Two adjacent circles

The line through the centers $K$ and $L$ of circles is parallel to the exterior tangent line and the point of intersection of the interior common tangent lines lies on $KL$ . Hence, the altitude $CH$ is congruent to the radius $KD$ , thus $CH=10$ . Consequently, for the side $s$ of the equilateral triangle we have $s=CH\cdot \csc 60{}^\circ \Rightarrow s=\frac{20}{\sqrt{3}}$ The area of the triangle is $\frac{{{s}^{2}}\sqrt{3}}{4}=\frac{100}{\sqrt{3}}$ For the answer, $a=100$ , $b=3$ , thus, $ab=\boxed{300}$ .

Each circle is an excircle of the equilateral triangle, which has a radius of $r_a = \sqrt{\cfrac{s(s - b)(s - c)}{s - a}}$ .

If each side of the equilateral triangle is $x$ , then $10 = \sqrt{\cfrac{\frac{3}{2}x(\frac{3}{2}x - x)(\frac{3}{2}x - x)}{\frac{3}{2}x - x}}$ , which solves to $x^2 = \cfrac{400}{3}$ .

Therefore, the area of the triangle is $A = \cfrac{\sqrt{3}}{4}x^2 = \cfrac{\sqrt{3}}{4}\cdot \cfrac{400}{3} = \cfrac{100}{\sqrt{3}}$ , so $a = 100$ , $b = 3$ , and $ab = \boxed{300}$ .

The radius is equal to the triangle's height if you rotate the 'hour hand' to 4.30 or 7.30. With 10 being its height, half of the base must be 1/√3 that, so area equal 10² x (1/√3) = 100/√3. Answer is 100 x 3 = 300.