there is a much simplier way. Since the top left triangle and the entire trangle are similar, the ratio between their sides is 3/4, i.e. r + 3r/4 = 3, so r = 12/7.
@antidro_XY look at the figure on 1:18 of the video. The top left triangle is similar to the entire triange, so ratio of sides will be the same. so the left side of the small top left triangle is (3/4) * r. Now look at the left side of the large triangle, the side 3 = r + left side of top left triangle.
As someone preping for JEE, watching this guy makes me not only watch educational content on free time but develop a Genuine interest in maths. Thanks. I occasionally watch videos like this and I actually feel my attention in classes increase. How interesting!
I found an easier method I think. I set the origin point to be the right angle so that the x and y axes aligned with the triangle sides. That would mean that the hypotenuse would have a slope of -3/4x + 3. So, the coordinates of the center point must be the point on the line that’s equidistant to the two axes, since the circle is tangent to both. Since the line y=x is always equidistant from both axes, you just have to solve for the point at which that and the hypotenuse intersect, which is (12/7,12/7), which is our radius.
I earned a bachelor's degree in agriculture in the 90's. The highest math I took was calculus. Now my daughter is in middle school and the math is still simple but will be ramping up. I want to be able to help her so I've been watching a lot of these videos and really enjoying the math.
For any such right triangle, the radius of the semicircle with diameter along the hypotenuse will always be: product of lengths of the legs divided by the sum of their lengths. In this case r = (3*4)/(3+4).
@paparmar ah I see, but then wouldn't that mean the completely quantized version shown in the video would be the same thing? yet (3*4)/(3+4) isn't the same solution as shown in the vid
@@cleargrits9117 : Sorry, I'm not following you - at 3:19 the video concludes r = 12/7, the same as the formula I gave. I'm stopping at the radius, rather than the area of the semi-circle.
A simpler solution: Draw the 2 radii from the center of the semicircle to the points of tangency and 1 to the right angle of the triangle. The sum of the areas of the 2 triangles created (both of which have height r!) can be written as 3*r/2+4*r/2, or 3.5 r. This should be equal to the area of the 3-4-5 triangle, or 6. 3.5r=6, and from this you get r=12/7. Therefore, πr^2/2=(144/9) π/2=72/9 π
Alternate solution: when the figure is reflected about the hypotenuse, we'd get a complete circle which will also be an incircle to the quadrilateral. Then radius of incircle is given by r = A/S where A is area and S is semi perimeter of polygon
My initial guess was to mirror the triangle along the hypotenuse, so the circle's diameter is approximately 3, resulting in 3π/2 = 4.71, but the actual answer is 4.61, which I consider to be good enough for just coming up with it in not even a twenty seconds
Thank you Andy, just love your videos, I'm not studying maths, and have been a long time out of school. I just love seeing how it all fits together and your approach clearly comes from a genuine love for maths.
Mirror the triangle about its hypotenuse, then we'll have deltoid/kite with inscribed circle. Its area is twice the area of triangle; on the other hand, tangential quadrilateral area is semi-perimeter multiplied by inradius, therefore r = 2S/p = 3·4/(3+4) = 12/7 and semicircle area is 72π/49.
I thought this, and concluded the circle would have diameter=3, but 1.5^2 * pi / 2 isn't equivalent to the other answers given and I don't understand why this doesn't work.
I had a slightly different approach, without a quadratic equation or similarity of angles: Because the circle is tangent to the x and y axes, the circle's center must have y=x. The linear formula for the hypothenuse is y=3-(3/4)x, so solving x=3-(3/4)x gives the center point x=y=12/7 and this is also the radius, resulting in an area of ½π(12/7)².
The equation of the hypotenuse is Y = 3 -3X/4 (point-slope by inspection). The center of the circle is at Y = X. So X = 3 -3X/4 => X = 12/7. All else are extraneous complications.
Bro which software do you use to make these kind of videos ? I mean the animation , writing equations and solving them !! I am a math teacher and I want to make videos like this for my students
If you stick with Euclidian geometry, quadratic eqn is not needed. Similarity tells us: (3-r)/r=3/4 and r/(4-r)=3/4, r can be solved even without a flip of the pen.
