Given:
[tex]\triangle ABC\sim\triangle PRT[/tex]
As the triangles are similar, the corresponding sides are proportional
[tex]\frac{AC}{PT}=\frac{BC}{RT}[/tex]
Given: AC = 15, BC = 12 , PT = 6
[tex]\begin{gathered} \frac{15}{6}=\frac{12}{RT} \\ \\ RT=\frac{6\cdot12}{15}=4.8 \end{gathered}[/tex]
We will find the area of the triangle ABC using the following ratio:
[tex]\frac{Area\triangle\text{ABC}}{Area\triangle\text{PRT}}=\frac{0.5\cdot AC\cdot BC\cdot\sin C}{0.5\cdot PT\cdot RT\cdot\sin T}[/tex]
The angle C is congruent to the angle T
So, substitute with the given values:
[tex]\begin{gathered} \frac{Area\triangle ABC}{24}=\frac{15\cdot12}{6\cdot4.8}=\frac{180}{28.8}=6.25 \\ \\ Area\triangle\text{ABC}=24\cdot6.25=150 \end{gathered}[/tex]
So, the answer will be:
The area of ΔABC = 150 mm²
