Find the area of $\triangle \mathrm{ABC}$ with vertices $\mathrm{A}(0,-1), \mathrm{B}(2,1)$ and $C(0,3)$. Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is $4: 1$

Home » Find the area of with vertices and . Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is

Find the area of $\triangle \mathrm{ABC}$ with vertices $\mathrm{A}(0,-1), \mathrm{B}(2,1)$ and $C(0,3)$ . Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is $4: 1$

Find the area of $\triangle \mathrm{ABC}$ with vertices $\mathrm{A}(0,-1), \mathrm{B}(2,1)$ and $C(0,3)$ . Also, find the area of the triangle formed by joining the midpoints of its sides. Show that the ratio of the areas of two triangles is $4: 1$

Let $A\left(x_{1}=0, y_{1}=-1\right), B\left(x_{2}=2, y_{2}=1\right)$ and $C\left(x_{3}=0, y_{3}=3\right)$ be the given points.

=> $\text { Area }(\Delta \mathrm{ABC})=\frac{1}{2}\left[\mathrm{x}_{1}\left(\mathrm{y}_{2}-\mathrm{y}_{3}\right)+\mathrm{x}_{2}\left(\mathrm{y}_{3}-\mathrm{y}_{1}\right)+\mathrm{x}_{3}\left(\mathrm{y}_{1}-\mathrm{y}_{2}\right)\right]$

$=\frac{1}{2}[0(1-3)+2(3+1)+0(-1-1)]$

$=\frac{1}{2} \times 8=4$ sq. units

=> the area of the triangle $\triangle \mathrm{ABC}$ is 4 sq. units

Let $D\left(a_{1}, b_{1}\right), E\left(a_{2}, b_{2}\right)$ and $F\left(a_{3}, b_{3}\right)$ be the midpoints of $A B, B C$ and $A C$ respectively Then