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$ABC$ is a right-angled triangle at $A$. $AB=3cm$, $BC=5cm$, $CD=1cm$. If $BE=EC$, then what is the area of the shaded region?

Diagram_Inage

I could solve some parts of this question but got stuck and was able to find the following:

$AC=4cm$, $AD=3cm$

I also did two constructions. They were:

Drawing a line $DG \parallel AB$ and $HE \parallel AB$. $HE$ intersects $BD$ at $O$. Through these and using similarity, I found the area of $\Delta DCG=\frac{3}{8} cm^2$ and quadrilateral $DGEO=\frac{5}{8} cm^2$.

This question was on an olympiad website. I am unable to find the area of $\Delta OFE$. Can someone please help me out? It would be very helpful. Thank you.

insipidintegrator
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Rishith Raj Raizada
  • 121

2 Answers2

4

Assign coordinates so $A$ is the origin, $B=(3,0)$ and $C=(0,4)$. Then $F=(3t,4t)$ satisfies $3t+4t=3$ or $F=(9/7,12/7)$.

The shoelace formula can now be used to find the shaded area, since $E=(3/2,2)$: $$\frac12\begin{vmatrix} 0&3\\ 9/7&12/7\\ 3/2&2\\ 0&4\end{vmatrix}=\frac12(18/7+6-27/7-18/7)=\frac{15}{14}$$

Parcly Taxel
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2

Let

Area of $\triangle AFB=S_1$

Area of $\triangle BEF=S_2$

Area of $\triangle ADF=S_3$

Area of quadrilateral $ CEFD=S_4$

Applying Menalaus Theorem: $\frac{AD}{AC}\times \frac{CE}{EB}\times \frac{BF}{FD}=1$

$\frac{BF}{FD}=\frac{4}{3}$

$S_1+S_2+S_3+S_4=6$

$S_1+S_2=3=S_3+S_4$

$S_1+S_3=\frac{9}{2}$ and $\frac{S_1}{S_3}=\frac{4}{3}$

$S_3=\frac{27}{14}$

$S_4=3-\frac{27}{14}=\frac{15}{14}$

Lion Heart
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