Question

15
in \\( \triangle a b c \\), point \\( p \in \overline{a b} \\) such that \\( a p: b p=1: 3 \\), and point \\( m \\) is the midpoint of segment \\( \overline{c p} \\). find the area of \\( \triangle a b c \\) if the area of \\( \triangle b m p \\) is equal to \\( 21 \mathrm{~m}^{2} \\).
answer: \\( a_{a b c}=\quad \mathrm{m}^{2} \\)

Explanation:

Step1: Find the area of $\triangle BPC$

Since $M$ is the mid - point of $CP$, then $S_{\triangle BMP}=S_{\triangle BMC}$ (triangles with the same base and height).
Given $S_{\triangle BMP} = 21\ m^{2}$, so $S_{\triangle BPC}=S_{\triangle BMP}+S_{\triangle BMC}=2\times21 = 42\ m^{2}$.

Step2: Find the area of $\triangle ABC$

Because $AP:BP = 1:3$, then $S_{\triangle APC}:S_{\triangle BPC}=1:3$ (triangles with the same height, the ratio of areas is equal to the ratio of bases).
Let $S_{\triangle APC}=x$, then $\frac{x}{S_{\triangle BPC}}=\frac{AP}{BP}=\frac{1}{3}$. Since $S_{\triangle BPC} = 42\ m^{2}$, then $x = 14\ m^{2}$.
And $S_{\triangle ABC}=S_{\triangle APC}+S_{\triangle BPC}$.

Answer:

$S_{\triangle ABC}=14 + 42=56\ m^{2}$