Fun Blue Semicircle

[@eu4um]

The 5 was just happy to be invited.

[@wither67]

it is 𝘬𝘪𝘯𝘥𝘢 useful since by the converse of Pythagorean you could determine that it is a 3-4-5 right angled triangle

[@eu4um]

@ The 5 will remember this compliment for many years to come.

[@jerwinjaudines6098]

​@@eu4umyou're RIGHT!

[@Ludaris1]

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

[@urquimedes4459]

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.

[@XxFALCONxX-]

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.

[@jeffreypauk679]

Yep, solved this in my head using this simple proportion.

[@baselinesweb]

@@jeffreypauk679 Special

[@antidro_XY]

Why its r+3r?

[@XxFALCONxX-]

​ @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.

[@antidro_XY]

Ohh i get it thx​@@XxFALCONxX-

[@torreyintahoe]

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.

[@jonaslundholm]

I know, right. I wish I had understood the beauty of math when I was in school.

[@n8blank]

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!

[@uvuvwevwevweossaswithglasses]

He is actually teaching how to think to solve a problem whereas our great maths teachers for JEE are absolutely useless

[@n8blank]

@uvuvwevwevweossaswithglasses are you high maybe ill informed or straight up dumb?

[@PrinceGupta-oc1tx]

Try watching 3blue1brown and even some math vids by veritasium

[@welcometosebs]

yeah it develops alot of thinking skills by watching these typa videos, i personally saw a big progress in my maths result of class 10! i have a deep interest in physics, chemistry is also so fun too ive been thinking to try out jee too, whenever ill be eligible

[@n8blank]

@welcometosebs You definitely should! I wish you best of luck & skills. it's kinda sad that many don't see pure science as fun & only as a "hard boring subject." However those subjects are one of the most interesting things to learn as a human.

[@frankstrawnation]

The animations are a really nice touch.

[@paparmar]

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).

[@cleargrits9117]

diameter doesnt span the whole hypotenuse

[@paparmar]

@@cleargrits9117 : understood - it's the biggest semi-circle you can have whose diameter lies along the hypotenuse and still fits inside the triangle.

[@cleargrits9117]

@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

[@paparmar]

@@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.

[@cleargrits9117]

@paparmar ohhhh my fault og I got you thank you 🙏🙏

[@ŚrīĆaitanyaŚikṣāmṛtam]

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

[@zeppa]

You just blew my mind

[@raghvendrasingh1289]

❤ yes , in fact we have to find r for a right kite which is ab/(a+b) = 12/7

[@ratandmonkey2982]

The new, reflected semi-circle won't fit in the rectangle when reflected. Another identical semi-circle will fit in the rectangle at the same time. However, the 2 identical semi-circles will not form one continuous circle and fit in the rectangle at the same time. A non-square rectangle can not have an incircle.

[@raghvendrasingh1289]

@ratandmonkey2982 after reflection we will get a right kite and kite is a tangential quadrilateral.

[@ŚrīĆaitanyaŚikṣāmṛtam]

@@ratandmonkey2982 hey, that's the exact reason why I refrained from using the word rectangle as after reflection we'd get a kite and not a rectangle as rectangle isn't symmetrical diagonally. Any rectangle which isn't a square cannot have an incircle as opposite sides are equal failing the property of tangents having equal length from the point of inception. Hope that clarifies, good point though!

[@Random_person_hehehe]

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

[@ghostracer_978]

Very nicely explained

[@dogspaghetti7118]

Andy, just wanna let you know. You, my good sir, are a fun one :) YOU, my good sir, are exciting

[@briogochill6450]

How exciting !

[@-wx-78-]

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.

[@TOT3m1c]

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.

[@-wx-78-]

@@TOT3m1c ½π·1.5² means that circle is inscribed in _rectangle_ (i.e. triangle is rotated, not reflected). That's the mistake.

[@astrolad293]

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.

[@Rev03FFL]

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.

