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.
All comments are subject to approval before they are published, so they will not appear immediately. Comments should be civil, relevant, and substantive. Anonymous comments are not allowed and will be unceremoniously deleted. For more on my comments policy, please see this entry on my other blog.
AND A NEW RULE (per this post): comments critical of Trump's lying must include criticism of Biden's or Kamala's or some prominent leftie's lying on a one-for-one basis! Failure to be balanced means your comment will not be published.
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.