

=) 6x-10y= 12...........(5)

=) 6x +9y= 183............(6)

So,

On subtracting eq. (6) from (5),we get:

=) -19y= -171

=) y= 171/19

=) y= 9

On substituting y= 9 in eq. (6), we get:

=) 6x + 9(9)= 183

=) 6x + 81 = 183

=) 6x= 183 -81

=) 6x= 102

=) x= 102/6

=) x= 17

Therefore,

The dimensions of the rectangle are:

Length(x)= 17 units

Breadth(y)= 9 units

Hope it helps ☺️

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