it is not necessary to remember 30 - 60 - 90 triangle just use tan60 value to find the base of the triangle tan60 = root 3 and tan60 = base/2*root(2). Therefore base = tan60 * 2*root(2) => base = root(3) * 2 * root(2) => base = 2 * root(6). Therefore base = 2 * root(6) which is also found using 30 - 60 - 90 triangle too.
@@stephenbeck7222 😅I understood what you mean. But I find it best to remember trigonometric angles instead of whole 30 - 60 - 90 triangle sides propositions and other triangle sides propositions.
I solved it diferently. I first used the Pythagorean theorem to find the height of the large triangle, which was 4sqrt(3). Then I used that to find the total area of the large triangle, which was 16sqrt(3). Then I divided by 2 to find the area of the yellow triangle, which was 8sqrt(3). Then I called the side length of the yellow triangle x and used the area I found to solve for its height in terms of x, which was (16sqrt(3))/x. Then I used that and the Pythagorean theorem to solve for x, which was 4*4^(1/4). Then I plugged that into the height formula I found above and calculated the height, which was (4sqrt(3))/(4^(1/4)). I knew the height of the yellow triangle was the radius of the semicircle, so I plugged that into the equation (1/2)(pi)r^2, and calculated the area, which was 12pi. Then I watched the video and found out I was right. How exciting!
3:00 Why isn't half of the triangle side length equal to 4 + (√2/2) ? Edit: writing it out solved it for me, the 4/1 and the 1/2 make 2, and that leaves √2/1 alone since there's no /√2 to cancel it out from the numerator 😄
To solve the problem, we find that there is a type of congruent triangles and another type of similar triangles. Using the given data, we find after calculation that the side of the square is a=(10√33)/11 and that the area of the yellow region is equal to (11a²/5)-30=60-30=30.
45 is the anwser to the last question, it actually equals to triple the overlap area. intresting question, you don't even need to calculate the radius or anything. you can solve it by rotating and rearranging triangle puzzle peices or simply notice that yellow plus purple area is four times the overlap area
Can you take it as a given that both triangles are similar? The diagram doesn't indicate that the base of the semicircle is parallel to the base of the larger triangle
Are these great geometry problems your own or taken from somewhere else? I must admit, your math and geometry skills are unbelievable. Math and geometry are stunning fields of science. Your skill is on level of great mathematician Presh Talwalkar.
Tomorrow's solution? I just spent an hour on it, used some similar triangles to get proportions of side lengths and came up with an equation in terms of s (side length of square) which was s(s/2) x 1/2 + 2sroot5 / 5 x 3sroot5 / 10 x 1/2 = 15, and that led to a total area formula of 2(3sroot5/10)(2sroot5/5) + 300/11 - 30 which came out to be 30 units squared, so exactly double the 15 given in the problem! How exciting!
@@JLvatronExactly. That's because the sides of the yellow equilateral triangle don't connect from the center of the semicircle to its circumference, which is why it can't be considered the radius. The height of the yellow equilateral triangle, on the other hand, does connect the center of the semicircle to its circumference, which is why it's equal to the radius.