Alternate solution. Start by drawing the radii from the tangents as in the video. This forms two triangles, each with a right angle and one of he angles of the original triangle. This means both triangles are similar to the original triangle and to each other since the remaining angle of each new triangle is the complement of the angle it shares with the original triangle, and the complement of that shared angle is the shared angle of the other triangle. The upper triangle has a length of r. The lower triangle has a height of r. The length of the lower triangle is 4/3 of it's height of r, because it's similar to the original triangle. The length of the 2 new triangles equal 4 because together they add up to the length of the original triangle. 4 = r +4/3 r = 7/3 r 12 = 7r r = 12/7 area = pi * 1/2 (12/7)^2
Alternative solution: you can treat the hypotenus like a line with the equation "y = -(3/4)x + 3" and since we know the center of the circle where be where "y = x" we just use substitution. The answer will be the radius.
Here’s what I did. Put the triangle on a coordinate system with the right angle being corner being the origin. The hypotenuse is then represented as the line y=-3/4x+3 and the center point of the circle lies along the line at point (r,r). Plug in r for x and y to get r=-3/4r+3 and solve for r getting r=12/7
Based on the fact that the circle was equal to pi, my immediate estimation was that the semi-circle was ~1.5pi. Cool to see my intuitive guess wasn't too far off!
It' s easier to write the similiarity between one of the two small triangles and the bigger 3-4-5 triangle. For example, (3-r)/r = 3/4 or r/(4-r) = 3/4. That way you have a first grade equation only
How exciting! Hey, do you have any videos on solving geometric progressions or sequences? Came up today in real life for me (had to solve for a common ratio for a series). It was kind of neat, but maybe not exciting 😅
I just connected the center and the right angle to make two triangles. One with a base of 3 and a height of r, and the other with a base of 4 and a height of r. I added them together to get the bigger triangle, which has an area of 3×4÷2=6 squared units. 3r÷2+4r÷2=7r÷2=6, r=12÷7
would it also be possible to find 12/7 by setting the sums of the 2 triangles (r(3-r) and r(4-r)) and one square (r*r) equal to the area of the whole triangle (3*4/2)?
You can also use Pythagoras theorem on each triangle as sqrt((3-r)^2+r^2)+sqrt(r^2+(4-r)^2)=5 but the calculation is rather tedious compared to the method shown.
I don't understand how we know/can prove that the lines from the point of tangency on each side both A) make a perfect radius at the hypotenuse and B) meet at the hypotenuse
It's because it's a semi-circle (i.e., exactly half of the circle) and not just a random portion of a circle. Since it's a semi circle, the center of the circle has to lie on the line cutting the circle in half, which is also the hypotenuse. The other parts are explained (tangent perpendiculars must go through the center of a circle).
Agree that the center of the full circle is on the hypotenuse, since it's a semi-circle. Still unsure why a line perpendicular to the point of tangency of the two sides are guaranteed to meet the hypotenuse at the center of the "circle" and that they both meet at the same point.
That's a rule about tangent lines. A line that's perpendicular to a tangent line at the point of tangency goes through the center of the circle, he kinda describes that at 0:43. So when you have two lines that are perpendicular to two tangents, they both pass through the center of the circle; therefore, the point of intersection is/must be the center of the circle.
@@josephulton I agree with you. I get the center is on the hypotenuse, I get the tangent lines are perpendicular from the sides, but I don’t get why the perpendicular lines have to meet at the hypotenuse. This was not explained.
upon reflecting on it more, I realize that these things must be true about a Semi-Circle described in the video: A. a line perpendicular to the ONLY (1) point of tangency will create a perfect Diameter, thus going through the center of that circle B. the center of a circle will be on the hypotenuse since it is listed as a semi-circle somehow didn't grasp that from the video, even though it is touched on
Oh, wow, turns out I got it, good for me. It was just somewhat irrational. I did it kind of a different way. We know the area of the whole triangle is bh/2, (3*4)/2, 6. And we know it's made up of the two triangles and the square. So r^2 (area of the square), plus the area of both triangles (r * (4-r)) / 2 and (r * (3-r)) /2 will all equal 6. If we distribute both r's, we can add the two chunks that are both over 2, and get r^2 + (7r-2r^2)/2. Multiply r^2 by 2/2 so it'll be over 2 so we can add the fraction, and the 2r^2-2r^2 cancels out, leaving us with 7r/2 = 6. Multiply both sides by 2/7 to get r alone, gives us r = 12/7. And with r just get the area of the semi-circle as he showed us.
