Question

If the area of a square is given by 4a²+ 12ab + 9b² Find the side of the square. ​

Answer 1

Solution :

The area of a square is given by 4a² + 12 ab + 9b² .

We need to find the side of this required square .

It is obvious that the sides of a square are always equal length , thus, the expression 4a² + 12 ab + 9b² is the square of a length represented by two variables a & b .

Let this be equal to t .

So ,

t² = 4a² + 12 ab + 9b²

> t² = ( 2a )² + (3b )² + 2(2a)(3b)

> t² = ( 2a + 3b )²

> t = | 2a + 3b |

Thus , the side of the required square comes out to be | 2a + 3b | . The modulus sign is given because the value of a and b can be in negative resulting in the value of the side comming out to be negative which we know isn't possible .

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Interesting Facts :

For a square and a rectangle having the same perimeter , the area of the square is always greater than the area of the rectangle .

This follows from the proof of the square inequality graphically .

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Answer 2

Area of square = (Side)²

4a²+ 12ab + 9b² = (Side)²

$\sqrt{4 {a}^{2} + 12ab + 9 {b}^{2} } = side$

$2a + 2 \sqrt{3} ab + 3b = side$