Find Area of shaded region.

euclidean-geometrygeometry

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 . figure given below

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.

Best Answer

Here is the picture for the situation:

Stackexchange picture 3968678

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.
  • $A_1+A_2+2A_3=[ABCD]=120$.
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