Please someone point to me where my reasoning is broken. First, mirror the triangle around the hypotenuse. Now you've got a rectangle with sides 3 and 4. Since the original triangle is a square triangle, the horizontal lines are parallel and are separated by 3 units. Since the circle is tangent to both the original horizontal line and the new mirrored horizontal line, then its diameter is 3, its radius is 3/2 and therefore the area is (pi . (3/2)^2) / 2 = 9/8 . pi.
If you mirror the triangle around the hypotenuse (i.e. the side of length 5 is your line of reflection) you get a kite as your quadrilateral, not a rectangle.
No; the circle of radius 12/7 has a diameter of 24/7, and so would not actually fit inside a rectangle of height 3, because 3=21/7 which is smaller than the diameter.
It seems you want to use the formula for the inner circle radius of a polygon; note that the area in that formular is that of the polygon, not that of the inner circle: Mirror all along the line with length 5, so you get a kite with circumference u_kite = 3+4+3+4 = 14 and area A_kite = 2*(0.5*3*4) = 12. Then the radius of the inner circle is r = 2*A_kite/u_kite = 2*12/14 = 12/7. ==> A_blue = 0.5*PI*r = 0.5*PI*(12/7)^2 = 72/49 PI (in square units).
Until you drew the second radius you didn’t know where the center is. Because we didn’t know if it was exactly a half circle. In fact you only assumed that the two radii were meeting exactly at the hypotenuse.
can you try to find the area above a circle in a right triangle, i got this question on my highschool entrance test. u found the area of the circle, but can you find the area above?
Allô 🎉if you put the right triangle in an orthogonal system il will be easier to find the center of the Demi circle which is (12/7,12/7) and you are done s=1/2*pi*(12/7)^2 🎉🎉🎉🎉🎉🎉
Hmm, eyeballing it I thought the diameter was going to somehow be exactly equal to 4, which would lead to a radius of exactly 2; 12/7 ended up being further off than I thought!!
They tell us it is a semicircle, but is there a way of know that it is really a semicircle? For right angle triangles, if you draw a portion of a circle such that it is tangent to the adjacent and opposite sides, the portion of the circle within the triangle will not always be a semicircle. How do we confirm it is a semicircle?
A portion of a circle that has tangents on the adjacent and opposite will always be a semi circle. We're told it's a semi circle SO THAT we know those points are tangents. As shown in the video, the two tangents will meet at a point on the hypotenuse. Two radii meeting at a point is always the centre of a circle. The hypotenuse is a line through the centre of the circle, which therefore makes the portion a semi circle.
the diagonal isn't the diameter of the circle; but a diameter of the (semi)circle must lie on the diagonal, as the center point of the (semi)circle must be a point on the diagonal. There is a difference.
the blue shape is a semi-circle. so the center point of the semi-circle (or more accurately, the center point of the circle that gives us the semi-circle), must be on the hypotenuse.
There are a number of ways to solve. This was maybe the most rigorous.But you could note that the two small similar triangles must be in length proportion 3:4. That's skipping some rigorous steps but an experienced solver would see that right away.
Idk... if we get another triangle and put it on the opposite side, would that finish the semicircle? Hence, because the height of square made by 2 triangles is 3, the diameter is 3, hence 3²=square area. Then circle: square area ratio is pi:4, implies area = 9pi/4 or (3/2)²pi..? No? Someone please explain..
This doesn't work because if you copy and flip the triangle around to make a rectangle the 2 semicircles won't line up with eachother. Notice that the semicircle is not centered on the hypotenuse, it's much closer to the top corner of the triangle than it is to the bottom corner. So when you flip the copied triangle upside down the flipped semicircle will be closer to the bottom corner than the top. Also consider that the copied semicircle will be touching both legs of the copied triangle, which means that the "completed" circle would somehow be 3 units tall and 4 units wide, which wouldn't be a circle.
how can there be a semi circle if there is a triangle?, you would need to steal from the triangle to create a semi circle, therefore negating the question....the object should be called the "Shaded area"
Haven't seen this yet... got an answer on my own but it can't possibly be right... I can't see the error in my math but it's a weird irrational number.
This ended up not even being relevant but how do we know the triangles formed a square just from the fact the quadrilateral had 4 right angles? Couldn’t it be a rectangle?
