Given that:
The triangles ABC and A1B1C1 are similar.
The ratio of two sides is 6:9
The sum of their areas is equal 52 (Sabc + Sa1b1c1 = 52)
Find the two areas
Sabc = ? Sa1b1c1 = ?
I tried to create a system, without any luck. I have no idea how to proceed any help is appriciated
Notice, the ratio of the areas of the similar triangle $\triangle ABC$ & $\triangle A_1B_1C_1$ is equal to the ratio of squares of their corresponding sides $$\frac{S_{\triangle ABC}}{S_{\triangle A_1B_1C_1}}=\left(\frac{6}{9}\right)^2=\frac{4}{9}$$ $$\implies S_{\triangle ABC}=\frac{4}{9}S_{\triangle A_1B_1C_1}$$ given that $$S_{\triangle ABC}+S_{\triangle A_1B_1C_1}=52$$, substituting the value $$\frac{4}{9}S_{\triangle A_1B_1C_1}+S_{\triangle A_1B_1C_1}=52$$ $$\color{red}{S_{\triangle A_1B_1C_1}}=\color{blue}{36}$$ & $$\color{red}{S_{\triangle ABC}}=52-36=\color{blue}{16}$$