Question

Find the area of the triangle formed by joining the mid points of the sides of the triangle, whose vertices are $(0,1);(2,1)$ and $(0,3)$. Find the ratio of this area to the area of the given triangle

Answer 1

Let $A(0,1),B(2,1),C(0,3)$ be the vertices of the triangle and $D,E$ and $F$ are mid-points of $AB,BC$ and $CD$ respectively.

Then, by midpoint formula

$D\equiv (\frac{0+2}{2},\frac{1+1}{2}),E\equiv (\frac{2+0}{2},\frac{1+3}{2}),$ and $F\equiv (\frac{0+0}{2},\frac{1+3}{2})$

$D\equiv (1,1,)E\equiv (1,2),$ and $F\equiv (0,2)$

Area of the$△DEF$ $=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$=\frac{1}{2}\times |1\times (2-2)+1\times (2-1)+0\times (1-2)|$

$=\frac{1}{2}\times |0+1+0|$

$=\frac{1}{2}$ square units.

Area of the $△ABC$$=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

$=\frac{1}{2}\times |0\times (1-3)+2\times (3-1)+0\times (1-1)|$

$=\frac{1}{2}\times |0+4+0|$

$=2$ square units.

$\frac{\text{Area of the}△DEF}{\text{Area of the}△ABC}=\frac{1}{2\times 2}=\frac{1}{4}$.