# Area of a triangle

8 posts / 0 new
Engr Jaydee
Area of a triangle

In square $ABCD$, $E$ is the midpoint of side $\overline{AB}$ and $F$ is a point of side $\overline{AD}$ such that $F$ is twice as near from $D$ as from $A$. $G$ is the intersection of the line segments $\overline{DE}$ and $\overline{CF}$. If $AB = 1\text{ cm}$, find the area of $\triangle CDG$.

There are several ways to solve this problem, please show your solutions.

alidoandrei28

I used analytic geometry to get the equation of the lines DE and CF. Thus,

DE: (0,0) (0.5,0.5)
equation: y = x

FC: (0,1/3) (1,0)
equation: y = 1/3 - (1/3)x

Then, I get the point of intersection; (0.25,0.25)

After that, I used integral calculus to solve for the area using horizontal strip.

### Plain text

• No HTML tags allowed.
• Lines and paragraphs break automatically.