There are three figures of areas x2,12x and z respectively, as shown. Figures A and C are squares while B is a rectangle, with the length of one side x. What should be the value of z so that all the 3 figures can be arranged to form a square of side (x+6)?
36
Using the algebraic identity (a+b)2=(a2+2ab+b2)
(x+6)2=x2+12x+36
Thus, 36 must be added to x2 and 12x to make it a perfect square.
To make a square of side x+6, a rectangle of area 6x must be added beside the square of area x2. But then it becomes a rectangle so another rectangle of area 6x must be added at the top of the square of area x2. Now to complete the square we need a square of area 62 i.e. 36 sq.units. So the area of the square available with us must be equal to 36 so as to form a square of side (x+6).