the length of a rectangle is greater than the breadth by 3 CM if the length is increased by 9 cm and the breadth is reduced by 5 cm the area remain the same find the dimension of the rectangle
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Let the breadth be x.cm
Length=(x+3)cm
Area of rectangle =x×(x+3)
=x^2+3x.
According to second condition,
Length=x+3+9=(x+12)cm.
Breadth=(x-5)cm.
=> (x+12)(x-5)=x^2+3x
= x^2-5x+12x-60=x^2+3x
= -5x+12x-60=3x
= 7x-60=3x
=7x-3x=60
=4x=60
x=60/4=15cm.
Breadth=x=15cm.
Length=x+3=15+3=18cm.
Length=(x+3)cm
Area of rectangle =x×(x+3)
=x^2+3x.
According to second condition,
Length=x+3+9=(x+12)cm.
Breadth=(x-5)cm.
=> (x+12)(x-5)=x^2+3x
= x^2-5x+12x-60=x^2+3x
= -5x+12x-60=3x
= 7x-60=3x
=7x-3x=60
=4x=60
x=60/4=15cm.
Breadth=x=15cm.
Length=x+3=15+3=18cm.
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