Figure this one out. I did it all in my head. Took maybe a minute.
My answer will appear in the comments. If you watch the video, you'll see that the teacher uses a different method that arrives at the same conclusion.
We have to calculate the area of the "blue" circle (it looks more blue-green on my screen). A chord, AB, has (1) one point on the rim of the large semicircle and the small, inscribed circle ("Point A") and (2) its other point on the rim of the circle at the 3 o'clock position ("Point B"), with the picture noting the large semicircle's center and diameter, and the small circle's center. Chord AB measures 4 units, and that's all the clues we're given.
Now imagine: a chord drawn from A (the top of the semicircle) to the other side of the semicircle's diameter—call that Point C—would also measure 4 units, and angle ABC would be 90 degrees, which means we're looking at a right triangle that is half of a square.
A square's diagonal is √2 times larger than the size of its edge, so we know the diameter of the semicircle is 4√2. The radius is therefore 2√2, which also happens to be the diameter of the "blue" circle.
If the blue circle's diameter is 2√2, then its radius must be just √2. The area of a circle is πr^2, so
π(√2^2) = 2π.
There's the area of the inscribed circle. QED.
The above can be done in your head. If a stroke-addled idiot like me can do it, anyone can.
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We have to calculate the area of the "blue" circle (it looks more blue-green on my screen). A chord, AB, has (1) one point on the rim of the large semicircle and the small, inscribed circle ("Point A") and (2) its other point on the rim of the circle at the 3 o'clock position ("Point B"), with the picture noting the large semicircle's center and diameter, and the small circle's center. Chord AB measures 4 units, and that's all the clues we're given.
ReplyDeleteNow imagine: a chord drawn from A (the top of the semicircle) to the other side of the semicircle's diameter—call that Point C—would also measure 4 units, and angle ABC would be 90 degrees, which means we're looking at a right triangle that is half of a square.
A square's diagonal is √2 times larger than the size of its edge, so we know the diameter of the semicircle is 4√2. The radius is therefore 2√2, which also happens to be the diameter of the "blue" circle.
If the blue circle's diameter is 2√2, then its radius must be just √2. The area of a circle is πr^2, so
π(√2^2) = 2π.
There's the area of the inscribed circle. QED.
The above can be done in your head. If a stroke-addled idiot like me can do it, anyone can.