the area of yellow part is 30 (hint - the center is also the mid point for sides of rectangle and square because distance of two chords of equal length is also equal from the center of the circle )
it would be 15 acc to me bcz 30 is the total area of square and rectangle we need to find the area of yellow part that will be 30 -15 (total - pink) @harshirwaqar8228
For me the problem was way easier using direct calculator substitutions. I know that area of an equilateral triangle is (S²*√3)/4, and its height is (S*√3)/2. From here I just do a series of direct substitutions to get x, area of the big triangle from the formula is 16√3, half of it is 8√3, equating that to (x²*√3)/4 we get x² = 8*4 = 32, x = 4√2. Height of small triangle= r, 4√2 * √3/2 = 2√6. Area of semicircle= π (2√6)²/2 = 12π
Answer to the next question: The rectangle is half of a square (put a mirror along the diameter of the semicircle and you’ll see), it’s length & width = L & L/2 Circle radius = r = 1/2 of the diagonal of a square with side length L = (√2)L/2 = L/√2 r = L/√2, L = r√2, W = (r√2)/2 Rectangle area = L(L/2) = (L^2)/2 = r^2 Small square area = (side length)^2 = W^2 + (r - L/2)^2 = (r^2)/2 + (r - r√2/2)^2 = (2 - √2)r^2 Bottom and left yellow triangles are similar with respective hypotenuse r & (r)√(2 -√2) (i.e. the small square side length) Left triangle area = W(r - L/2)/2 = ((r√2)/2)(r - (r√2)/2)/2 = (r^2)((√2)/2 - 1/2)/2 = (r^2)(√2 - 1)/4 Bottom triangle area = left triangle area ÷ (2 -√2) = (r^2)(√2 - 1)/[4(2 -√2)] = (r^2)(√2)/8 Red area = 15 = small square area - the combined area of the two yellow triangles = (2 - √2)r^2 - (r^2)(√2 - 1)/4 - (r^2)(√2)/8 = (r^2)[(2 - √2) - (√2 - 1)/4 - (√2)/8] = (r^2)(16 - 8√2 - 2√2 + 2 - √2)/8 = (r^2)(18 - 11√2)/8 r^2 = 15(8)/(18 - 11√2) = 15(8)(18 + 11√2)/82 = 15(4)(18 + 11√2)/41 = (18 + 11√2)(60/41) Yellow area = small square area + rectangle area - 2(red area) = r^2 + (2 - √2)r^2 - 30 = (3 - √2)r^2 - 30 = (3 - √2)(18 + 11√2)(60/41)- 30 = (32 + 15√2)(60/41)- 30 = (120 + 900√2)/41
Yellow triangle is b=2r/{3}, h=r Blue-yellow triangle is b=8, h=4{3} Yellow area is bh/2=r^2/{3} Blue area is bh/2-yellow=16{3}-r^2/{3} Blue equals yellow so 16{3}=2r^2/{3} => r^2=24 => r=2{6} (Yellow area = 8{3}) Semicircle is pi*r^2/2=12pi
The yellow triangle is similar to the larger triangle, and their angles are congruent, i.e. they are all 60°. Therefore, the yellow triangle is equilateral.
How did you do it? I did it the long way by findin the height of the triangle. Then I worked out the areas - height of small triangle and then that’s the radius.
![I.N "HALLUCINATION" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/n5B5q1Hwt_U/mqdefault.jpg)
30 - 60 - 90 triangle is the MVP of December
it is not necessary to remember 30 - 60 - 90 triangle just use tan60 value to find the base of the triangle tan60 = root 3 and tan60 = base/2*root(2). Therefore base = tan60 * 2*root(2) => base = root(3) * 2 * root(2) => base = 2 * root(6).
Therefore base = 2 * root(6) which is also found using 30 - 60 - 90 triangle too.
@@Ramu9119 I am not saying that is it necessary to remember, I am just saying this because of the use of this by Andy math in these aggvents
@@Ramu9119but how do you know tan(60) = sqrt(3)?
@@stephenbeck7222 😅I understood what you mean. But I find it best to remember trigonometric angles instead of whole 30 - 60 - 90 triangle sides propositions and other triangle sides propositions.
@@TotallyNotJ4denn Even I agree with you and it is best to know different ways to solve a problem.
Brother Andy Math: I know you will catch up by New Year's. Nobody now remembers that the Statue of Liberty was months late and over budget.
I do, it was a pr disaster. I remember hearing the statue was late and thinking "boy howdy by golly, my taxes at work huh!"
I would solve these problems with Geogebra or Desmos. Its only one possibility for those, who are not so skilled in math as Andy is.
I’m a year 7 student, your videos have got me to love math and today I actually got it myself!!!
Good job!
Same!!
Gee it's a lot easier when you know stuff. Ratio of square of similars was a new idea to me, thanks!
These puzzles are so much fun! I hope we will continue with this throughout the next year!
