The length of a rectangle exceeds its breadth by 8cm. If the length was halved and the breadth increased by 6cm, the area would be decreased by 36 cm^2. Find the length andperimeter of the original rectangle.
We let the length be represented by Land the breadth by B.
The length of the rectangle exceeds its breadth by 8cm. This can be translated to a mathematical expression:
L = B + 8
Now, the length was halved and the breadth was increased by 6. This means that the new length is L/2 and the new breadth is B + 6.
The area of the original rectangle is:
`A_(old) = L *B = (B+8)*B = B^2 + 8B`
The area of the new rectangle is:
`A_(n e w) = L/2 * (B+6) = (B+8)/2 * (B+6) = (B^2 + 14B + 48)/2`
Now, we know that the new rectangle has a lower area then the old:
`A_(old) - A_(n e w) = 36`
This means that:
`B^2 + 8B - (B^2 + 14B + 48)/2 = 36`
This is just a quadratic equation that can easily be solved. First, we simplify it:
`2B^2 + 16B - B^2 - 14B - 48 = 72`
`B^2 + 2B - 120 = 0`
Using the quadratic formula:
`(-2 pm sqrt(2^2 - 4*1*-120))/2`
`(-2 pm sqrt(484))/2`
The roots are 10 and -12
Hence, `B=10` (the other root is negative, and hence, is not a reasonable answer, since lengths are positive).
We know that in the original rectangle, the length is 8cm more than B. Hence, L = 18cm.
The perimeter of the original rectangle then, is P = 18*2 + 10*2 = 56 cm.