Question

Find the area of triangle formed by joining the mid point sides of the $△ABC$ whose vertices are A (0,-1) B (2,1) C(0,3) Find the ratio of this area to the area of $△ABC$.

Answer 1

Let $P,Q$ and $R$ be the mid-points on the sides $AB,BC$ and $CA$ respectively.

Therefore, by mid-point formula,

Coordinate of $P=(\frac{0+2}{2},\frac{-1+1}{2})=(1,0)$

Coordinate of $Q=(\frac{0+2}{2},\frac{3+1}{2})=(1,2)$

Coordinate of $R=(\frac{0+0}{2},\frac{-1+3}{2})=(0,1)$

Now,

$ar\left(△ABC\right)=\frac{1}{2}\times \left|\begin{array}{ccc}0& -1& 1\\ 2& 1& 1\\ 0& 3& 1\end{array}\right|$

$\Rightarrow ar\left(△ABC\right)=\frac{1}{2}\times [0-2(-1-3)+0]=4\text{sq. units}$

Again,

$ar\left(△PQR\right)=\frac{1}{2}\times \left|\begin{array}{ccc}1& 0& 1\\ 1& 2& 1\\ 0& 1& 1\end{array}\right|$

$\Rightarrow ar\left(△PQr\right)=\frac{1}{2}\times [1(2-1)-0+1(1-0)]=1\text{sq. units}$

Therefore,

$\frac{ar\left(△PQR\right)}{ar\left(△ABC\right)}=\frac{1}{4}$

Thus area of triangle formed by joining the mid point sides of the $△ABC$ is $1$ sq. unit and the ratio between the triangle formed by joining the mid point sides of the $△ABC$ and area of $△ABC$ is $1:4$.