I solved it diferently. I first used the Pythagorean theorem to find the height of the large triangle, which was 4sqrt(3). Then I used that to find the total area of the large triangle, which was 16sqrt(3). Then I divided by 2 to find the area of the yellow triangle, which was 8sqrt(3). Then I called the side length of the yellow triangle x and used the area I found to solve for its height in terms of x, which was (16sqrt(3))/x. Then I used that and the Pythagorean theorem to solve for x, which was 4*4^(1/4). Then I plugged that into the height formula I found above and calculated the height, which was (4sqrt(3))/(4^(1/4)). I knew the height of the yellow triangle was the radius of the semicircle, so I plugged that into the equation (1/2)(pi)r^2, and calculated the area, which was 12pi. Then I watched the video and found out I was right. How exciting!
i also did it the same way
You can almost catch a glimpse of exhaustion at the end of his "How Exciting". Hang in there Andy, just 10 more to go!
Video length 4:20 = nice
Really interesting problem sets this month! Thanks for sharing 🙌
Blue triangle / 2 = Yellow Triangle
(half base * height)
4 * 4√3 / 2 = (r /√3 ) * r
8√3 = r² /√3
8√3 * √3 = r²
24 =r² => Semicircle 12π
3:00 Why isn't half of the triangle side length equal to 4 + (√2/2) ?
Edit: writing it out solved it for me, the 4/1 and the 1/2 make 2, and that leaves √2/1 alone since there's no /√2 to cancel it out from the numerator 😄
To solve the problem, we find that there is a type of congruent triangles and another type of similar triangles. Using the given data, we find after calculation that the side of the square is a=(10√33)/11 and that the area of the yellow region is equal to (11a²/5)-30=60-30=30.
We need to schedule double headers the rest of the way! It's a great day for puzzle solving - let's play two!
45 is the anwser to the last question, it actually equals to triple the overlap area. intresting question, you don't even need to calculate the radius or anything. you can solve it by rotating and rearranging triangle puzzle peices or simply notice that yellow plus purple area is four times the overlap area
30 is the answer then as per your own logic
Can you take it as a given that both triangles are similar? The diagram doesn't indicate that the base of the semicircle is parallel to the base of the larger triangle
why am I watching this first thing in the morning.. . still waiting for my tea even...
Are these great geometry problems your own or taken from somewhere else? I must admit, your math and geometry skills are unbelievable. Math and geometry are stunning fields of science. Your skill is on level of great mathematician Presh Talwalkar.
wait can anyone tell me how we can say that the yellow triangle is equilateral? (just a lil confused)
Tomorrow's solution?
I just spent an hour on it, used some similar triangles to get proportions of side lengths and came up with an equation in terms of s (side length of square) which was s(s/2) x 1/2 + 2sroot5 / 5 x 3sroot5 / 10 x 1/2 = 15, and that led to a total area formula of 2(3sroot5/10)(2sroot5/5) + 300/11 - 30 which came out to be 30 units squared, so exactly double the 15 given in the problem! How exciting!
New achievement unlocked: ENHANCE!🤩
Isn't the side of the yellow equilateral triangle also equal to the radius
No, the height was, but the side wasn’t
@@JLvatronExactly. That's because the sides of the yellow equilateral triangle don't connect from the center of the semicircle to its circumference, which is why it can't be considered the radius. The height of the yellow equilateral triangle, on the other hand, does connect the center of the semicircle to its circumference, which is why it's equal to the radius.
the area of yellow part is 30 (hint - the center is also the mid point for sides of rectangle and square because distance of two chords of equal length is also equal from the center of the circle )
it would be 15 acc to me
bcz 30 is the total area of square and rectangle we need to find the area of yellow part that will be 30 -15 (total - pink) @harshirwaqar8228
For me the problem was way easier using direct calculator substitutions.
I know that area of an equilateral triangle is (S²*√3)/4, and its height is (S*√3)/2.
