Question

There are three figures of areas $x^2,12x$ and $z$ in the below figure. Figures A and C are squares while figure B is a rectangle, with length of one side as $x.$ What should be the value of $z$ so that all the 3 figures can be arranged to form a square of side (x +6)?

Answer 1

Using the algebraic identity $(a+b{)}^2=(a^2+2ab+b^2),$ we have
$(x+6{)}^2=x^2+12x+\mathrm{36....}(1)$
Thus, 36 must be added to $x^2$ and $12x$ to make to complete the square.

In mathematical terms, we have
$x^2+12x+z=(x+6{)}^2$
$⟹x^2+2\left(x\right)\left(6\right)+z=(x+6{)}^2$
$⟹z=\mathrm{36...}\text{(Using (i))}$

Thus, the value of $z$ so that all the 3 figures can be arranged to form a square of side (x +6) is 36.