Circles Inscribed In a Triangle

Zoom into the last row:

As we can see, from the center of the first circle to the center of the last circle, we can form a rectangle (or 8 squares). This is because the side of the triangle is tangent to the circle, and the angle between the side of the triangle and the radius of the circles is $90^{\circ}$ .

Now, a unit circle is a circle with a radius of $1$ unit, therefore the rectangle's length is $8$ units.

The only thing left is to find the end parts of the side of the triangle.

Two lines, which are tangent to the circle, meet at the triangle's vertex. This implies that there are two triangles with the same length and angles formed there (as illustrated in the diagram).

From the diagram,

$\tan 30^{\circ} = \dfrac{1}{a}\\ \dfrac{1}{\sqrt{3}} = \dfrac{1}{a}$

$a=\sqrt{3}$ units

The side of the triangle $= 8 + 2\sqrt{3}$ units.

We can then use the formula to find the area of the triangle:

Area of $\triangle$

$=\dfrac{1}{2}(8 + 2\sqrt{3})^2\sin 60^{\circ}\\ =\dfrac{1}{2}(64 + 32\sqrt{3} + 4(3))\left(\dfrac{\sqrt{3}}{2}\right)\\ =\dfrac{\sqrt{3}(76 + 32\sqrt{3})}{4}\\ =\sqrt{3}(19+8\sqrt{3})\\ =19\sqrt{3} + 8(3)\\ =\boxed{19\sqrt{3} + 24 \text{ units}^2}$

Non-trig solution:

We can solve this just using 30-60-90 right triangle ratios.

The area of a triangle is $\dfrac{1}{2}bh$ . In this case, we have $\dfrac{\dfrac{1}{2}(8\sqrt{3}+6)(8+2\sqrt{3})}{2}=\boxed{19\sqrt{3}+24}$