The diagram shows a regular 12-sided polygon inscribed in a circle and a right-angled triangle PQR with length $QR = 6$ cm. It is given that $\tan \angle QPR = \dfrac{3}{4}$. The ratio of the area of triangle $ABC$ : triangle $PQR$ is 1 : 6. Find the radius of the circle.

1. Find the length $PR$ using the tangent ratio in triangle $PQR$.

In right-angled triangle $PQR$,

$\tan \angle QPR=\frac{QR}{PR}$

Given $\tan \angle QPR=\frac34$ and $QR=6$,

$\frac{6}{PR}=\frac34$

$PR=8\text{ cm}$

2. Find the area of triangle $PQR$.

Since $PQR$ is right-angled at $R$,

$\text{Area of }\triangle PQR=\frac12(6)(8)=24\text{ cm}^2$

3. Use the given area ratio to find the area of triangle $ABC$.

The ratio of the areas is

$\triangle ABC : \triangle PQR=1:6$

So

$\text{Area of }\triangle ABC=\frac16\times 24=4\text{ cm}^2$

4. Use the area formula for triangle $ABC$ to find the radius of the circle.

Triangle $ABC$ is formed by two radii and one side of the regular 12-sided polygon, so

$AB=AC=r$

The central angle is

$\angle BAC=\frac{360^\circ}{12}=30^\circ$

Using $\text{Area}=\frac12 ab\sin C$,

$4=\frac12(r)(r)\sin 30^\circ$

$4=\frac12r^2\left(\frac12\right)$

$4=\frac14r^2$

$r^2=16$

$r=4\text{ cm}$