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Geometry

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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

#1
+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
+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! :)

NotThatSmart Jun 6, 2024