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 the length of one side is $x$. What should be the value of $z$ so that all the 3 figures can be arranged to form a square of side $(x+6)$?

• 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.