Mr Ahuja has two square plots of land which he utilises for two differ

1.	Mr Ahuja has two square plots of land which he utilises for two different purposes— one for providing free education to the children below the age of 14 years and the other to provide free medical services for the needy villagers. The sum of the areas of two square plots is 15425 m2. If the difference of their perimeters is 60 m, find the sides of the two squares.Which qualities of Mr Ahuja are being depicted in the question?
Answer» Let x be the length of the side of first square and y be the length of side of the second square. Then, x²+ y²= 15425 ...(i) Let x be the length of the side of the bigger square. 4x – 4y = 60 ⇒ x – y = 15 or x = y + 15 ...(ii) Putting the value of x in terms of y from equation (ii), in equation (i), we get (y + 15)² + y²= 15425 ⇒ 2y² + 30y – 15200 = 0 or y² + 15y – 7600 = 0 or (y + 95)(y – 80) = 0 ⇒ y = – 95, 80 But, sides cannot be negative, so y = 80 Therefore, x = 80 + 15 = 95 Hence, sides of two squares are 80 m and 95 m.

Mr Ahuja has two square plots of land which he utilises for two different purposes— one for providing free education to the children below the age of 14 years and the other to provide free medical services for the needy villagers. The sum of the areas of two square plots is 15425 m2. If the difference of their perimeters is 60 m, find the sides of the two squares.Which qualities of Mr Ahuja are being depicted in the question?

Answer»

Let x be the length of the side of first square and y be the length of side of the second square.
Then, x²+ y²= 15425 ...(i)
Let x be the length of the side of the bigger square.
4x – 4y = 60
⇒ x – y = 15 or x = y + 15 ...(ii)
Putting the value of x in terms of y from equation (ii), in equation (i), we get
(y + 15)² + y²= 15425 ⇒ 2y² + 30y – 15200 = 0
or y² + 15y – 7600 = 0 or (y + 95)(y – 80) = 0 ⇒ y = – 95, 80

But, sides cannot be negative, so y = 80
Therefore, x = 80 + 15 = 95
Hence, sides of two squares are 80 m and 95 m.

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