[@MegaGandhiman]

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.

[@rezashiri6093]

If you get the any of small triangles similar to the real original triangle, it’s become easier as r^2 does not even form. That was great. Good on you

[@loopscoop3175]

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.

[@eeedee3001]

Alternative method ✓easier...

[@Caine61]

I'm glad you put a box around it at the end.

[@afelix]

How exciting.

[@newperve]

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

[@maxfuentes5435]

The radius is also the harmonic mean shown geometrically. The harmonic mean is the reciprocal of the sum of the reciprocals of the two numbers. 1/((1/a)+(1/b)) It simplifies to ab/(a+b) or 12/7 in this case

[@SpeckyYT]

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

[@idromano]

yeah, I had almost the same thought process: just mirror the whole thing :P

[@hajovonta6300]

@@idromano I did this too, but then I was thinking whether is it a given that the center of the circle is on the hypotenuse? I wasn't sure, but it wasn't proven in the video either.

[@floppy4everyone913]

@@hajovonta6300 im pretty sure that because the problem states that the blue area is a semi circle, its safe to assume that the hypotenuse has the full circle's center on it

[@hinaftouseef2803]

Hi, i thought the same! But it’s not right. 5 is the diameter of the semicircle

[@gargleballss]

@@hinaftouseef28035cm is the length of the hypotenuse

[@travisstoll3582]

This was excellent. You explained everything so clearly, logically, and visually. Subscribed.

[@stromboli183]

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.

[@divym0]

This is a highly assumptious solution HIGHLY ASSUMPTIOUS.

[@something7419]

No

[@Echoes_act_3378]

How

[@lovenottheworld5723]

@@Echoes_act_3378 It's not a semicircle.

[@Echoes_act_3378]

@@lovenottheworld5723I guess he did assume that hmmm

[@bluenewton5951]

@@lovenottheworld5723 because he did two perpendicular lines on the circle he confirmed its center. since a line perpendicular to a circle will intersect the center, the point where two of those overlap is the center.

[@Khashayarissi-ob4yj]

👏👏👏 Good luck. Hoping for more videos.

[@matthewgrumbling4993]

Thanks for this concise explanation.

[@drpjmtaylor]

Damn it you are so great, love the video man don't ever stop

[@vanshagrawal1841]

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

[@chrishelbling3879]

What program or app do you use to animate & process tour graphics? They're very well done, like a bad steak.

[@asdf242]

probably powerpoint? and he’s clicking through slides.

[@seandevice238]

You can use the Area of triangles to solve it. Total area of Triangle = Triangle with 3 as base +Triangle with 4 as base. The height of both triangles are the radius of the circle. => 6 = 3r/2 + 4r/2 the radius is 12/7.

[@JeffY-ri2nj]

You assumed in your solution that the center of the circle fell on the triangle hypotenuse. That assumption needs proven or stated as the given.

[@kj3-o7m]

it is essentially stated as given as the question said it’s a semicircle which automatically implies that the centre lies on the hypotenuse

[@Katherine-qs8ws]

@@kj3-o7mhow is this possible? If it was truly a semicircle the sides of the triangle should be equal

[@leocomerford]

@@kj3-o7mAnd at 0:36 he states that it is a diameter.

[@leocomerford]

@@Katherine-qs8wsThat’s not so. If you draw a circle of radius (spoiler) 12/7 and then draw a vertical tangent line to its left and a horizontal tangent line underneath it, there is any number of top-left-to-bottom-right diagonal lines which you can draw through the centre of the circle. Only the line with a slope of -1 will make the triangle isosceles.

[@sergiofernandez116]

a slightly faster proportion would be to compare the large triangle 3,4,5 to either of the small r, 4-r or 3-r, r. For example, using the r, 4-r triangle the proportion would be 3/r=4/4-r yielding r=12/7

[@blazingfire7517]

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

[@senhual_v]

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

[@TheVaivoda]

Great video and animations. Thank you

[@CipherX]

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)?

