Question

Jonny is playing with some combined geometric shapes. He makes a geometric figure as shown below. Find the length of AC if the radius of the circle with the center O is 6 ft.

• A. $\sqrt{130}ft$
• B. $6\sqrt{5}ft$
• C. $6\sqrt{2+\sqrt{3}}ft$
• D. $\sqrt{120}ft$

Answer 1

The correct option is C $6\sqrt{2+\sqrt{3}}ft$

It is given that the radius of the circle is 6 ft.

Therefore, $OA=OB=OC=6ft$

Also, $\angle AOB={60}^{\circ }$ (given)

$⟹OA=OB=AB=6ft$,
which makes AOB an equilateral triangle.

From the figure, we can see that OD is perpendicular to AB ($\angle D={90}^{\circ }$).

$AD=DB=\frac{AB}{2}=\frac{6}{2}=3$

From $\Delta ODC,$

AD = 3 ft and OA = 6 ft

By using the Pythagorean theorem, we have:

$O{A}^2=A{D}^2+O{D}^2$

$⟹6^2=3^3+O{D}^2$

$⟹36=9+O{D}^2$

$⟹36–9=O{D}^2$

$⟹O{D}^2=27$

$⟹OD=\sqrt{27}$


Let us consider $\Delta ADC$.

$\text{AD = 3 ft}$
$\text{and DC = (DO + OC)}=(\sqrt{27}+6)ft$
$=3\sqrt{3}+6ft$

By using the Pythagorean theorem, we have:

$A{C}^2=A{D}^2+D{C}^2$

$⟹A{C}^2=3^2+(6+3\sqrt{3}{)}^2$

$⟹A{C}^2=9+36+27+36\sqrt{3}$

$⟹A{C}^2=72+36\sqrt{3}$

$⟹AC=\sqrt{72+36\sqrt{3}}=6\sqrt{2+\sqrt{3}}$

The length of AC is $6\sqrt{2+\sqrt{3}}ft$.