Hey what do all circles have in common? The gradient of the degrade of the forever degrading. Edge is the same. That means the distance between devian is equal in all circles. Now you may not realize this, but that says something. It says we may be subject to some b******* Art, when it comes to pie because that formula, it would seem the minimal possible sum is 4. And those. 2 things put together equal some pretty sad s*** I don't know. I just know the result isn't mysterious at all.
But Hey, yo. I'm kind of older so like real deal when I was in elementary school. Calculating the circumference of a circle using Pi was new and shotty as f*** It particularly didn't help you understand the calculation or the concept area. So they had both options and get this. The teacher never instructed us cause she couldn't go into it further with Pi. So yeah, that s*** was new in 1980, so if your books are saying different. Yeah, there are some b******* going around. People said there was money flying around to destroy Western education through the 70s. I therefore I am.
Hey, you wanna see something real crazy? I don't know if you'll be able to notice. What the f*** you're looking at but just s*** is crazy. Okay. So what you do is take point 5. 1.5 basically every point 5. And see how many of these Results in a prime number and look at the interval sets between prime numbers and it's progressive distancy.
Hey, in my opinion, that's just as dazzling as pie. Cause you know, produce in architects too. If they were to use which state regulations determine, they can't use Pi. When they use that formula in the field back in the day they used to modify it to = 4.If it were applied to a circle, then double that number that's not up to building code, though, so don't use that.But when they use Here I'm gonna pick it up for you.Yeah the real world's known that was bullshit the whole time
The 5 was just happy to be invited.
it is 𝘬𝘪𝘯𝘥𝘢 useful since by the converse of Pythagorean you could determine that it is a 3-4-5 right angled triangle
@ The 5 will remember this compliment for many years to come.
@@eu4umyou're RIGHT!
it shows that it is a right angle ig (although there is the 90 degrees sign)
btw im pretty sure you can work it out using the 5 if you wanted to
Because the 5 doesn't exist without the 3, the 4 and the right angle. It it was any other number, the triangle wouldn't exist.
there is a much simplier way. Since the top left triangle and the entire trangle are similar, the ratio between their sides is 3/4, i.e. r + 3r/4 = 3, so r = 12/7.
Yep, solved this in my head using this simple proportion.
@@jeffreypauk679 Special
Why its r+3r?
@antidro_XY look at the figure on 1:18 of the video. The top left triangle is similar to the entire triange, so ratio of sides will be the same. so the left side of the small top left triangle is (3/4) * r. Now look at the left side of the large triangle, the side 3 = r + left side of top left triangle.
Ohh i get it thx@@XxFALCONxX-
As someone preping for JEE, watching this guy makes me not only watch educational content on free time but develop a Genuine interest in maths. Thanks. I occasionally watch videos like this and I actually feel my attention in classes increase.
How interesting!
He is actually teaching how to think to solve a problem whereas our great maths teachers for JEE are absolutely useless
@uvuvwevwevweossaswithglasses are you high maybe ill informed or straight up dumb?
Try watching 3blue1brown and even some math vids by veritasium
The animations are a really nice touch.
I found an easier method I think. I set the origin point to be the right angle so that the x and y axes aligned with the triangle sides. That would mean that the hypotenuse would have a slope of -3/4x + 3. So, the coordinates of the center point must be the point on the line that’s equidistant to the two axes, since the circle is tangent to both. Since the line y=x is always equidistant from both axes, you just have to solve for the point at which that and the hypotenuse intersect, which is (12/7,12/7), which is our radius.
I earned a bachelor's degree in agriculture in the 90's. The highest math I took was calculus. Now my daughter is in middle school and the math is still simple but will be ramping up. I want to be able to help her so I've been watching a lot of these videos and really enjoying the math.
For any such right triangle, the radius of the semicircle with diameter along the hypotenuse will always be: product of lengths of the legs divided by the sum of their lengths. In this case r = (3*4)/(3+4).
diameter doesnt span the whole hypotenuse
@@cleargrits9117 : understood - it's the biggest semi-circle you can have whose diameter lies along the hypotenuse and still fits inside the triangle.
@paparmar ah I see, but then wouldn't that mean the completely quantized version shown in the video would be the same thing? yet (3*4)/(3+4) isn't the same solution as shown in the vid
@@cleargrits9117 : Sorry, I'm not following you - at 3:19 the video concludes r = 12/7, the same as the formula I gave. I'm stopping at the radius, rather than the area of the semi-circle.
