Find the area of the shaded $\triangle MON$ in the figure below

Question

In the figure shown, if ABCD is an isosceles trapezoid, calculate the area of the triangular region MON.(Answer: $\frac{3\sqrt{15}}{4}$)

I have calculation of the base of the triangle. I would need to find the radius of the inscribed circle.

As for the base $MN$ , we find the missing length $x$ through harmonic division since $M, E$ ($MN$ meeting $AD$), $F$($NM$ meeting $CD$) and $N$ form a harmonic quadruple .

So: $$\frac{x}{1}=\frac{(x +4)}{3}$$$$ \implies x=2\\ \therefore MN=6$$

Answer 1

The solution can be correct with following conditions:

1-ON is perpendicular on CD, Mark it's intersection with CD as H.

2-OE is parallel with CD, so $OE\bot ON$

3-$\frac {OH}{ON}=\frac 14$

4- $OE=DE=1$

5- The intersection of MN and CD, G is midpoint of MN , so $ME=2$

6- points M, O and C are colinear.

7- The extension of side BA meets extension of NE at M such that $ME=2 EG=2$

In this case we have:

$ON=\sqrt {4^2-1^2}=\sqrt {15}$

$A_{OEN}=\frac 12 ON\times OE=\frac{\sqrt{15}}2$

The altitude of right triangle OEN from O is :

$ OF=OH=\frac {\sqrt{15}}4$

$A_{OEM}=\frac 12 ME \times OF=\frac 12 \times 2\times\frac {\sqrt{15}}4=\frac{\sqrt{15}}4 $

finally :

$A_{MON}= A_{OEN}+A_{OEM}=\frac{\sqrt{15}}2+\frac{\sqrt{15}}4=\frac{3\sqrt {15}}4$

Answer 2

Let $a$ be the radius of the incircle, $2b=AD$ and $2c=BC$. From right triangle $ABO$ we get $c=a^2/b$.

If $x=ED$ then $CN=3x$ and from $BN=3AE$ we get: $x=b-c/3$. If moreover $\alpha=\angle BAD$, we get: $$ \cos\alpha={b-c\over b+c}={b^2-a^2\over b^2+a^2}. $$ We can then apply the cosine rule to triangle $MAE$, and after some simplifications we obtain: $$ a^2+9b^2=12b. $$ Any pair of numbers $(a,b)$ satisfying the above equation and with $b\ge a$ gives a valid solution to the problem, leading to an area of $3a$.

Note that $(a,b)$ can be seen as the coordinates of a point on an ellipse and $0<a\le6/5$, $6/5\le b<4/3$. Hence the area can take any value between $0$ and $18/5$.

EDIT.

If $K$ is the intersection between $NO$ and $CD$, $H$ the midpoint of $BC$ and making the additional assumption that $CD\perp ON$, from Pythagoras' theorem applied to triangles $NOH$ and $NCK$ we get: $$ ON^2=(3x+c)^2+a^2=9b^2+a^2=12b $$ and $$ (ON-a)^2=(3x)^2-c^2. $$

Combining these equations with the relations found above, one gets $b=5/4$ and $a=\sqrt{15}/4$, whence the solution $3\sqrt{15}/4$ for the area.