+1067 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.
+9664 The diameter AB is 9 + 16 + 9 = 34. Therefore, the radius is half the diameter, which is \(\dfrac{34}2 = 17\).
EDIT: There is one detail I missed, that is, it is not possible for PQDC to be a square. Simple calculations gives \(QD = \sqrt{9(9 + 16)} = 15\), not 16. The quadrilateral PQDC is a rectangle with side lengths 16 and 15.
+98 @MaxWong, you're right about the radius is in fact 16.
For part b,
Like you already said, the dimater of the halfcircle is 34. We already know the height of the trapezoid since PQDC is a square. We also know DC is 16.
Now, we have everything needed for the area of the trapezoid, so let's begin!
We know the area of the trapezoid is \(\frac{(b1+b2)h}{2}\) where b1 and b2 are the bases and the h is the height.
We have \(\frac{(25+16)16}{2} = (31)8 = 248.\)
So the area of the trapezoid is 248!
Thanks!
EDIT: Yes, Max Wong is right about the fact that PQDC cannot be a square. Plugging the new numbers in gets us an area of 232.5