@paparmar ohhhh my fault og I got you thank you 🙏🙏
A simpler solution: Draw the 2 radii from the center of the semicircle to the points of tangency and 1 to the right angle of the triangle. The sum of the areas of the 2 triangles created (both of which have height r!) can be written as 3*r/2+4*r/2, or 3.5 r. This should be equal to the area of the 3-4-5 triangle, or 6. 3.5r=6, and from this you get r=12/7. Therefore, πr^2/2=(144/9) π/2=72/9 π
Very nicely explained
Alternate solution: when the figure is reflected about the hypotenuse, we'd get a complete circle which will also be an incircle to the quadrilateral. Then radius of incircle is given by r = A/S where A is area and S is semi perimeter of polygon
You just blew my mind
❤ yes , in fact we have to find r for a right kite which is ab/(a+b) = 12/7
My initial guess was to mirror the triangle along the hypotenuse, so the circle's diameter is approximately 3, resulting in 3π/2 = 4.71, but the actual answer is 4.61, which I consider to be good enough for just coming up with it in not even a twenty seconds
yeah, I had almost the same thought process: just mirror the whole thing :P
Andy, just wanna let you know. You, my good sir, are a fun one :) YOU, my good sir, are exciting
How exciting !
Thank you Andy, just love your videos, I'm not studying maths, and have been a long time out of school. I just love seeing how it all fits together and your approach clearly comes from a genuine love for maths.
Mirror the triangle about its hypotenuse, then we'll have deltoid/kite with inscribed circle. Its area is twice the area of triangle; on the other hand, tangential quadrilateral area is semi-perimeter multiplied by inradius, therefore r = 2S/p = 3·4/(3+4) = 12/7 and semicircle area is 72π/49.
I thought this, and concluded the circle would have diameter=3, but 1.5^2 * pi / 2 isn't equivalent to the other answers given and I don't understand why this doesn't work.
@@TOT3m1c ½π·1.5² means that circle is inscribed in _rectangle_ (i.e. triangle is rotated, not reflected). That's the mistake.
I'm glad you put a box around it at the end.
How exciting.
I had a slightly different approach, without a quadratic equation or similarity of angles: Because the circle is tangent to the x and y axes, the circle's center must have y=x. The linear formula for the hypothenuse is y=3-(3/4)x, so solving x=3-(3/4)x gives the center point x=y=12/7 and this is also the radius, resulting in an area of ½π(12/7)².
The equation of the hypotenuse is Y = 3 -3X/4 (point-slope by inspection). The center of the circle is at Y = X. So X = 3 -3X/4 => X = 12/7. All else are extraneous complications.
Yep, how I did it. good ol' y = mx + b, create a line equation for the hypotenuse, slope is -3/4. Then solve for where x = y to get the radius.
What program or app do you use to animate & process tour graphics? They're very well done, like a bad steak.
probably powerpoint? and he’s clicking through slides.
This was excellent. You explained everything so clearly, logically, and visually. Subscribed.
Bro which software do you use to make these kind of videos ?
I mean the animation , writing equations and solving them !!
I am a math teacher and I want to make videos like this for my students
Thanks for this concise explanation.
If you stick with Euclidian geometry, quadratic eqn is not needed. Similarity tells us:
(3-r)/r=3/4 and r/(4-r)=3/4, r can be solved even without a flip of the pen.
❤
The explanation is perfect.
🎉
Alternate solution.
Start by drawing the radii from the tangents as in the video. This forms two triangles, each with a right angle and one of he angles of the original triangle. This means both triangles are similar to the original triangle and to each other since the remaining angle of each new triangle is the complement of the angle it shares with the original triangle, and the complement of that shared angle is the shared angle of the other triangle.
The upper triangle has a length of r.
The lower triangle has a height of r.
The length of the lower triangle is 4/3 of it's height of r, because it's similar to the original triangle.
The length of the 2 new triangles equal 4 because together they add up to the length of the original triangle.
4 = r +4/3 r
= 7/3 r
12 = 7r
r = 12/7
area = pi * 1/2 (12/7)^2
Alternative solution: you can treat the hypotenus like a line with the equation "y = -(3/4)x + 3" and since we know the center of the circle where be where "y = x" we just use substitution. The answer will be the radius.
Here’s what I did. Put the triangle on a coordinate system with the right angle being corner being the origin. The hypotenuse is then represented as the line y=-3/4x+3 and the center point of the circle lies along the line at point (r,r). Plug in r for x and y to get r=-3/4r+3 and solve for r getting r=12/7
Based on the fact that the circle was equal to pi, my immediate estimation was that the semi-circle was ~1.5pi. Cool to see my intuitive guess wasn't too far off!
