Wow, this is fun. A lot of geometry in one problem. Great for a teacher, (which i was once).if you let students have notes for tests, this could be a good problem to build test problems on.
The valley between the two peaks being equal to 60 degrees will always cut a chord equal to the length of the radius, no matter where it is along the diameter. I forget how i know this, but it is a fact and that's how i did it lol.
I used Sen to determine the side of the big triangle and then use Pitagoras theorem to get the height of the big triangle (which in this case is the radius) and then just use the same formula described in the video
Good skills man ! Love to see you solve these. Was a bit buffled on how you took for granted the segment cutting the angle in half. I mean I know a right angle on the opposite base of an equilateral triangle also cuts its corner in half but why did we take the segment between the two points as disecting the angle for granted? Thanks!
I solved it by assuming both triangles have side length 2 so that the distance between the peaks is 2. Then the radius of the quarter circle is also 2, from which it follows that the area is pi.
Hey Andy, i have a request for you. Can you please prove(in two different ways) that the midline in a trapezoid is parallel to the bases I proved it with a weird isosceles triangle and rhombus
Consider a trapezoid ABCD where AB and CD are the bases, and EF is the midline parallel to bases AB and CD, with E and F being the midpoints of sides AD and BC, respectively. Let's focus on triangles ABE and CDF. Since E and F are midpoints of sides AD and BC, we have: AE = ED (since E is midpoint of AD) BF = FC (since F is midpoint of BC) Now, consider that ∠ABE = ∠DCF and ∠BAE = ∠BCF (alternate interior angles). Therefore, by AA (Angle-Angle) similarity criterion, triangles ABE and CDF are similar. In similar triangles, corresponding sides are proportional. Hence, AB/CD = BE/DF. Since BE = DF (because E and F are midpoints), it follows that AB/CD = BE/DF = 1. Therefore, AB = CD, which means EF is parallel to bases AB and CD.
I solved it by assuming the size of the second triangle is 0, which forces the first triangle to have sides length 2, if we're maintaining the distance between the peaks. That makes the base of the first triangle and the radius of the quarter circle the same length, i.e. 2
Love your videos. It's like a mathematical espresso in the morning! He's just proved that a finite shape has an irrational and infinite squared value. Have I missed something? Lol
You used a couple esoteric theorems or wherever talking about "subtended angles" or whatever, and you didnt give much explaination, just blew right through them. This doesnt help someone following along to learn what you did. Please slow down a bit on this more esoteric stuff and give a better explanation.
Thank you for the feedback! I agree. In retrospect, I totally could've made a cool animation with the inscibed angles and another one with central angles. I'll be better next time!
