In \(\triangle ABC\), point \(P \in AC\) such that \(AP:PC = 3:2\). If the area of \(\triangle BPC = 14 \text{ m}^2\), find the area of \(\triangle ABP\) and the area of \(\triangle ABC\).

The area of triangle ABP is 21 m² and the area of triangle ABC is 35 m², computed using the ratios provided and the given area of triangle BPC.

Explanation:

The ratio of AP to PC is given as 3:2, meaning that line AC is divided into 5 equal parts where AP is 3 parts and PC is 2 parts. The areas of triangles ABP and BPC are proportional to the lengths of line segments AP and PC respectively. As the area of triangle BPC is given as 14 m², we can divide this by 2 (the number of parts that make up line segment PC) to find the area of one part, which is 7 m². Thus, the area of triangle ABP is 7 m² * 3 (the number of parts that make up line segment AP), which equals 21 m². Add together the areas of the two triangles ABP and BPC (21 m² + 14 m²) to find the area of whole triangle ABC, which equals 35 m².

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