Trapezoid Made into 2 Kites

Draw point $E$ so that segment $AE\perp ED$ and draw point $F$ so that $BF\perp CF$ . Now let $BF=ED=h$ . Then, because $AD=20$ , $EA=\sqrt{400-h^2}$ . Likewise, because $CB=13$ , segment $CF=\sqrt{169-h^2}$ . Since quadrilateral $EBFD$ is a rectangle, it must be that $EB=DF$ , or that $1+\sqrt{400-h^2}=12+\sqrt{169-h^2}$ . Thus, $h=12$ , and the area of the trapezoid is $\frac{12+1}{2}\cdot 12=\boxed{78}$ .

First of all, let us duplicate the trapezoid image so that it forms a parallelogram with twice the area, as the following:

Obviously, we can observe that $BCEF$ is not only a parallelogram, but also a rhombus because all four sides are congruent and the opposite sides are parallel. It is also natural to conclude that this rhombus has twice the area of the trapezoid, as the conventional formula of the trapezoid has portrayed.

Then we will join all the blue lines connecting the points $R, D, A, Q$ and red diagonal lines, as shown below:

With triangles $ABQ$ and $RED$ congruent by $AB = BQ = RE = ED$ and $\angle ABQ = \angle RED$ by trapezoid tiling, we can also deduce that the triangles $PRD$ and $PAQ$ are also congruent because $RD = AQ; DP = PA; and \angle RPD = \angle APQ$ . Thus, $RP = PD$ .

By similar methods, we could prove that $RA = DQ$ and both diagonals $RQ$ and $DA$ intersect at their midpoints, making $RAQD$ a rectangle with point $P$ as its centroid.

Then by having $RF = FA$ and $RE = AB$ , we can conclude that $RA || EB$ , and similarly, $EB || DQ$ , where $EB$ passes point $P$ , halving the rectangle $RAQD$ .

Thereby, we can prove triangles $REP$ and $PBQ$ are congruent because $RE=BQ$ , $RP=PQ$ and $\angle RPE = \angle BPQ$ . Thus, $EP = PB$ . That means the point $P$ is also the centroid of the rhombus.

Hence, another diagonal $FC$ will intersect $EB$ at this point $P$ as the perpendicular bisectors, due to the properties of the rhombus.

Therefore, $BPQ$ is a right triangle with half the area of the trapezoid because it is formed by halves of the adjacent kites.

Then let $BP = x$ and $PC = y$ . By Pythagorean theorem, $x^2 + y^2 = (1+12)^2 = 169$ .

Then by Stewart's theorem , $10^2 = \dfrac{1x^2 + 12y^2}{1+12} - 1\times 12 = \dfrac{ x^2 + 12y^2}{13} - 12$

Substituting $x^2 = 169 - y^2$ , we will obtain:

$10^2 = \dfrac{ 169 - y^2 + 12y^2}{13} - 12 = \dfrac{ 169 +11y^2}{13} - 12$

$112 \times 13 = 169 + 11y^2$

$11y^2 = 99\times 13$

$y^2 = 9\times 13$

Then $x^2 = 4\times 13$ , and so $x = 2\sqrt{13}$ and $y = 3\sqrt{13}$ .

Finally, the area of the trapezoid $ABCD$ = twice the area of the triangle $BPC$ = $2\times \dfrac{1}{2} \times 2\sqrt{13} \times 3\sqrt{13} = \boxed{78}$ .

Bonus

When drawing the height of this trapezoid/rhombus, we will obtain the height of $12$ and extended base of $5$ as shown below:

Draw segments $AQ$ and $DQ$ . Since $PA = PQ = PD$ , they are all radii of a circle with center $P$ and diameter $AD$ , so by Thales' Theorem $\angle AQD = 90°$ . Also drop the perpendicular from $B$ to $CD$ to meet at $R$ .

Let $x = \angle ABQ$ . Then since $AB || CD$ , $\angle BCD = 180° - x$ and $\angle BCR = x$ .

By the law of cosines on $\triangle ABQ$ and $\triangle CDQ$ , $AQ^2 = 2 + 2 \cos x$ and $DQ^2 = 288 - 288 \cos x$ .

By Pythagorean's Theorem on $\triangle ADQ$ , $20^2 = 2 + 2 \cos x + 288 - 288 \cos x$ , which solves to $\cos x = \frac{5}{13}$ .

Since $\sin^2 x + \cos^2 x = 1$ and $\cos x = \frac{5}{13}$ , $\sin x = \frac{12}{13}$ . Then by solving $\triangle BCR$ , $BR = 12$ , which is the height of the trapezoid.

The area of trapezoid $ABCD$ is then $A_{ABCD} = \frac{1}{2}(12 + 1)12 = \boxed{78}$ .

Extend the line segments DA and CB, so that they meet at point O. Now, we have two similar triangles, OAB and OCD (as AB and CD are parallel, as they are the two bases of a trapezium).

The ratio of their side lengths is AB : CD = 1 : 12, therefore the ratio of their areas is 1 : 144, and the ratio of the area of the trapezium ABCD to the area of OCD triangle is 143 : 144.

Also, AD : OD = BC : OC = 11 : 12

Now, it is easy to calculate the side lengths of the OCD triangle:

$OD = (10 + 10) ÷ \frac {11}{12} = \frac {240}{11}$

$OC = (1 + 12) ÷ \frac {11}{12} = \frac {156}{11}$

$CD = 12$

At this point, we can calculate the area of the triangle either by calculating an angle by using the cosine rule and then applying the area formula, or from the side lengths directly, by using Heron's formula:

$A_{ \triangle ABC} = \sqrt { s(s -a)(s-b)(s-c) } \text { , where } s = 0.5 × (a + b + c)$

$s = 0.5 × ( \frac {156}{11} + \frac {240}{11} + 12) = 24$

$A_{ \triangle OCD} = \sqrt { 24(24 - \frac {156}{11})(24 - \frac {240}{11})(24 -12) } = \frac {864}{11}$

Hence:

$A_{ trapezium ABCD} = \frac {864}{11} × \frac {143}{144} = \boxed {78}$