+743 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.
+1926 The diameter of AB is \(9 + 16 + 9 = 34.\) This means the radius is \(34/2 = 17\)
However, let's note something.
\(QD = \sqrt{9(9 + 16)} = 15\), but in the problem, it says the square sidelenghth is 16. This means that PQRS is NOT a square, but a rectange with sidelengths 16 and 15.
Like I said, the diameter is 34. We already know the height of the trapezoid, which we got earlier as 15. DC is 16, since PQRS is a rectangle.
We have everything we need to solve for the area of the trapezoid!
The area of a trapezoid is \(\frac{(b1+b2)h}{2}\) where b1 and b2 are the bases and h is the height.
We have \(\frac{(25+16)15}{2} = (31)15/2 = 232.5\)
so the area of the trapezoid is 232.5.
Thanks! :)
+1926 The diameter of AB is \(9 + 16 + 9 = 34.\) This means the radius is \(34/2 = 17\)
However, let's note something.
\(QD = \sqrt{9(9 + 16)} = 15\), but in the problem, it says the square sidelenghth is 16. This means that PQRS is NOT a square, but a rectange with sidelengths 16 and 15.
Like I said, the diameter is 34. We already know the height of the trapezoid, which we got earlier as 15. DC is 16, since PQRS is a rectangle.
We have everything we need to solve for the area of the trapezoid!
The area of a trapezoid is \(\frac{(b1+b2)h}{2}\) where b1 and b2 are the bases and h is the height.
We have \(\frac{(25+16)15}{2} = (31)15/2 = 232.5\)
so the area of the trapezoid is 232.5.
Thanks! :)
