# 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

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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)

=> L^2-5L+48=L^2+L-6

=> 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`