Question

The area of a square is equal to the expression (4x² + 6x + 6x + 9) sq. units. Find an algebraic expression for the length of the side of the square

Answer 1

GIVEN :-

• Expression for area of square = ( 4x² + 6x + 6x + 9 ) units².

TO FIND :-

• An algebraic expression for the length of the side of the square.

SOLUTION :-

As we know that,

$\implies \boxed{\boxed{\sf \: (side) ^{2} = Area}}$

Now substitute the values,

$\implies \sf \: (side) ^{2} = 4x ^{2} + 6x + 6x + 9$

$\implies \sf \: side = \sqrt{4x ^{2} + 6x + 6x + 9}$

$\implies \sf \: side = \sqrt{2x(2x + 3) + 3(2x + 3)}$

$\implies \sf \: side = \sqrt{(2x + 3)(2x + 3)}$

$\implies \sf \: side = \sqrt{(2x + 3) ^{2} }$

$\implies \underline{\boxed{\sf \: side =(2x + 3) \: units}}$

Answer 2

Answer:

Given :-

• Area of square = (4x² + 6x + 6x + 9) sq. units.

To Find :-

Side of square in algebraic expressions

Solution :-

We know that

Area of square = side²

Side = √Area

$\implies$ Side = √4x² + 6x + 6x + 9

• Splitting middle term

$\implies$ Side = √2x(2x + 3) + 3(2x + 3)

• Taking 2x and 3 as common

$\implies$ Side = √(2x + 3) (2x + 3)

$\implies$Side = √(2x + 3)²

$\implies$Side = 2x + 3

Verification :-

Now,

We know that

Area of square = side²

4x² + 6x + 6x + 9 = (2x + 3)(2x + 3)

4x² + 6x + 6x + 9 = (2x × 2x) + (2x × 3) + (2x × 3) + (3 × 3)

4x² + 6x + 6x + 9 = 4x² + (2x × 3) + (2x × 3) + (3 × 3)

4x² + 6x + 6x + 9 = 4x² + 6x + 6x + (3 × 3)

4x² + 6x + 6x + 9 = 4x² + 6x + 6x + 9

Hence verified

Diagram :

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(4,0){2}{\line(0,1){4}}\multiput(0,0)(0,4){2}{\line(1,0){4}}\put(-0.5,-0.5){\bf D}\put(-0.5,4.2){\bf A}\put(4.2,-0.5){\bf C}\put(4.2,4.2){\bf B}\put(1.5,-0.6){\bf\large 2x + 3\ cm}\put(4.4,2){\bf\large 2x + 3\ cm}\end{picture}$