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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.

 

 Jun 6, 2024

Best Answer 

 #1
avatar+759 
+1

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! :)

 Jun 6, 2024
 #1
avatar+759 
+1
Best Answer

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! :)

NotThatSmart Jun 6, 2024

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