Question

A pentagon is made up of an equilateral $△ABC$ of side length $2$ $cm$ on top f a square $BCDE$ Circumscribe a circle through points $A,D$ and $E.$ The radius of the circle is:

• A. $1+\frac{\sqrt{3}}{2}$
• B. $5-2\sqrt{3}$
• C. $2$
• D. $1+\sqrt{3}$

Answer 1

The correct option is C $2$

$EF=1$

$AB=2$

$AG=\frac{AB\sqrt{3}}{2}=\frac{2\sqrt{3}}{2}=\sqrt{3}$

Let H be the circumcentre of $\Delta ADE$

$GH=x$

$\therefore HF=2-x$

$HE=x+\sqrt{3}$

In $\Delta HEF$, By Pythagoras theorem,

$H{E}^2=H{F}^2+E{F}^2$

$3+2\sqrt{3}x+x^2=(2-x{)}^2+1^2$

$3+2\sqrt{3}x+x^2=4-4x+x^2+1$

$2\sqrt{3}x+4x=5-3$

$(4+2\sqrt{3})x=\frac{5}{4}$

$x=\frac{2}{4+2\sqrt{3}}=\frac{1}{2+\sqrt{3}}\times \frac{2-\sqrt{3}}{2-\sqrt{3}}$

$=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$

Circum radius $=HE=2-\sqrt{3}+\sqrt{3}=2$.