A logo, as shown below, has a circle with centre O and radius 2 cm, an equilateral triangle ABC of side 5 cm and an isosceles triangle DEF with OD = OF. DEF is a tangent to the circle at E and angle DOF = 2.4 radians. Half of the circle lies inside triangle ABC. Calculate the area of the shaded region. Answer ___ cm^2

1. Find the area of the equilateral triangle and subtract the semicircle inside it.

Area of equilateral triangle $ABC$
$=\frac{\sqrt3}{4}(5)^2=\frac{25\sqrt3}{4}$

Since half of the circle lies inside triangle $ABC$,
$\text{area of semicircle}=\frac12\pi(2)^2=2\pi$

So the shaded area in the upper triangle is
$\frac{25\sqrt3}{4}-2\pi.$

2. Find the area of the lower isosceles triangle using the tangent-radius perpendicular relationship.

The tangent at $E$ is perpendicular to the radius $OE$, so $OE\perp DF$.

Since $OD=OF$, $OE$ bisects $\angle DOF$.
$\angle DOE=\frac{2.4}{2}=1.2\text{ radians}$

In right triangle $ODE$,
$\tan 1.2=\frac{DE}{OE}$
$DE=2\tan 1.2$

Hence
$DF=2DE=4\tan 1.2$

Area of lower isosceles triangle
$=\frac12\times DF\times OE$
$=\frac12\times 4\tan 1.2\times 2$
$=4\tan 1.2.$

3. Subtract the circular sector inside the lower triangle and combine both shaded parts.

Area of sector inside the lower triangle:
$\frac12r^2\theta=\frac12(2)^2(2.4)=4.8$

Total shaded area
$=\left(\frac{25\sqrt3}{4}-2\pi\right)+\left(4\tan1.2-4.8\right)$
$=10.0309\ldots$

$\approx 10.0\text{ cm}^2$