Question 1110236: A triangle is inscribed in a semi-circle. The triangle is created by the line segments PQ, QR, and PR as shown below. The points P, Q, and R are on the circle. If the length of the line segment PQ = 44 and the length of the line segment PR = 54, then find the area of the shaded region. The triangle PQR is a right triangle. Round your answer to at least 2 decimal places.
Hi Everybody! I'm currently in precalculus 142, doing trigonometry. This problem is posted on my online portion, and my book doesn't really cover this nor did we do it in class. Every answer I come up with doesn't match up with the answer. I've tried using law of sines and law of cosines. I've tried to find the third length and my area is always way larger for some reason. I would really appreciate some help, especially how to set up this problem. I'm a visual learner so any steps or advice would be so helpful! Thank you for taking the time to read this problem and respond :)
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! I'm assuming the line segments are of the triangle, not of the arc of the circle. I am assuming the shaded area is the area outside the triangle.
If one leg is 44 and the hypotenuse (PR) is 54, then the third leg is (54^2-44^2)^(1/2)=31.305 (I won't round until next step).
Because it is a right triangle, the area is (1/2) b*h, and b and h are the base and altitude or sides
This would give a triangular area of 688.71 (rounding here).
The area of the semicircle is (1/2)*pi*27^2, as radius is 27, or 1145.110
The shaded area or area(s) I am assuming are not filled by the triangle. Their total area would be 1145.110 minus the area of the triangle or 456.40 sq units.
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