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Question 54
The area of a square and a rectangle are equal. The length of the rectangle is greater than the side of square by 9 cm and its breadth is less than the side of square by 6 cm. What will be the perimeter of the rectangle?
Solution
Let the side of the square be S cm.
Length of the rectangle be l cm and breadth of the rectangle be b cm.
Given, l = s+9 cm and b = s-6 cm
Given, $$lb = s^2$$
$$(s+9)(s-6) = s^2$$
=> $$s^2-6s+9s-54 = s^2$$
=> $$3s-54 = 0$$
=> $$3s = 54 => s = 18$$
Then, l = s+9 = 18+9 = 27 cm
b = s-6 = 18-6 = 12 cm
Therefore, Perimeter of the rectangle = 2(l+b) = 2(27+12) = 2*39 = 78 cm
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