[@jacobom2141]

Yep it’s possible just remember to use half of r(3 - r) & r(4 - r)

[@KipIngram]

Ok, this is at least a little interesting. When you push a circle into a right angle corner like this, the two tangent points have to be equal distances from the origin. We'll call that distance u. So the center of the circle will be at (u, u) and the circle's area will be pi*u^2. The semicircle will have area 0.5*pi*u^2. So we just need to find u. Since it's a semicircle, the center of the circle will be on the hypotenuse of the triangle. The equation of that hypotenuse line is y = 3 - 3*x/4. There will be only one point along the hypotenuse where x = y: x = 3 - 3*x/4 x + (3/4)*x = 3 (7/4)*x = 3 x = 12/7 = u Semicircle_Area = 0.5*pi*u^2 = (72/49)*pi,

[@ZachHixsonTutorials]

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.

[@kaideng2571]

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

[@johndoe-ow2ns]

How do you know that the center of the circle lies on the hypotenuse?

[@timblebug]

I tried to solve the problem in my head before watching the video for fun. If you are like me, and you tried to quickly flip the image along the hypotenuse in your head and assumed _r = 1.5_, *this doesn't work because the circle would be broken up if you flipped the image along the hypotenuse*. If you actually complete the image (ie triangle goes to a 3x4 rectangle and semi-circle to circle) you'd see the circle with r = 12/7 sticking outside of the rectangle. Here's visual proof if you're like me and you wanted to see it: Open Desmos and graph y=3-0.75x (for the triangle, the line will be the hypotenuse and the x and y axis will be the opposite and adjacent sides), and then graph (x - 12/7)^2 + (y-12/7) = (12/7)^2 (for the circle). Graph some guiding lines (y = 12/7, x = 12/7, and y = 3) to prove to yourself that the center of the circle is on the hypotenuse, and the area of the total circle after mirroring it extends beyond the y=3 line. Now change the circle equation to (x - 1.5)^2 + (y-1.5) = (1.5)^2 and the guiding lines to y = 1.5 and x = 1.5. This will show the center of the circle is no longer on the hypotenuse, and the area of the circle in the triangle is not a semi-circle.

[@TheHunch8ack]

Very nice work! What are you using for the animations? Is it just PowerPoint/Keynote?

[@mehrzadm8899]

Thanks for the videos.

[@komalkasyap3365]

Amazing 🤩

[@josephulton]

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

[@bradramsay8299]

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).

[@josephulton]

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.

[@bradramsay8299]

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.

[@powguma]

⁠​⁠@@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.

[@josephulton]

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

[@Jahwobbly]

How do you know the center of the circle is on the triangle's 5-unit sized side? Are you assuming it?

[@teb7426]

At 41 secs, how do you know that line will intersect the centre of the circle. Please could you explain that bit. The rest is calculable for the layman after those initial assumptions are made. Thank you v much

[@rhaining]

What I want to know is, how can you be certain there actually is a true semicircle with a midpoint on the hypotenuse that is tangent to both sides of the triangle? Imagine you had a much pointier right triangle -- say the 5 12 13 right triangle or something even vastly pointier. Is it always possible to inscribe a true semi-circle within any right triangle? That seems intuitive wrong or at least possibly impossible so to speak, but of course one can't always trust one's intuition on such things.

[@Hanzi2103]

What program is he using to illustrate?

[@lazarduke6596]

what a great video !!

[@AndyPanda9]

I don't know a thing about math etc. but I really enjoy your videos and it makes me want to learn :) I wondered why you assumed the circle was exactly half but Google confirmed "semicircle" does, in fact, mean half. Why isn't it a "hemicircle" since in regular language "hemi" always means half? "semi" and "demi" usually mean partial.

[@AndyPanda9]

A further Google search tells me that "semicircle" comes from Latin where it means half and "hemisphere" comes from Greek for half.

[@salatielgenerale6053]

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?

