Solution:
Let the triangle’s vertices be A (0, -1), B (2, 1), and C (3, 1). (0, 3).
Let D, E, and F be the midpoints of the triangle’s sides.
D, E, and F coordinates are given by
D .= (0+2/2, -1+1/2 ) = (1, 0)
E = ( 0+0/2, -1+3/2 ) = (0, 1)
F = ( 0+2/2, 3+1/2 ) = (1, 2)
The area of a triangle
![\[=~1/2\text{ }\times \text{ }[{{x}_{1}}({{y}_{2}}~\text{ }{{y}_{3}})\text{ }+\text{ }{{x}_{2}}({{y}_{3}}~\text{ }{{y}_{1}})\text{ }+\text{ }{{x}_{3}}({{y}_{1}}~\text{ }{{y}_{2}})]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-74a46eca3fe5cf31dbfa469ef08096a0_l3.png "Rendered by QuickLaTeX.com")
The area of ΔDEF
![\[=\text{ }1/2\text{ }\left\{ 1\left( 2-1 \right)\text{ }+\text{ }1\left( 1-0 \right)\text{ }+\text{ }0\left( 0-2 \right) \right\}\text{ }=\text{ }1/2\text{ }\left( 1+1 \right)\text{ }=\text{ }1\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-56849b40695b29f1472ab801a0a1d3ec_l3.png "Rendered by QuickLaTeX.com")
Area of ΔDEF is 1 square units
The area of ΔABC
![\[=\text{ }1/2\text{ }\left[ 0\left( 1-3 \right)\text{ }+\text{ }2\left\{ 3-\left( -1 \right) \right\}\text{ }+\text{ }0\left( -1-1 \right) \right]\text{ }=\text{ }1/2\text{ }\left\{ 8 \right\}\text{ }=\text{ }4\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-78217850049327008df5a1ddf4d060ee_l3.png "Rendered by QuickLaTeX.com")
Area of ΔABC is 4 square units
As a result, 1:4 is the required ratio.