This is the Important question based on Linear Equations in two Variables Chapter of R.S Aggarwal book for

ICSE & CBSE Board.

Question Number 52 Exercise 3E of RS Aggarwal Solution.

In this Question following Points are given.

(i)The sum of the areas of two squares

(ii) the difference of their perimeters

You have to find the sides of squares.

# The sum of the areas of two squares is 157m^2. If the sum of their perimeters is 68m, find the sides of the squares.

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Let the sides of the two squares be x and y.

Their areas will be :

x² and y² respectively.

Now their perimeters will be :

4x and 4y respectively.

We do this since the sides of a square are equal.

From the question we can do the substitution as follows:

Sum of the areas:

x² + y² = 157………..1)

4x + 4y = 68……2)

Divide equation 2 all through by 4 to get :

x + y = 17

We need to solve for x and y.

By substitution we have :

x = 17 – y

Replace this in equation 1 as follows:

(17 – y)² + y ² = 157

289 – 34y + y² + y² = 157

Collecting the like terms together we have :

2y² – 34y + 132 = 0

Solving the quadratic equation:

Divide through by 2 to get :

y² – 17y + 66 = 0

The roots are :

-11 and -6

We expand the equation as follows:

y² – 11y – 6y + 66 = 0

y(y – 11) – 6(y – 11) = 0

(y – 6)(y – 11) = 0

y = 6 or 11

When y is 11 x is (17 – 11) = 6

So the values can either be 6 or 11 for y or vice versa for x.

The sides are thus :

11 meters and 6 meters