10.354. Point $M$ lies inside an equilateral triangle $A B C$. Calculate the area of this triangle, given that $A M=B M=$ $=2 \text{ cm}, \text{a} C M=1 \text{ cm}$.
Solution.

Let the side of the triangle be $a$ and draw $M D \perp A B$ (Fig. 10.135). Since $A M=B M$, points $C, M, D$ lie on the height $C D$. In $\triangle A C D$ and $\triangle A M D$, we have $(1+M D)^{2}=a^{2}-\frac{a^{2}}{4}, M D^{2}=4-\frac{a^{2}}{4}$. Then $a^{2}=4\left(4-M D^{2}\right)$ and we get the quadratic equation $(1+M D)^{2}=$ $=3\left(4-M D^{2}\right)$ or $4 M D^{2}+2 M D-11=0$, from which $M D=\frac{3 \sqrt{5}-1}{4}$ (the second root is not suitable). Next, we find $a^{2}=4\left(4-M D^{2}\right)=16-\frac{46-6 \sqrt{5}}{4}=$ $=\frac{9+3 \sqrt{5}}{2}$ and, therefore,

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0812.jpg?height=518&width=567&top_left_y=76&top_left_x=112)

Fig. 10.136 .1

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0812.jpg?height=526&width=586&top_left_y=68&top_left_x=674)

Fig. 10.136 .2

$$
S=\frac{a^{2} \sqrt{3}}{4}=\frac{(9+3 \sqrt{5}) \sqrt{3}}{2}=\frac{9 \sqrt{3}+3 \sqrt{15}}{8} \approx 3.4 \text{ cm}^{2}
$$

Answer: $\frac{9 \sqrt{3}+3 \sqrt{15}}{8} \approx 3.4 \text{ cm}^{2}$.