+1768 Trapezoid $ABCD$ is inscribed in the semicircle with diameter $\overline{AB}$, as shown below. Find the radius of the semicircle. Find the area of ABCD.
PQDC is a square.
+129850 radius = (9 + 16 + 9) / 2 = 17
Let O be the center of the circle
QO = 8 OD = 17
Triangle QOD is right with OD= the hypotenuse
So
DQ = sqrt ( 17^2 - 8^2) = sqrt ( 225) = 15
So ABCD is a trapezoid with height = 15 and bases of 34 and 16
[ ABCD ] = (1/2) ( 15) ( 34 + 16) = 375
