63

The sum of the areas of two squares is 640 m^{2}. If the difference in their perimeters be 64 m, find the sides of the two squares.

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

According to the question;

x^{2} + y^{2} = 640 – – – – (1)

Also 4x – 4y = 64

x – y = 16

x = 16 + y

Putting the value of x in(1) we get

(16 + y)^{2} + y^{2} = 640 using (a + b)^{2} = a^{2} + 2ab + b^{2}

256 + 32y + y^{2} + y^{2} = 640

2y^{2} + 32y – 384 = 0

y^{2} + 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

y^{2} + 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.

63

64

The length of a rectangle is thrice as long as the side of a square. The side of the square is 4 cm more than the width of the rectangle. Their areas being equal, find their dimensions.

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