[@TimMaddux]

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.

[@derwolf7810]

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).

[@WhatsDaveUpTo]

The circle has height 3 because its bounded by a 3x4 rectangle. So the diameter is 3. Find the area of a circle with radius 1.5 and then divide by 2.

[@ramennoodle9918]

the height isn’t bounded by 3 but it is close

[@WhatsDaveUpTo]

@@ramennoodle9918 sure it is. copy the image and fold it over to make a rectangle with diagonal 5. Otherwise its not a semi circle. which is by definition half a circle

[@krrothwell1]

@@WhatsDaveUpTo folding does not make a rectangle

[@professordogwood8985]

0:50 OK, so a tangent point is where a circle intersects with line of the triangle? (There's a lot of geometry that was kept from us in math)

[@FM_MC]

Nah it wasn't kept from you bro if you didn't know this you weren't paying attention

[@professordogwood8985]

@@FM_MC Mister, I'm not kidding you. In my high-school geometry there was never a lecture about how sin, cos, and tangent had to do with triangles within a circle. They just wanted us to know what to plug into our TI-83 calculators.

[@RecycleBin0]

ARE YOU HAVING FUN THERE TIMOTHY! :) NICE RALPH WIGGUM HAIRCUT BTW

[@FranklinFranklin-kb8ej]

How about using Pythagoras Theorem to find r like this 5^2=(3−r)^2+(4−r)^2 and then find the area of half of the circle after solving for the area of the circle?

[@ridgmont61]

Nice explanation

[@orionred2489]

Was there something saying the center of the circle is on the hypotenuse?

[@braveuser8250]

How exciting!

[@Youtuber-ku4nk]

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.

[@OctavianUngureanu]

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

[@Alessandro-1977]

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

[@SaumayaMahaseth]

Can it be done this way :- The sides of 3 units and 4 units are tangents to the semicircle, so we draw 2 radii to the point of contact of both the tangents and we get a square . Now ,we Break the base into 2 parts y and 4-y ,where 4-y is the side of the square and do the same with the side of 3 units and write the side of the square to be equal to 3-x and the rest to be x . Now we get 2 right angled triangles, one with the base and height of 3-x and x respectively and the other with the base and height of 4-y and y respectively. Now the sum of the hypotenuse of the triangles is 5 so now we make an equation where we add the square of the bases and heights of both the triagles and write it to be equal to 25 (5^2) . And in the place of y we write x+1( bc 3-x and 4-y are the radii of the semicircle and hence equal and if we simplify it we get y-x=1) and then we can find the value of x plug it in the'equation '3-x = radius of thr circle' and get the radius and find the area of the semicircle

[@digbycrankshaft7572]

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.

[@richyglitched]

I'm just happy to be here.

[@cyruschang1904]

(3 - x)(5/3) + (4 - x)(5/4) = 5 20(3 - x) + 15(4 - x) = 5(12) 35x = 60 x = 12/7 radius = (3 - 12/7)(4/3) = 4 - 16/7 Semicircle area = (π/2)(4 - 16/7)^2 = (2π)(6/7)^2 = 72π/49

[@johnmarley8268]

Why did you do the extra steps? Your sub-triangles are both similar to the entire triangle. (3-r)/r = 3/4 --> 7r = 12. That was a lot less work.

[@enlightenment3777]

Definition of a semicircle, is exactly half a circle. Reflect along hypotenuse, you get rectangle with full circle, touching short side in centre, therefore r = 1.5.

[@bryanscott8675]

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 😅

[@thetaomegatheta]

I did it with coordinates this time. We know the function f: [0; 4] -> [0; 3] that describes the coordinates of the point (x; f(x)) on the hypotenuse of the triangle in a Cartesian system of coordinates, and we know that r = f(r) for the center of the semicircle. After that, we just solve a simple linear equation, get the radius, and leave it to engineers to find the area of the semicircle.