I found that you could also find the r by dividing two vertical and horizontal sides of the similar triangles, so that r/3=(4-r)/4; r=12/7
Very nice work! What are you using for the animations? Is it just PowerPoint/Keynote?
Great video and animations. Thank you
It' s easier to write the similiarity between one of the two small triangles and the bigger 3-4-5 triangle. For example, (3-r)/r = 3/4 or r/(4-r) = 3/4. That way you have a first grade equation only
ARE YOU HAVING FUN THERE TIMOTHY! :)
NICE RALPH WIGGUM HAIRCUT BTW
Damn it you are so great, love the video man don't ever stop
3 is to 4 as R is to (4-R) is simpler.
With areas: The area of the triangle = (3x4)/2= (rx3)/2 + (rx4)/2 - the areas of the smaller triangles. r=12/7, Area of semicircle = π x 144/ 49
How exciting!
Hey, do you have any videos on solving geometric progressions or sequences? Came up today in real life for me (had to solve for a common ratio for a series). It was kind of neat, but maybe not exciting 😅
I just connected the center and the right angle to make two triangles. One with a base of 3 and a height of r, and the other with a base of 4 and a height of r.
I added them together to get the bigger triangle, which has an area of 3×4÷2=6 squared units.
3r÷2+4r÷2=7r÷2=6, r=12÷7
would it also be possible to find 12/7 by setting the sums of the 2 triangles (r(3-r) and r(4-r)) and one square (r*r) equal to the area of the whole triangle (3*4/2)?
Yep it’s possible just remember to use half of r(3 - r) & r(4 - r)
Great explanation but The u squared is not necessary and will confuse people.
You can also use Pythagoras theorem on each triangle as sqrt((3-r)^2+r^2)+sqrt(r^2+(4-r)^2)=5 but the calculation is rather tedious compared to the method shown.
I don't understand how we know/can prove that the lines from the point of tangency on each side both A) make a perfect radius at the hypotenuse and B) meet at the hypotenuse
It's because it's a semi-circle (i.e., exactly half of the circle) and not just a random portion of a circle. Since it's a semi circle, the center of the circle has to lie on the line cutting the circle in half, which is also the hypotenuse. The other parts are explained (tangent perpendiculars must go through the center of a circle).
Agree that the center of the full circle is on the hypotenuse, since it's a semi-circle. Still unsure why a line perpendicular to the point of tangency of the two sides are guaranteed to meet the hypotenuse at the center of the "circle" and that they both meet at the same point.
That's a rule about tangent lines. A line that's perpendicular to a tangent line at the point of tangency goes through the center of the circle, he kinda describes that at 0:43. So when you have two lines that are perpendicular to two tangents, they both pass through the center of the circle; therefore, the point of intersection is/must be the center of the circle.
@@josephulton I agree with you. I get the center is on the hypotenuse, I get the tangent lines are perpendicular from the sides, but I don’t get why the perpendicular lines have to meet at the hypotenuse. This was not explained.
upon reflecting on it more, I realize that these things must be true about a Semi-Circle described in the video:
A. a line perpendicular to the ONLY (1) point of tangency will create a perfect Diameter, thus going through the center of that circle
B. the center of a circle will be on the hypotenuse since it is listed as a semi-circle
somehow didn't grasp that from the video, even though it is touched on
In general, r = ab/(a+b) so
A = πa^2b^2/(2(a+b)^2).
r=12/7 , A= 6 - 4.6 = 1.4 Sq.units approx
Inscribe a square with side length the radius of the semicircle
3-r / r = r / 4-r
(3-r)(4-r) = r²
12 - 7r + r² = r²
r = 12/7
½pir² = 72pi/49
Love your vids. Grow big
How exciting!
There's ones a legend says "Imagination life is ur creation"
Oh, wow, turns out I got it, good for me. It was just somewhat irrational.
I did it kind of a different way. We know the area of the whole triangle is bh/2, (3*4)/2, 6. And we know it's made up of the two triangles and the square. So r^2 (area of the square), plus the area of both triangles (r * (4-r)) / 2 and (r * (3-r)) /2 will all equal 6. If we distribute both r's, we can add the two chunks that are both over 2, and get r^2 + (7r-2r^2)/2. Multiply r^2 by 2/2 so it'll be over 2 so we can add the fraction, and the 2r^2-2r^2 cancels out, leaving us with 7r/2 = 6. Multiply both sides by 2/7 to get r alone, gives us r = 12/7. And with r just get the area of the semi-circle as he showed us.
