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.

Let the breadth of a rectangle be x cm

According to the question;

Side of square = (x + 4) cm

Length of a rectangle = [3(x + 4)] cm

Area of rectangle and square are equal – –

3(x + 4)x = (x + 4)^{2}

3x^{2} + 12x = (x + 4)^{2}

3x^{2} + 12x = x^{2} + 8x + 16 { using (a + b)^{2} = a^{2} + 2ab + b^{2}}

2x^{2} + 4x – 16 = 0

x^{2} + 2x – 8 = 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 = 2 c = – 8

= 1. – 8 = – 8

And either of their sum or difference = b

= 2

Thus the two terms are 4 and – 2

Difference = 4 – 2 = 2

Product = 4. – 2 = – 8

x^{2} + 4x – 2x – 8 = 0

x(x + 4) – 2(x + 4) = 0

(x + 4) (x – 2) = 0

⇒ x = – 4 or x = 2

x = 2 (width cannot be negative)

Thus the breadth of a rectangle = 2 cm

Length of a rectangle = [3(x + 4)] = 3(2 + 4) = 18 cm

Side of square = (x + 4) = 2 + 4 = 6cm

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