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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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

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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.

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rimmery | Student, Undergraduate | (Level 1) Honors

Posted on

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.

In this particular question, 
Let the width of the rectangle be `w` .
Let the length be `l=w+5` 
Using the second statement, we can get the following formula:
`(w+3)((w+5)+3)=w(w+5)+48` (Subbing in `w+5` for `l` )
`(w+3)(w+8)=w(w+5)+48` (Simplify)
So from here, it's simply a matter of expanding and solving the equation.
`w^2+11w+24=w^2+5w+48` (Expand both sides)
`6w=24` (Simplify)
`w=4` 
And from there, `l=w+5=4+5=9`
Wiggin42's profile pic

Wiggin42 | Student, Undergraduate | (Level 2) Valedictorian

Posted on

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`

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