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You didn't need to use a.b=c.d formula in order to find r. I hate formulas. You can clearly see it at 1:43. Just draw another radius, create a 3-3-r triandle and notice it's a 45-45-90 triangle. :)
Saw the same thing as above. Could use Pythagorean theorem to come up with √18 or the known ratio of a 45 45 90 triangle to come up with 3√2 which also equals √18.
you don't have to use the intersecting chords theorem; as soon as you have the side lengths of 3 you can extend a radius that becomes the hypotenuse of a right triangle and then use Pythagorean theorem
Wow nice observation. This is exactly why I love math, there may be a million ways to get a problem wrong, but there are also a million ways to get a problem right.
Then continue watching his videos, and support his channel. A person doesn't have to be your math teacher in order for you to learn math from them. That goes for any subject.
I ended up doing this a different way. Since you don't know how the chord is angled within the square, then it must be true that any chord that cuts the square in half works. Therefore, choose a chord that extends from the upper-left corner of the square to the bottom-right corner. In this case, each side of the square is simply the radius of the semicircle. With the chord as the hypotenuse of a 45-45-90 right triangle, we can then use the equation x^2 + x^2 = 6^2, where x is the length of the side of the square. This gives us a length (x) equal to 3sqrt(2), which is also the radius of the semicircle. From there you can get the area of the semicircle.
The chord length 6 splitting the square in half is given no other specifications. So I chose a convenient chord, the diagonal of the square. This accomplishes two things. First, it splits the square into two triangles and you can solve for the side length using the Pythagorean theorum, 2s²=6², 2s²=36, s²=18, s=3√2. Second, the points of the chord touching the edge of the circle also correspond to the corners of the square. So the sides of the square are also radii of the semi-circle. Now you can solve for the Area of the semi-circle. A=πr²/2, A=(3√2)²π/2, A=(18π)/2, A=9π.
With the center point and the point which the chord intersects the semicircle, you have a right hand triangle with both sides of three with the hypotenuse as the radius.
As always great video andy!! I realy enjoy watching videos and this one is no different.. but i have question.. You have easily solved it on your tablet or computer by showing us how the chord is at the centre of square... However when this question is printes on paper the visual proof becomes less helpful... I meant to say when questions are printed on paper you never really know wether the chord would be passing through the centre of square or not
I just did a special case version. Since the chord bisects the square, I decided to just move the square so that the bottom left vertex was the center of the circle and the bottom right vertex was the right endpoint of the semicircle. The chord is still 6 units, but now it’s forced to make a 45-45-90 triangle. Boom Pythagorus.
6 units is the hypotenuse of a right triangle with the other two sides being the radius. so 2r^2 = 36 >>> r=sqrt(18), so A = pi*r^2 >>> A = 9*pi. Not sure how to prove the right triangle bit though, it just seems... right (sorry!)... Thanks again Andy, another fun one.
Make logic reasonings! Then one sees that the intersections on the ☐square from the chord "6" make equal lengths on the left and right ☐square sides, the upper left length equals the lower right length and vice versa. I call the short one x and the long one y. Due to symmetry the perpendicular bisector of the chord "6" goes through the circle's center and intersects the ☐square in the same manner as above, leaving the same lenthgs (short x/long y). The triangle △Circlecenter-Intersect1-Intersect2 must be an isosceles RIGHT-angled triangle! A quick calculation gives 1. (x+y)/2 · (x+y) = ☐/2 (trapezoid area!) 2. r² = x² + y² 3. ☐/2 = 2·(x·y)/2 + 6·(r/√2)/2 So setting equation 1 = equation 3 leads to (x+y)/2 · (x+y) = 2·(x·y)/2 + 6·(r/√2)/2 |·2 x² + 2xy + y² = 2xy + 6·(r/√2) |-2xy x² + y² = 6·(r/√2) introducing equation 2 → r² = 6·(r/√2) , dividing by r>0 gives r = 6/√2 units length, whence the area of the semicircle is r²π/2 = 18π/2 = 9π ≈ 28.27 square units.
If you draw two lines from the origin to both ends of the chord, you create a right-angled isosceles triangle. Since it is a right isosceles triangle, the positive angle is 45 degrees, which means r is the square root of 18. It's so easy, right? ps. It was translated from Korean, so the sentences may sound unnatural.
3:33 If the video was correct, don't delete it. If people don't believe that it's correct, make a new video showing WHY it's correct. Put that video back up!
So you proved that the problem generalizes for any position of the square, and that you can find the solution from the general condition. But that was not the question. The quickest solution took 2 lines to solve. Ignore the diagram, and reposition the square based upon the problem description. Place one corner of the square at the center of the circle, and the other corner at the corner of the semicircle. Draw a diagonal of the square from the center of the semicircle. This makes the radius r=6/√2, and the result is reached by plugging this r into πr²/2.
I mean, you couldve just connected the radii of the circle to the spots on the square where the semicircle and the square cross. You would figure out it was a right triangle, and actually have the radius in like 30 seconds. Thats how long it took me with this method.
