A circumscribed right-angled trapezoid 
A right-angled trapezoid is circumscribed about a circle $k(O;r=6)$. If the leg that is not perpendicular to the bases is $d=14$, find the area of the trapezoid.
  

We know that the area of every circumscribed polygon is $S=pr$. So I am trying to find the semi-perimeter. Let the trapezoid be $ABCD$, $p=\dfrac{AB+CD+AD+BC}{2}$. Can you give me some hints?
 A: 
Label the points on the trapezoid as shown above (i.e., the circle tangents & trapezoid vertices, starting at the upper left and going clockwise, as $A$, $B$, $C$, $D$, $E$, $F$, $G$ and $H$), with $O$ being the circle center, plus add the lines also shown (i.e., $OB$, $OC$, $OD$, $OE$, $OF$ and $OH$). 
Since $OB = OD = r$, $OC$ is a common line segment and $\angle OBC = \angle ODC = 90^{\circ}$ due to the lines being tangent to the circle, you have $\triangle OBC \cong \triangle ODC$, so $BC = CD$ (note you also can get this by the Pythagorean theorem). Similarly, also by congruent triangles of $\triangle ODE \cong \triangle OFE$, you have $DE = EF$. Thus, $BC + FE = CD + DE = 14$.
In addition, $AB = FG = GH = HA = r = 6$. Using this all together gives that $p = \frac{4(6) + 2(14)}{2} = 26$. Thus, the trapezoid area is $26 \times 6 = 156$.
A: Hint: Break up the trapezium using a line through the centre of the circle, orthogonal to the bases. The trapezium will be broken into a rectangle ($12\times 6$) on the left, and a smaller trapezium on the right, of which you know the height ($12$) and the sum of the bases ($14$ - the same as the longer leg: why?). That lets you calculate both of their areas and therefore the area of the whole trapezium.
