129. In a parallelogram, there are two circles of radius 1, each touching the other and three sides of the parallelogram. It is also known that one of the segments of a side of the parallelogram from a vertex to the point of tangency is $\sqrt{3}$. Find the area of the parallelogram.
129. $\frac{4}{3}(2 \sqrt{\overline{3}}+3)$. 130. $\frac{2 R^{2} \sin ^{3} \alpha \sin \beta}{\sin (\alpha+\beta)}$. 131. $\frac{3 \sqrt{\overline{3}}(\sqrt{\overline{1}}-1)}{32 \pi}$. 132. $1,1, \quad 133 . \frac{a^{2}}{16 R}$. 134. $90^{\circ}$. 135. $30^{\circ}$. 136. $\frac{a \sqrt{ } \overline{7}}{4}$. 137. $\frac{R(3-2 \sqrt{2})}{3}, 138.4 \sqrt{\frac{1-\cos \beta}{3-\cos \beta}}$. 139. $\frac{a b \operatorname{tg} \alpha}{\sqrt{a^{2} \operatorname{tg}^{2} \alpha+(b-a)^{2}}}$. (In triangle $O N P$, segments $K P$ and $N M$ are altitudes, so $O A$ is also an altitude.)