Nothing irrational there, everything can be expressed as a ratio of integers 😅
I just know that the semicircle fits into the Square hole!
I love the -how exited at the end
I thought the answer was gonna be 2pi just eyeballin' it, which actually wasn't too far off!
Please someone point to me where my reasoning is broken. First, mirror the triangle around the hypotenuse. Now you've got a rectangle with sides 3 and 4. Since the original triangle is a square triangle, the horizontal lines are parallel and are separated by 3 units. Since the circle is tangent to both the original horizontal line and the new mirrored horizontal line, then its diameter is 3, its radius is 3/2 and therefore the area is (pi . (3/2)^2) / 2 = 9/8 . pi.
If you mirror the triangle around the hypotenuse (i.e. the side of length 5 is your line of reflection) you get a kite as your quadrilateral, not a rectangle.
@rbrown2948 True, basic mistake on my part.
r : 4-r = 3 : 4 4r = 3(4-r) 4r = 12 - 3r 7r = 12 r = 12/7
Area of BlueSemicircle = 12/7 * 12/7 * π * 1/2 = 72π/49
Cross multipy makes me cross - explain this correctly to avoid confussion
I just guessed it would be sqrt(2)*pi for no particular reason and I’m surprised at how close that actually was, haha.
Dude would make a very good Pirat.
r
Is it okay to just assume that it was a circle inside the rectangle and get the area of the circle then divide it by 2?
No; the circle of radius 12/7 has a diameter of 24/7, and so would not actually fit inside a rectangle of height 3, because 3=21/7 which is smaller than the diameter.
It seems you want to use the formula for the inner circle radius of a polygon; note that the area in that formular is that of the polygon, not that of the inner circle:
Mirror all along the line with length 5, so you get a kite with circumference u_kite = 3+4+3+4 = 14 and area A_kite = 2*(0.5*3*4) = 12.
Then the radius of the inner circle is r = 2*A_kite/u_kite = 2*12/14 = 12/7.
==> A_blue = 0.5*PI*r = 0.5*PI*(12/7)^2 = 72/49 PI (in square units).
Until you drew the second radius you didn’t know where the center is. Because we didn’t know if it was exactly a half circle.
In fact you only assumed that the two radii were meeting exactly at the hypotenuse.
r/4 = (3-r)/3, 3r = 12-4r, r = 12/7.
can you try to find the area above a circle in a right triangle, i got this question on my highschool entrance test. u found the area of the circle, but can you find the area above?
Allô 🎉if you put the right triangle in an orthogonal system il will be easier to find the center of the Demi circle which is (12/7,12/7) and you are done s=1/2*pi*(12/7)^2 🎉🎉🎉🎉🎉🎉
This is reupload right?
Are you using green screen?
how exciting indeed...
Hmm, eyeballing it I thought the diameter was going to somehow be exactly equal to 4, which would lead to a radius of exactly 2; 12/7 ended up being further off than I thought!!
It goes into the square hole
They tell us it is a semicircle, but is there a way of know that it is really a semicircle?
For right angle triangles, if you draw a portion of a circle such that it is tangent to the adjacent and opposite sides, the portion of the circle within the triangle will not always be a semicircle. How do we confirm it is a semicircle?
A portion of a circle that has tangents on the adjacent and opposite will always be a semi circle. We're told it's a semi circle SO THAT we know those points are tangents. As shown in the video, the two tangents will meet at a point on the hypotenuse. Two radii meeting at a point is always the centre of a circle. The hypotenuse is a line through the centre of the circle, which therefore makes the portion a semi circle.
Could you tweak the problem to change from solving for area of semi circle to finding area of everything other than blue?
Or triangle - semi circle.
How exciting
curious to see the result was (3*4)/(3+4)
(3-r)/r=3/4 is easier to get r
how did you know the diagonal was the diameter of the circle?
the diagonal isn't the diameter of the circle; but a diameter of the (semi)circle must lie on the diagonal, as the center point of the (semi)circle must be a point on the diagonal. There is a difference.
Anything special about a 345 triangle?
