Q. 634.7( 6 Votes )

The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Answer :

Let the length of first and second square be x and y respectively

According to the question;


x2 + y2 = 640 – – – – (1)


Also 4x – 4y = 64


x – y = 16


x = 16 + y


Putting the value of x in(1) we get


(16 + y)2 + y2 = 640 using (a + b)2 = a2 + 2ab + b2


256 + 32y + y2 + y2 = 640


2y2 + 32y – 384 = 0


y2 + 16y – 192 = 0


Using the splitting middle term – the middle term of the general equation is divided in two such values that:


Product = a.c


For the given equation a = 1 b = 16 c = – 192


= 1. – 192 = – 192


And either of their sum or difference = b


= 16


Thus the two terms are 24 and – 8


Difference = 24 – 8 = 16


Product = 24. – 8 = 192


y2 + 24y – 8y – 192 = 0


y(y + 24) – 8(y + 24) = 0


(y + 24) (y – 8) = 0


(y + 24) = 0 (y – 8) = 0


y = 8 or y = – 24


y = 8 (y cannot be negative)


x = 16 + 8 = 24m


Hence the length of first square is 24m and second square is 8m.


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