Triangle In Pentagon

Since the triangle BDE is isosceles, the angels EBD and BDE are both equal to 45 degree. By drawing an axis of symmetry of the figure, we can easily obtain two isosceles triangles, BNE and DNE. So, NE = BN = DN = 6. It can be easily check that MN = 2.5.

The shaded region can be divided into two triangles, AME and DME. Both of them have height 6 and base length 8.5. Thus the shaded region has area 2 x 1/2 x 8.5 x 6 = 51.

Vectors are very useful for this problem.

Let point D be the origin of the reference frame, then

$DA = -12 i + 5 j$

$DE = -6 i - 6 j$

Area = $\frac{1}{2} | DA \times DE |$

$= \frac{1}{2} | (-12 i + 5 j) \times (-6 i - 6 j)|$

$= \frac{1}{2} | ( 72 k + 30 k ) | = 102 / 2 = 51$

Let $\angle ADB = \theta$ . We know that $\Delta BED$ is an isosceles right triangle, right angled at E. Therefore $\angle BDE = \angle DBE = 45$ .

Now let $DE = BE = x$ . Since ABDC is a rectangle $AC = BD = 12, AB = DC = 5$ , using pythagoras in ABD $AD = 13$

$\text{In }\Delta DBE \\ BE^2 + DE^2 = DB^2 \\ x^2 + x^2 = 12^2 \\ 2x^2 = 12 \times 12 \\ x^2 = 12 \times 6 \\ x = 6 \sqrt[]{2}$

Now we know that Area of a triangle is equal to $\dfrac{1}{2}\ ab \ \sin(\theta)$ where $\theta$ is the angle between $a$ and $b$ .

$\therefore \mathbb{ar}(DBE) = \dfrac{1}{2} \ 6 \sqrt[]{2} \ 13 \ \sin(\theta + 45) \\ \boxed{\sin \theta = \dfrac{5}{13}, \cos \theta = \dfrac{12}{13}, \sin 45 = \cos 45 = \dfrac{1}{\sqrt[]{2}}} \\ \implies \mathbb{ar}(DBE) = 3 \sqrt[]{2} \times 13 \times \left(\sin\theta \cos 45 + \cos \theta \sin 45 \right) \\ = 3 \sqrt[]{2} \times 13 \times\left(\dfrac{5}{13} \dfrac{1}{\sqrt[]{2}} + \dfrac{12}{13} \dfrac{1}{\sqrt[]{2}} \right) \\ = 3 \times (5 + 12) = 3 \times 17 \\ \boxed{\therefore \mathbb{ar}(DBE) = 51}$