How do you know it is exactly half of the circle?
It’s given
Great
bro always changing place
I don't know how you can assume that the hypotenuse crosses the centre of the circle?
the blue shape is a semi-circle. so the center point of the semi-circle (or more accurately, the center point of the circle that gives us the semi-circle), must be on the hypotenuse.
Andy you miss to see a simp!e way to solve this
There are a number of ways to solve. This was maybe the most rigorous.But you could note that the two small similar triangles must be in length proportion 3:4. That's skipping some rigorous steps but an experienced solver would see that right away.
Idk... if we get another triangle and put it on the opposite side, would that finish the semicircle? Hence, because the height of square made by 2 triangles is 3, the diameter is 3, hence 3²=square area. Then circle: square area ratio is pi:4, implies area = 9pi/4 or (3/2)²pi..? No? Someone please explain..
This doesn't work because if you copy and flip the triangle around to make a rectangle the 2 semicircles won't line up with eachother. Notice that the semicircle is not centered on the hypotenuse, it's much closer to the top corner of the triangle than it is to the bottom corner. So when you flip the copied triangle upside down the flipped semicircle will be closer to the bottom corner than the top. Also consider that the copied semicircle will be touching both legs of the copied triangle, which means that the "completed" circle would somehow be 3 units tall and 4 units wide, which wouldn't be a circle.
how can there be a semi circle if there is a triangle?, you would need to steal from the triangle to create a semi circle, therefore negating the question....the object should be called the "Shaded area"
how do we know for sure that it is a square and not a rectangle?
Super disappointing that the answer isn't something straightforward like a whole number multiple of pi.
...or 69, or 420.
Not every answer is exciting.
@@chrishelbling3879 heh
Ok😊
the way i would simply put pi/2
Haven't seen this yet... got an answer on my own but it can't possibly be right... I can't see the error in my math but it's a weird irrational number.
This ended up not even being relevant but how do we know the triangles formed a square just from the fact the quadrilateral had 4 right angles? Couldn’t it be a rectangle?
it has 4 right angles plus both the height and width are r
Hey what do all circles have in common? The gradient of the degrade of the forever degrading. Edge is the same. That means the distance between devian is equal in all circles. Now you may not realize this, but that says something. It says we may be subject to some b******* Art, when it comes to pie because that formula, it would seem the minimal possible sum is 4. And those.
2 things put together equal some pretty sad s*** I don't know. I just know the result isn't mysterious at all.
But Hey, yo. I'm kind of older so like real deal when I was in elementary school. Calculating the circumference of a circle using Pi was new and shotty as f*** It particularly didn't help you understand the calculation or the concept area. So they had both options and get this. The teacher never instructed us cause she couldn't go into it further with Pi. So yeah, that s*** was new in 1980, so if your books are saying different. Yeah, there are some b******* going around. People said there was money flying around to destroy Western education through the 70s. I therefore I am.
Hey, you wanna see something real crazy? I don't know if you'll be able to notice. What the f*** you're looking at but just s*** is crazy. Okay. So what you do is take point 5. 1.5 basically every point 5. And see how many of these Results in a prime number and look at the interval sets between prime numbers and it's progressive distancy.
Oh, I'm saying to multiply it by 2. By 0.5 × 2 = one. The first prime number 1.5 × 2 = 3 another prime number.
Hey, in my opinion, that's just as dazzling as pie. Cause you know, produce in architects too. If they were to use which state regulations determine, they can't use Pi. When they use that formula in the field back in the day they used to modify it to = 4.If it were applied to a circle, then double that number that's not up to building code, though, so don't use that.But when they use Here I'm gonna pick it up for you.Yeah the real world's known that was bullshit the whole time
Damn. Too bad my semicircle is red…😔
I tried a method that was 5 =√r²+(3-r)² + √(4-r)²+r² but idk if it works
Thanks, not the neat answer (maybe 2*pi) I was expecting. Nicely explained.
I'm sorry for being inappropriate in a past comment.
Hi room in the bg ....
4^x^2 + x = 8
Hello yeah
There are easier ways
The answer is meaningless.
I got 1 for radius tho
Did not get a pretty answer.
Just reflect the triangle to make a rectangle and now you got a circle of diameter 3
The only faulty logic I saw was "Since it has 3 right angles, it HAS to be a square". Any rectangle has four right angle.
1 pair of adjacent sides were both equal to r.
geometry super boring 😂