\[2x\]

. Draw a line from the centre of the circle - which is also the centre of the triangles - to a vertex of the triangle as shown.This line will bisect the angle at the vertex, producing an angle of 30 degrees. From the centre of the circle draw a line down to meet the base of the triangle at right angles. We can use simple trigonometry to find the radius of the circle.

\[cos 30 = \frac{x}{r} \rightarrow r= \frac{x}{cos 30}= \frac{x}{2/ \sqrt{3}} = \frac{2x}{ \sqrt{3}}\]

Now drop a line from the centre of the circle to the centre of the base of the large triangle to form a right angle triangle as show. We use simple trigonometry to find the base of the;large triangle.

\[tan 30 = \frac{r}{half \: the \: base} \rightarrow base =\frac{r}{1/2 tan 30}= \frac{2x / \sqrt{3}}{1/2 \times 1 / \sqrt{3}}= 4x\]

Then sides of the big triangle are

\[\frac{4x}{2x} = 2\]

times the sides of the small triangle and the are of the big triangle is the square of this ( equals 4) times the area of the small triangle.