Question

In ΔABC ,P is a point on side BC such that BP = 4cm and PC = 7 cm. A(ΔAPC):A(ΔABC)=

Answer 1

$\textbf{Given:}$

$\text{In \triangle ABC, BP=4 cm and PC=7 cm}$

$\textbf{To find:}$

$\text{Area( \triangle APC):Area( \triangle ABC)}$

$\textbf{Solution:}$

$\text{Draw AD \perp BC}$

$\textbf{Area of \triangle APC}$

$=\dfrac{1}{2}{\times}b{\times}h$

$=\dfrac{1}{2}{\times}BP{\times}AD$

$=\dfrac{1}{2}{\times}4{\times}AD$

$\textbf{Area of \triangle ABC}$

$=\dfrac{1}{2}{\times}BC{\times}AD$

$=\dfrac{1}{2}{\times}11{\times}AD$

$\text{Now}$

$\dfrac{\text{Area( \triangle APC)}}{\text{Area( \triangle ABC)}}$

$=\dfrac{\dfrac{1}{2}{\times}4{\times}AD}{\dfrac{1}{2}{\times}11{\times}AD}$

$=\dfrac{4}{11}$

$\implies\textbf{Area( \triangle APC):Area( \triangle ABC)}\bf=4:11$