1.42. The sides of a triangle are 13, 14, and 15 cm. Determine the areas of the triangles into which the given triangle is divided by its medians.
1.42. First, let's prove that all the triangles mentioned in the problem are equal in area. We have \( S_{\triangle A O M} = S_{\triangle C O M} \) (Fig. 1.34), since

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-433.jpg?height=475&width=483&top_left_y=151&top_left_x=101)

Fig. 1.34

these triangles have the same heights and the same bases, equal to \(\frac{1}{2} A C\). Similarly, \( S_{\triangle A O N} = S_{\triangle B O N} \) and \( S_{\triangle B O K} = S_{\triangle C O K} \). But \( S_{\triangle B O K} = S_{\triangle B O N} \), since \( S_{\triangle B O K} = S_{\triangle D O K} + S_{\triangle B K D} \), \( S_{\triangle B O N} = S_{\triangle D O N} + S_{\triangle B N D} \), and \( S_{\triangle D O K} = S_{\triangle D O N} \) and \( S_{\triangle B K D} = S_{\triangle B N D} \) (the bases \( K D \) and \( N D \) are equal, as well as the heights dropped from them). Thus, the area of each triangle is \(\frac{1}{6} S_{\triangle A B C}\). Now, using Heron's formula, we find \( S_{\triangle A B C} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = 84 \, \text{cm}^2 \), from which \( S_{\triangle A O M} = 14 \, \text{cm}^2 \).

Answer: \( 14 \, \text{cm}^2 \).

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-433.jpg?height=396&width=513&top_left_y=1108&top_left_x=55)

Fig. 1.35