From here I just do a series of direct substitutions to get x, area of the big triangle from the formula is 16√3, half of it is 8√3, equating that to (x²*√3)/4 we get x² = 8*4 = 32, x = 4√2.
Height of small triangle= r, 4√2 * √3/2 = 2√6.
Area of semicircle= π (2√6)²/2 = 12π
Answer to the next question:
The rectangle is half of a square (put a mirror along the diameter of the semicircle and you’ll see), it’s length & width = L & L/2
Circle radius = r = 1/2 of the diagonal of a square with side length L = (√2)L/2 = L/√2
r = L/√2, L = r√2, W = (r√2)/2
Rectangle area = L(L/2) = (L^2)/2 = r^2
Small square area = (side length)^2 = W^2 + (r - L/2)^2 = (r^2)/2 + (r - r√2/2)^2 = (2 - √2)r^2
Bottom and left yellow triangles are similar with respective hypotenuse r & (r)√(2 -√2) (i.e. the small square side length)
Left triangle area = W(r - L/2)/2 = ((r√2)/2)(r - (r√2)/2)/2 = (r^2)((√2)/2 - 1/2)/2 = (r^2)(√2 - 1)/4
Bottom triangle area = left triangle area ÷ (2 -√2) = (r^2)(√2 - 1)/[4(2 -√2)] = (r^2)(√2)/8
Red area = 15 = small square area - the combined area of the two yellow triangles = (2 - √2)r^2 - (r^2)(√2 - 1)/4 - (r^2)(√2)/8 = (r^2)[(2 - √2) - (√2 - 1)/4 - (√2)/8] = (r^2)(16 - 8√2 - 2√2 + 2 - √2)/8 = (r^2)(18 - 11√2)/8
r^2 = 15(8)/(18 - 11√2) = 15(8)(18 + 11√2)/82 = 15(4)(18 + 11√2)/41 = (18 + 11√2)(60/41)
Yellow area = small square area + rectangle area - 2(red area) = r^2 + (2 - √2)r^2 - 30 = (3 - √2)r^2 - 30 = (3 - √2)(18 + 11√2)(60/41)- 30 = (32 + 15√2)(60/41)- 30 = (120 + 900√2)/41
Yellow triangle is b=2r/{3}, h=r
Blue-yellow triangle is b=8, h=4{3}
Yellow area is bh/2=r^2/{3}
Blue area is bh/2-yellow=16{3}-r^2/{3}
Blue equals yellow so 16{3}=2r^2/{3} => r^2=24 => r=2{6}
(Yellow area = 8{3})
Semicircle is pi*r^2/2=12pi
Can also observe that blue area is half of total area, so blue must equal 8{3} and set that equal to yellow area for same result
How am I always so close yet I get it wrong every time 😅😢
How do you know the yellow triangle is equilateral?
We can assume that based on the fact that it came from a larger equilateral triangle.
The yellow triangle is similar to the larger triangle, and their angles are congruent, i.e. they are all 60°. Therefore, the yellow triangle is equilateral.
we will catch up!
Probably…
w andy chat, lets get some w for him
I GOT THIS MYSELF!!!!
I was so close! I did it my own way and just forgot to divide by two at the end, so I found the area if the semicircle was a full circle.
How did you do it? I did it the long way by findin the height of the triangle. Then I worked out the areas - height of small triangle and then that’s the radius.
@ I think I did the same thing.
It hapens to everyone!
W
day22
5x^2/4 +3x^2/2 =15
→11x^2/4 =15
→A=11x^2/2 =30😊
WTF is x?
... Speaking of Brilliant?
👏🤝👍
The answer of last question is 45
Wow
I'm somewhat early
Im also early
hi somewhat early
Same
@khalief_.hi khalief
I wonder why its december 27 and were still at day 21
We're catching up!
Life happens
Christmas yk
Dude, just be thankful Andy is giving us these great math vids!
21 does not equal 27 if you're sticking to base 10. Is one of the challenges to work out what base number system Andy is using?
No. The challenge is to keep him catching up from falling behind earlier this month.