SOLUTION: Point R is chosen on the circumference of a circle and arc PQ is drawn with R as its center. If P and Q are joined, the line segment formed is a diameter of the circle with length

Algebra ->  Circles -> SOLUTION: Point R is chosen on the circumference of a circle and arc PQ is drawn with R as its center. If P and Q are joined, the line segment formed is a diameter of the circle with length       Log On


   



Question 1209532: Point R is chosen on the circumference of a circle and arc PQ is drawn with R as its center. If P and Q are joined, the line segment formed is a diameter of the circle with length 2r cm. Find the area of the shaded region in terms of r, in square centimeters.
Link to diagram: https://ibb.co/GQFdy2SV

Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959) About Me  (Show Source):
You can put this solution on YOUR website!
Here's how to find the area of the shaded region:
1. **Area of the Semicircle:** Since PQ is the diameter, the radius of the circle is r. The area of the entire circle is πr², so the area of the semicircle (the region above PQ) is (1/2)πr².
2. **Area of Triangle PQR:** Triangle PQR is inscribed in a semicircle with PQ as the diameter. A fundamental theorem states that any triangle inscribed in a semicircle with the diameter as one side is a right triangle. Therefore, angle PRQ is a right angle.
* The base of triangle PQR is PQ (the diameter), which has a length of 2r.
* The height of triangle PQR is the perpendicular distance from R to PQ. Since R is on the circle's circumference, this height is equal to the radius, r.
* Area of triangle PQR = (1/2) * base * height = (1/2) * 2r * r = r²
3. **Area of the Shaded Region:** The shaded region is the area of the semicircle *minus* the area of triangle PQR.
Shaded Area = Area of Semicircle - Area of Triangle PQR
Shaded Area = (1/2)πr² - r²
4. **Simplify:** We can factor out r²:
Shaded Area = r² * [(π/2) - 1]
Therefore, the area of the shaded region is **r²[(π/2) - 1]** square centimeters.

Answer by ikleyn(52772) About Me  (Show Source):
You can put this solution on YOUR website!
.
Point R is chosen on the circumference of a circle and arc PQ is drawn with R as its center.
If P and Q are joined, the line segment formed is a diameter of the circle with length 2r cm.
Find the area of the shaded region in terms of r, in square centimeters.
Link to diagram: https://ibb.co/GQFdy2SV
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The solution in the post by @CPhill is TOTALLY WRONG.

It is totally wrong, since it is conceptually incorrect.

Indeed, it says "The shaded region is the area of the semicircle *minus* the area of triangle PQR".


                                            It is not so.


The shaded region is the area of the original semicircle "minus" the area of the segment of
the greater circle of the radius r%2Asqrt%282%29, which segment is formed by the arc PQ.


                                See my full solution below.


Half the area of the original circle is  %281%2F2%29%2Api%2Ar%5E2.


The area of the segment of the greater circle is  1/4  of the area of the greater circle,

        %281%2F4%29%2Api%2A%28r%2Asqrt%282%29%29%5E2 = %281%2F4%29%2Api%2Ar%5E2%2A2 = %281%2F2%29%2Api%2Ar%5E2,

MINUS  the area of the right-angled isosceles triangle with the legs  r%2Asqrt%282%29.


Thus,  the area of the shaded region is

        %281%2F2%29%2Api%2Ar%5E2 - %28%281%2F2%29%2Api%2Ar%5E2+-+%281%2F2%29%2A2r%5E2%29 ] = r%5E2.


The final answer is:   The area of the shaded region is  r%5E2.

Solved correctly.

For the peace in mind,  ignore the post by  @CPhill.