Question

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

• A. 16
• B. 36
• C. 18
• D. 24

Answer 1

The correct option is B

36

Using the algebraic identity $(a+b{)}^2=(a^2+2ab+b^2)$
$(x+6{)}^2=x^2+12x+36$
Thus, 36 must be added to $x^2$ 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 $x^2$. But then it becomes a rectangle so another rectangle of area $6x$ must be added at the top of square of area $x^2$. Now to complete the square we need a square of area $6^2$ 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 $(x+6).$