A square and rectangle have same perimeter. They differ in area by 1 square cm. The length of the rectangle exceeds its breadth by

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Abhilash Singh · Accepted Answer

Answer:  Let us assume a as side of square and l and b as length and breadth of the rectangle respectively.  It is given that:  perimeter of square = perimeter of rectangle  4a = 2(l + b)  a = (l + b)/2 ....(1)  It is also given that:  Difference between area of square and area of rectangle is 1 cm2  Let say, area of square is bigger  Area of Square - Area of Rectangle = 1  a2 - lb = 1 ....(2)  Substitute eq. 1 in eq. 2  (l + b)2/4 - lb = 1  (l2 + b2 + 2lb - 4lb)/4 = 1  l2 + b2 - 2lb = 4  (l - b)2 = 4  l - b = 2  Therefore, the length of the rectangle exceeds the breadth by 2 cm.  Note: If a square and rectangle have the same perimeter, then the area of the square is higher or we can also say that the area of the rectangle is highest for the rectangle which has the same length and breadth as the perimeter by 4. If you assume, the area of the rectangle to be higher, you will get the difference between length and breadth as a complex number.  Hence, (b) is the correct answer.

A square and rectangle have same perimeter. They differ in area by 1 square cm. The length of the rectangle exceeds its breadth by

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A square and rectangle have same perimeter. They differ in area by 1 square cm. The length of the rectangle exceeds its breadth by

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