1.40. Point $M$ lies inside an equilateral triangle $A B C$. Calculate the area of this triangle, given that $A M=$ $=B M=2 \text{ cm}, \text{a} C M=1 \text{cm}$.
1.40. Let the side of the triangle be denoted by $a$ and draw $M D \perp A B$ (Fig. 1.32). Since $A M = B M$, the points $C, M$, and $D$ lie on the same line - 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 $a^{2} = 4(4 - M D^{2})$ and we obtain the quadratic equation

$$
(1 + M D)^{2} = 3(4 - M D^{2}), \text{ or } 4 M D^{2} + 2 M D - 11 = 0
$$

from which $M D = \frac{3 \sqrt{5} - 1}{4}$ (the second root does not satisfy the condition). Next, we find

$$
a^{2} = 4(4 - M D^{2}) = 16 - \frac{46 - 6 \sqrt{5}}{4} = \frac{9 + 3 \sqrt{5}}{2}
$$

and, therefore,

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

Answer: $\approx 3.4 \mathrm{~cm}^{2}$.