Saved Bookmarks
| 1. |
Find the area of the triangle formed by joining the mid point of the sides |
| Answer» Let A → (0, -1), B → (2, 1) and C→ (0, 3) be the vertices of the triangle ABC. Let D, E and F be the mid-points of sides BC, CA and AB respectively. Then,\xa0{tex}D \\to \\left( {\\frac{{2 + 0}}{2},\\frac{{1 + 3}}{2}} \\right){/tex}{tex} \\Rightarrow D \\to \\left( {1,2} \\right){/tex}{tex}E \\to \\left\\{ {\\frac{{0 + 0}}{2},\\frac{{3 + ( - 1)}}{2}} \\right\\}{/tex}{tex}\\Rightarrow E \\to (0,1){/tex}{tex}F \\to \\left\\{ {\\frac{{2 + 0}}{2},\\frac{{1 + ( - 1)}}{2}} \\right\\}{/tex}{tex}\\Rightarrow F \\to (1,0){/tex}{tex}\\therefore{/tex} Area of the triangle DEF{tex}= \\frac{1}{2}{/tex}[1(1 -\xa00) + 0(0 -\xa02) + 1(2 - 1)]{tex}= \\frac{1}{2}{/tex}[1 + 0 + 1]= 1 square unit.Again, area of the triangle ABC{tex}= \\frac{1}{2}{/tex}[0(1 -\xa03) + 2 {3 -\xa0(-1)} + 0(-1 -1)]= 4 square units{tex}\\therefore{/tex} Ratio of the area of the triangle formed to the area of the given triangle = 1 : 4 | |