Question

All the three quadrilaterals ADEC, ABIH and BCGF are squares and $\angle ABC={90}^{\circ }$. If the area of ADEC $=x^2$ and the area of AHIB $=y^2(x^2>y^2)$, then the area of BCGF is:

• A. (x + y) (x - y)
• B. $(x+y{)}^2$
• C. $(x-y{)}^2$
• D. None of these

Answer 1

The correct option is A (x + y) (x - y)

In square ADEC, area $=x^2$
So, side AC = x
In square AHIB, area $=y^2$
So, side AB = y
As $x^2>y^2$ so, x > y
and in $\Delta ABC(\angle ABC={90}^{\circ })$
Applying Pythagoras Theorem,
Area of square BCGF = $Sid{e}^2=B{C}^2=A{C}^2-A{B}^2=x^2-y^2$.