Question

Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are $(2,2)$, $(4,4)$ and $(2,6)$.

• A. 5
• B. 3
• C. 1
• D. 0

Answer 1

Answer: (C)

1

Sol. Let $A(2,2)$, $B(4,4)$ and $C(2,6)$ be the vertices of a given $\Delta$ ABC.
Let D, E and F be the midpoints of AB, BC and CA respectively.
Then, the coordinates of D, E, and F are given as
$D\left(\begin{array}{c}\frac{2+4}{2},\frac{2+4}{2}\end{array}\right)$, $E\left(\begin{array}{c}\frac{4+2}{2},\frac{4+6}{2}\end{array}\right)$
and $F\left(\begin{array}{c}\frac{2+2}{2},\frac{2+6}{2}\end{array}\right)$
i.e., $D(3,3)$, $E(3,5)$ and $F(2,4)$
$\therefore (x_1=3,y_1=3),(x_2=3,y_2=5)and(x_3=2,y_3=4)$
Area of $\Delta DEF=\frac{1}{2}\cdot |x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$
$=\frac{1}{2}\cdot |3(5-4)+3(4-3)+2(3-5)|$
$=\frac{1}{2}\cdot |3\times 1+3\times 1+2\times (-2)|=\frac{1}{2}\times |3+3-4|$
$=\left(\begin{array}{c}\frac{1}{2}\times 2\end{array}\right)=1$ sq.units