Geometry is fun

Let the triangle be $\triangle ABC$ , $A$ being at the top, $BC$ the base, and the center of the unit radius incircle be $O$ . Then $AO$ , $BO$ and $CO$ divide internally the $\angle A$ , $\angle B$ and $\angle C$ respectively. Let the incircle touches $BC$ at $M$ . Due to symmetry, $M$ is the midpoint of $BC$ .

The area of the triangle $= \dfrac {1} {2} \times BC \times AM$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad = \dfrac {1}{2} \left( 2\times BM \right) \left( AO+OM \right)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad = \dfrac {1}{2} \left( 2 \times \dfrac {1} {\tan {30^\circ} } \right) \left( 1+\dfrac {1}{\sin{30^\circ} } \right)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad = \dfrac {1}{2} \left( \dfrac {2} {\frac {1}{\sqrt{3}} } \right) \left( 1+\dfrac {1}{\frac {1}{2} } \right)$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad = \dfrac {1}{2} \left( 2\sqrt{3} \right) \left( 1+2 \right) = 3\sqrt{3} \approx \boxed {5.196}$

You can divide the triangle into 6 congruent 30-60-90 triangles. 1 being the side opposite to the 30 degree angle. Hence, the base of the smaller triangle will be sqrt(3). Using this, we can find the area of the smaller triangle, [1xsqrt(3)]/2=0.866~. We multiply the area by 6, since we divided the triangle into 6 congruent parts, getting the answer 6x0.866=5.196.

If we add the top of the triangle with the center of the circle we will get another triangle there.In this case we know that the hypotenuse is twice the radius . The radius of the circle is 1 so, the hypotenuse is $2\times 1$ now use the theorem of the pithagoras $x^2+1^2=2^2$ ;[Base=x] so, x= $\sqrt{3}$ now, the length of equilateral triangle is $2 \times x$ so, $2\times\sqrt{3}$ =3.46) here a=3.46 ;[a is the length of the equilateral triangle] now, we know the formula of the area of equilateral triangle

by calculating we will get the area of the equilateral triangle so, the answer is 5.2