Let the length of the rectangle be represented by L and the width of the rectangle be represented by W. As the length is 5 cm more than the width it is equal to L-5. The area of the rectangle is L* W = L*(L-5).
If the length and breadth are both increased by 3 the area of there rectangle is given by the expression (L+3)*(W+3) = (L+3)*(L-2). This is greater than the area of the original rectangle by 48. This gives the equation L*(L-5)+48=(L+3)*(L-2)
=> 6L = 54
=> L = 9
The dimensions of the rectangle are 9 cm and 4 cm.
The most important thing in solving these types of questions are setting the variables for the missing dimensions and creating the right formula using them.
The length of a rectangle is 5 cm longer than its width.If both the length and width are increased by3cm the area is increased by 48 sq. what are dime
We should write the information algebraically. Let L = length and W = width.
L = 5 + W
( L + 3 )( W + 3 ) = LW + 48
(5 + W + 3)(W + 3) = (5 + W)W + 48
`(w +8)(w +3) = w^2 + 5w + 48`
`w^2 + 11w + 24 = w^2 + 5w + 48`
`6w = 24`
`w = 4`
`l = 5 + w`
`l = 5 + 4 = 9`