Question

$ABCD$ is a square inscribed in a circle of radius $14$ cm. $E,F,G$ and $H$ are the midpoints of the sides $DA,AB,BC$ and $CD$ respectively. The area of the square $EFGH$ will be equal to

• A. $89$ $c{m}^2$
• B. $196$ $c{m}^2$
• C. $98$ $c{m}^2$
• D. $392$ $c{m}^2$

Answer 1

The correct option is B $196$ $c{m}^2$

Join $BD.$ $BD$ is diameter of a circle.

$\therefore$ $BD=2\times 14cm=28cm$

In $△ABD,$

$\angle DAB={90}^o$ [ Angle of square ]

$AB=AD$ [ Sides of square ]

$\Rightarrow$ $(BD{)}^2=(AB{)}^2+(AD{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(28{)}^2=(AB{)}^2+(AB{)}^2$

$\Rightarrow$ $784=2(AB{)}^2$

$\Rightarrow$ $(AB{)}^2=392$

$\Rightarrow$ $AB=14\sqrt{2}cm$

$\Rightarrow$ $AE=AF=\frac{14\sqrt{2}}{2}cm$ [ Since $E$ and $F$ are midpoints of $AD$ and $AB$ ]

In $△AEF,$

$(EF{)}^2=(AE{)}^2+(AF{)}^2$ [ By Pythagoras theorem ]

$\Rightarrow$ $(EF{)}^2={\left(\frac{14\sqrt{2}}{2}\right)}^2+{\left(\frac{14\sqrt{2}}{2}\right)}^2$

$\Rightarrow$ $(EF{)}^2=\frac{392}{4}+\frac{392}{4}$

$\Rightarrow$ $(EF{)}^2=196$

$\Rightarrow$ $EF=14cm$

$\Rightarrow$ Area of a square $EFGH=(Side{)}^2$

$=(EF{)}^2$

$=(14cm{)}^2$

$=196c{m}^2$