If the length of a rectangle is reduced by 3 units and its breadth is increased by 2 units, the area reduces by 6 square units. If the length is increased by 4 units and the breadth is reduced by 1 unit, the area increases by 16 square units. Find the area of the rectangle.
Answers
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Let the length and breadth of the rectangle be x unit and y unit.
Now the area of the rectangle, A=xyA=xy.
Now, According to the statement given in the question, we have
The length is reduced by 5 units and the breadth is increased by 3 units. And the area has decreased by 9 units.
So , the new length will be =x−5x−5
So , the new breadth will be =y+3y+3
New area = xy−9xy−9
So, the new the relation will be
(x−5)(y+3)=xy−9⇒3x−5y−6=0→eqn(1)(x−5)(y+3)=xy−9⇒3x−5y−6=0→eqn(1)
Also, when the length is increased by 3 units and the breadth is increased by 2 units, then the area Is increased by 67 units.
New length = x+3x+3
New breadth will be =y+2y+2
New area = xy+67xy+67
So, the new the relation will be
(x+3)(y+2)=xy+67⇒2x+3y−61=0→eqn(2)(x+3)(y+2)=xy+67⇒2x+3y−61=0→eqn(2)
We know, cross – multiplication method
a1x+b1y+c1=0a2x+b2y+c2=0a1x+b1y+c1=0a2x+b2y+c2=0
Then they can be solved by using the relation
xb1c2−b2c1=ya2c1−c2a1=1a1b2−a2b1x=b1c2−b2c1a1b2−a2b1,y=a2c1−c2a1a1b2−a2b1xb1c2−b2c1=ya2c1−c2a1=1a1b2−a2b1x=b1c2−b2c1a1b2−a2b1,y=a2c1−c2a1a1b2−a2b1
⇒x305−(−18)=y−12−(−183)=19−(−10)⇒x323=y171=119⇒x=17,y=9⇒x305−(−18)=y−12−(−183)=19−(−10)⇒x323=y171=119⇒x=17,y=9
Therefore, the length of the rectangle = 17 units.
The breadth of the rectangle = 9 units.
Hence, the correct option is D. None of these.
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