Question

P is any point on base BC of $△ABC$ and D is teh mid point of BC. De is drawn parallel to PA to meet AC at E. If ar $\left(△ABC\right)=12c{m}^2,$ then find area of $△EPC$

Answer 1

P is any point on base of $△ABC$
D is mid point of BC
DE || PA drawn which meet AC at E
$ar\left(△ABC\right)=12c{m}^2$
AD and PE are joined

$\because D$ is mid point of BC
$\therefore AD$ is median
$\therefore ar\left(△ABD\right)=ar\left(ACD\right)$
$=\frac{1}{2}\left(△ABC\right)=\frac{1}{2}\times 12=6c{m}^2$
$\because △PEDand△ADE$ to both sides,
$\therefore ar\left(△PED\right)+ar\left(△DCE\right)=ar\left(△ADE\right)+ar\left(△DCE\right)ar\left(△EPC\right)=ar\left(△ACE\right)\Rightarrow ar\left(△EPC\right)=ar\left(△ABD\right)=6c{m}^2\therefore ar\left(△EPC\right)=6c{m}^2$