Question

A (6,1), B(8,2), C (9,4) are the three vertices of a parallelogram ABCD. if E is the mid point of BD, find the area of /ABE.

Answer 1

ans is 4.25 square units ..pls mark it as branliest..hope this helps u..

Answer 2

Area of triangle ADE = $\bold{\frac{3}{4}}$ sqr unit

Solution:

As the question states that, A (6, 1), B (8, 2) and C (9, 4) are three vertices of a parallelogram ABCD. Now let us take fourth vertex of parallelogram as (x, y).  

We all know that, diagonals of parallelogram intersect each other.

Now as we know that E bisects DC we get

To find the coordinates of D:

Midpoint of BD = Midpoint of AC

$\begin{array}{l}{\frac{8+x}{2}, \frac{2+y}{2}=\frac{6+9}{2}, \frac{1+4}{2}} \\ \\{\frac{8+x}{2}, \frac{2+y}{2}=\frac{15}{2}, \frac{5}{2}} \\ \\{\frac{8+x}{2}=\frac{15}{2} ; \frac{2+y}{2}=\frac{5}{2}}\end{array}$

x=7,y=3  

Therefore, the coordinates of vertex D is (7, 3)

Midpoint of DC i.e. coordinates of E is $\frac{9+7}{2},\frac{3+4}{2}=\left(8, \frac{7}{2}\right)$

E co-ordinates are $\left(8, \frac{7}{2}\right)$

Area of triangle ADE = $\frac{1}{2}\left(6\left(3-\frac{7}{2}\right)+7\left(\frac{7}{2}-1\right)+8(1-3)\right)=\left|-\frac{3}{4}\right|=\frac{3}{4}$