Find Area of shaded region. In the given figure $AB$ = $9cm$, $AD$ = $BC$ = $10cm$, $DB$ = $AC$ = $17cm$. If area of the shaded
region is ${m\over n}$, where ($m, n$) = $1$, then evaluate m + n . 
My Work :-
Well initially I had ideas of joining $D$ and $C$ and then maybe using something related to similar triangles. Well then I had another idea of maybe finding the height of the trapezium $ABCD$ and then maybe find height of the shaded triangle and similar ones. However whenever I am approaching either ways, I am just getting stuck in variables.
So I need an advice as to how to approach this problem.
 A: 
Let the intersection point of $AC,BD$ be denoted by $O$. Let $V$ be the point on $AB$ such that $OV\perp AB$. Since $\triangle DAB\cong\triangle CBA(SSS),\triangle AOB$ is isosceles and $AV=BV=4.5$.
Then $\tan\angle OAB=\frac{OV}{AV}$ which gives the required area $\Delta=OV\times AV=4.5^2\tan\angle OAB$.
Let $X$ be the point on $AC$ such that $BX\perp AC$. Then $BX^2=9^2-AX^2=10^2-(17-AX)^2$ giving $AX=\frac{135}{17}$.
Now, $\tan\angle OAB=\frac{BX}{AX}$. Can you complete?

$\tan\angle OAB=0.5\bar3$ giving $\Delta=10.8=\frac{54}5.$

A: Here is the picture for the situation:

One possibility to "see the height" is as follows.
Of course, we have an isosceles trapezium $ABCD$.
In the triangle $\Delta ABC$ the sides are known, $9, 10, 17$, the semi-perimeter is $(9+10+17)/2=18$, so Heron gives its area $[ABC]=A_1+A_3$ as
$$
[ABC]=A_1+A_3=\sqrt{18(18-9)(18-10)(18-17)}=36\ .
$$
Since $AB=9$, the heights $DD'$ and $CC'$ of the isosceles trapezium have both length $8$. This gives then $AC'=BD'=15$, since there is the pythagorean triple $(8, 15, 17)$. We get $BC'=AD'=6$ finally. (And one can take this as a start).
Let us now denote by $A_1$ and $A_2$ the areas of the similar triangles $\Delta XAB$ and $\Delta XCD$. It is easy to compute all areas now. We may also want to compute the position of $X$, the intersection of the diagonals, on the diagonals:
$$
\begin{aligned}
A_1 + A_3 &= [ABC] = 36 \ ,\\
\frac {A_1}{A_3} &= \frac{XA}{XC} =\frac{AB}{CD}=\frac {9}{21}=\frac37\ ,\\
\frac {A_1}{A_1+A_3} &= \frac3{3+7}=\frac3{10}\ ,\\
A_1 &= (A_1+A_3)\cdot \frac3{10}=36\cdot  \frac3{10}=\frac{54}5\ . 
\end{aligned}
$$
$\square$

@Note: Checks, and other possibilities to get $A_1$:
The area $[ABCD]$ is $\frac 12\cdot 8(9+21)=120$, and $A_1=54/5$, $A_3=126/5$, $A_2=294/5$. We have

*

*$A_1+A_3=36$, as it should be $\frac 12\cdot 9\cdot 8$,

*$A_2+A_3=84$, as it should be $\frac 12\cdot 21\cdot 8$,

*$A_1:A_2=9/49=(3/7)^2=(9/21)^2=(AB/CD)^2$, as it should be from the similarity.

*$A_1+A_2+2A_3=[ABCD]=120$.

A: Given the triangle’s base $9$ and its rational area $\frac mn$, its height is rational; so are likely the height and the base of the triangle DAB per similar triangles. Note that $(8,15,17)$ and $(6,8,10)$ are Pythagorean’s triplets. It is determined that $8$ and $15$ fit its height and base, respectively, which leads to $\frac{\frac 92}{15}\cdot 8=\frac{12}{5}$ as the height of the shaded triangle. Thus, its area is $\frac12\cdot\frac{12}5\cdot 9=\frac{54}5 $.