[@bethichspeedrun]

Love your vids. Grow big

[@omnarasi9637]

that was nice illustration and explanation. I spent about 3 minutes before watching the video trying to do it without paper and pencil, but could not. I guess, I am a little rusty 😞

[@DKSahasi]

❤ The explanation is perfect. 🎉

[@amauta5]

Once you have the sides can’t you just find r and then half the area of a circle?

[@lboy9889]

What if you got the triangle and circle and mirrored them along the hypotenuse to form a rectangle with sides 3 and 4, and a complete circle inscribed inside. Wouldn’t the diameter of the circle be 3 so the area would be 9*pi/8? Can someone explain why this wouldn’t work?

[@jim2376]

Cool beans. Well done.

[@rboothby]

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?

[@brandonm8901]

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.

[@CrDa-i7e]

Assuming the centre of the semi-circle is on the Hypotenuse

[@shoyo4ever15]

It’s so simple at the end of the video. At the beginning I was … no way I could’ve figured it out

[@darreltimothy9241]

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?

[@paradox5389]

How do you know it is exactly half of the circle?

[@zeppa]

It’s given

[@munionr]

I wondered the same thing. It is simply the definition of a semi circle. A semi circle half of a circle. I was thinking it could simply be part of a circle. Nope. Half of a circle.

[@Orillians]

This is reupload right?

[@gaylespencer6188]

3 is to 4 as R is to (4-R) is simpler.

[@peteranon8455]

Simply put, the blue area is not a semi circle on a 3,4,5 triangle.

[@eve_the_eevee_rh]

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

[@camjua00]

how did you know the diagonal was the diameter of the circle?

[@rbrown2948]

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.

[@F1_letsgo]

Anything special about a 345 triangle?

[@LeoSnyder-ef1bl]

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.

[@blade2368]

how do we know for sure that it is a square and not a rectangle?

[@jimiwills]

I used the areas of the little triangles being equal to the big triangle minus the square. Ended up with 6=r^2+r(3-r)/2+r(4-r)/2, so after multiplying everything by 2, exactly the same from there on.

[@supriyomandal1641]

Can't we just put 22/7 in place of pie ?

[@Trey_Vu]

I'm confused. Wouldn't the solution he presented be the area of the entire CIRCLE as opposed to a semi-circle?

[@shivrajsharma7891]

Are you using green screen?

[@Richard_AKL]

I don't know how you can assume that the hypotenuse crosses the centre of the circle?

[@rbrown2948]

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.

[@AkemiZhuang]

I had the exact same question for my 9th grade semi-finals however I solved it a little differently, sin theta one is 3/5 and cos theta one is 4/5. Radius is 2.5 which is the angle opposite to theta one, solve for r and use pirsq./2

[@DrAmanpreetSingh-tt7ef]

This was actually really simple

[@jerrybrown6169]

Not watched the video but I would measure the diameter along that hypotenuse divide by 2 to find the radius, then use pr^2 and divide by 2.

[@GreenMeansGOF]

In general, r = ab/(a+b) so A = πa^2b^2/(2(a+b)^2).

[@It.s-just-me]

You state that a quadrilateral with three right angles has to be a square. I would agree that it has to be a rectangle. I'd agree that it could be a square. Please tell me what it is about having three right angles, that makes it necessarily a square. I'm always paying particular attention to hear the rules that aren't evident in a diagram, but which are known to a regular practitioner of geometry. Those are like keys, so I listen closely, and duplicate them.

[@miki_the_little198]

0:30 how do we know that

[@IPlayzGames9]

Tangent lines are always perpendicular with a radius, and since it is a semicircle, the middle would be somewhere in the five side. The only place the radius and side collide is in that point

[@welabird4389]

r=12/7 , A= 6 - 4.6 = 1.4 Sq.units approx

[@grupocelebremos1]

I love the -how exited at the end

[@redknight07_]

I noticed something, the r = 12/7 is also 4x3/4+3 I dont know where this comes from but its interesting