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.
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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.
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