In the similar triangles
[tex]\begin{gathered} \frac{P_1}{P_2}=\frac{s_1}{s_2} \\ \frac{A_1}{A_2}=(\frac{s_1}{s_2})^2 \end{gathered}[/tex]
Since the ratio between the perimeters of 2 similar triangles, ABC and A1B1C1 is 4/5
By using the first rule above, the ratio between their sides is 4/5
[tex]\begin{gathered} \frac{P_{ABC}}{P_{A_1B_1C_1}}=\frac{4}{5} \\ \frac{s_{ABC}}{s_{A_1B_1C_1}}=\frac{4}{5} \end{gathered}[/tex]
Since the sum of their area is 410 cm^2, then
[tex]\begin{gathered} A_{ABC}+A_{A_1B_1C_1}=410 \\ A_{A_1B_1C_1}=410=A_{ABC} \end{gathered}[/tex]
By using the 2nd rule above
[tex]\frac{A_{ABC}}{A_{A_1B_1C_1}}=(\frac{4}{5})^2=\frac{16}{25}[/tex]
Substitute Area of triangle A1B1C1 by 410 - ABC
[tex]\frac{A_{ABC}}{410-A_{ABC}}=\frac{16}{25}[/tex]
By using the cross-multiplication
[tex]\begin{gathered} 25\times A_{ABC}=16\times(410-A_{ABC}) \\ 25A_{ABC}=6560-16A_{ABC} \end{gathered}[/tex]
Add 16A (ABC) to both sides
[tex]\begin{gathered} 25A_{ABC}+16A_{ABC}=6560-16A_{ABC}+16A_{ABC} \\ 41A_{ABC}=6560 \end{gathered}[/tex]
Divide both sides by 41
[tex]\begin{gathered} \frac{41A_{ABC}}{41}=\frac{6560}{41} \\ A_{ABC}=160cm^2 \end{gathered}[/tex]
Subtract it from 410 to find the area of triangle A1B1C1
[tex]\begin{gathered} A_{A_1B_1C_1}=410-160 \\ A_{A_1B_1C_1}=250cm^2 \end{gathered}[/tex]
The areas of the 2 triangles are 160 cm^2 and 250 cm^2
