Chapter 14.3, Problem 18PSC

a.

To determine

To prove that the altitude of triangle PBC is one third of the altitude of triangle PBC.

Explanation of Solution

Given:

The given figure is-

  

Calculation:

Draw the altitudes PQ and AR.

  

Proof:

  $△ABC$ has median AM with centroid P.

  $\begin{array}{l}AP=\frac{2}{3}AM\\ AM-PM=\frac{2}{3}AM\\ \frac{PM}{AM}=\frac{2}{3}AM\\ \angle MQP\cong \angle MRA\\ \angle PMQ\cong \angle AMR\\ △PMQ\sim △AMR\\ \Rightarrow \frac{PQ}{AR}=\frac{PM}{AM}\\ \Rightarrow \frac{PQ}{AR}=\frac{1}{3}\\ \Rightarrow PQ=\frac{1}{3}AR\end{array}$

b.

To determine

To find the ratio of the area of triangle PBC to the area of triangle ABC.

Answer to Problem 18PSC

The area of trianglePBCis one third of the area of triangle ABC.

Explanation of Solution

Given:

The given figure is-

  

Calculation:

  $\frac{ar(△PBC)}{ar(△ABC)}=\frac{\frac{1}{2}PQ\cdot BC}{\frac{1}{2}AR\cdot BC}=\frac{\frac{1}{3}AR}{AR}=\frac{1}{3}$