This Geometry Challenge Took a While

[@saminko0791]

Andy: invents cure for cancer Andy: looks important lets put a box around it

[@TauBlitzPlayz]

fr fr

[@JordanBiserkov]

Andy: "How exciting." in the most flat tone of voice he can manage. ;-)

[@medoostuf]

FBI OPEN UP

[@IamAlive-zk8jw]

i am in grade 9 and solved without looing the solution

[@genaroperez8325]

"This geometry challenge took awhile" sounds scarier than all of the other challenges i've seen in your channel

[@konradyearwood5845]

You could also do it by scale drawing! 😃Once you bisect the three angles the bisectors would intersect at the centre of the circle. The perpendiculars to the sides from the centre of the circle would then give you the radius. At school we did it by drawing before we did the mathematical proof as it was a good method of reinforcing the theorems and trigonometrical identities at play.

[@kenhaley4]

@@konradyearwood5845 Good approach to get a close estimate...as long as you remember it's an estimate--not exact. But using it to verify the math derivation or proof is a great idea.

[@unverifieduser69]

Indians already cracked it in grade 10 mathematics 😂

[@goodshiro10]

Only those with brains, unlike me😂😂​@@unverifieduser69

[@Rainoverse]

1​@@konradyearwood5845

[@citizenwriter2540]

Using Heron's formula for the triangle - A=sqrt{s*(s-a)*(s-b)*(s-c)}, where s= (a+b+c)/2, we have A =sqrt(21*8*7*6)=84 sq units. Now adding the areas of the three triangles, we get A=1/2*R*(13+14+15)=21R. Hence 21R=84 or R=4. Hence area is 16 pi sq units.

[@dkaloger5720]

Absolutely, I was typing this until I saw your comment .There also a formula for inscribed circles(basically the same as yours ) , A = s*r

[@Cahangir]

Exactly how i solved it, takes much less time.

[@danmat65752]

Same!

[@normalbattlecat8088]

I ended up memorizing the 13-14-15 identity after seeing it like 5 different times in math contests

[@nextwenxd4777]

i just did a=s*r so 84=21r

[@jerrypolverino6025]

I’m a retired airline pilot educated in Aerospace Science with a BS. 77 years old. I haven’t been doing math since college. I got the first few of your videos wrong. That was a wake up call for my brain and an ego killer. Ouch! Now, I’m finally getting most of them right although it takes me awhile to get the cobwebs out of my head. Love your videos man. How exciting! lol

[@troybaxter]

Same for me except I have only been an Automation Engineer for 2 years. It has really helped me get back into that mathematical mindset as I got so used to plug and chug, and only doing math that I deemed helpful to me for engineering. Andy's videos have started to make me fall in love with math again.

[@leolaudann3173]

@@troybaxter Looks important, let's put a box around it. How exciting

[@IamAlive-zk8jw]

i am in grade 9 and solved without looing the solution

[@abhinayragam8078]

Man I love people in STEM coming back to roots. Problem solving is a beautiful art , I hope we get the joy of solving more problems lol ( although I'm just a sophomore in Computer science and engineering lol )

[@paparmar]

By breaking the triangle into 3 "kites", you can quickly show that 7r + 6r + 8r = 21r is the same as the area of the triangle, which Heron's formula gives as 84. Hence, r = 4. This way you don't have to bring angles into it at all (other than recognizing the right angle where each radius meets the sides of the triangle - and thereby forms two sides of each 4-sided kite).

[@deniseockey6204]

Where do you get the 7r and 6r? The only way to know for sure is if the triangle were drawn to scale. Being that the sides are 14,13, and 15 where could you draw the kite?

[@paparmar]

Freeze the video at 6:42. Apply x = 6 to the diagram and you can see where the 7 and 8 come from. Break the triangle into 3 right-angled kites (draw the perpendicular from the incenter to each side of the triangle - Andy did this for just one of them). The top kite has sides 6, r, r, and 6, which means since it is a right-angled kite, its area is 6r. Similarly, the kite on the left has sides 7, r, r, 7, with area 7r, and the kite on the right has sides 8, r, r, 8, with area 8r. Thus the area of the triangle is the combined area of the 3 kites, or 21r.

[@CrustyCheapster]

I don’t have much interest in math, but I watch these videos almost exclusively for the satisfying payoff when he says “how exciting.”

[@brianglendenning1632]

These videos are rekindling my long lost interest in mathematics.

[@alineharam]

@@brianglendenning1632 how exciting, me too.

[@robinlydian4452]

That was a totally wild ride, really satisfying to use every high school trig rule at your disposal to solve such a weird-looking problem

[@kevinsmith9385]

That's a wonderful approach, Andy. I especially like the use of trig to find sides and side ratios, not angles. And really, no calculator needed. How exciting indeed!

[@alineharam]

No calculator or tables required. I lost my calculator and I eat off my table so it works for me.

[@AdsCoulter]

What about using Heron’s formula? Semiperimeter (s): General formula: s = (a + b + c) / 2. With values: s = (13 + 14 + 15) / 2 = 21. Area (A) of the triangle (using Heron’s formula): General formula: A = sqrt(s * (s - a) * (s - b) * (s - c)). With values: A = sqrt(21 * (21 - 13) * (21 - 14) * (21 - 15)) = sqrt(21 * 8 * 7 * 6). Radius (r) of the incircle: General formula: r = A / s. Substituting the area and semiperimeter, we find r that simplifies to the formula for calculating the area of the circle. Area of the circle: General formula: π * r^2. With the calculated r, the area of the circle simplifies to 16π square units.

[@AdsCoulter]

And I just read the comments, apparently everyone else had the same idea.

[@HoSza1]

With Heron's formula the derivation is more straithforward, no need to use the law of cosines: A² = s(s-a)(s-b)(s-c) where s = (a+b+c)/2. Also A = rs and equating both and multiplying both sides by 𝜋, gives the area of the circle which is r²𝜋 = 𝜋(s-a)(s-b)(s-c)/s, and s = 21 ⇒ r²𝜋 = 16𝜋. End of story.

[@suryavardhansinghshekhawat865]

Never heard of that formula before. Thanks for the new info

[@qwerty112311]

Nerd. His solution was exciting, your one is just bleh

[@denissecalle9457]

@@qwerty112311 Hater. There's no need to be so bitchy, he just gave his own solution get tf out of here.

[@filip_1432]

I was about to say this. The radius of the circle inscribed in a triangle is the area of the triangle over the semi perimeter of it

[@HoSza1]

@@filip_1432 exactly!

[@barryomahony4983]

This was a fun one. I originally went down the trig rabbit hole, but ended up using the Base*Height/2 formula. Computed the Height of the whole triangle to find its area. The big triangle is also made up of 3 smaller triangles of total area 13r/2 + 14r/2 + 15r/2. Then just solve for r.

[@mcjohngd3583]

Its insane how you make math so exciting, seriously, I love this channel

[@starboy001]

This was the most beautiful and useful geometric and overall problem, that I enjoyed. And the beauty of telling us the part of calculation on screen and not skipping it makes your videos 100% worth getting Abel price for best teaching method. ❤❤❤❤❤

[@LosOnTheCoast]

This problem takes a lot of concepts from Geometry to solve, requiring a pretty thorough understanding of the subject. Great job! You made me want to touch up on my Geometry skills 😁

[@SkiroGaming]

Though you can solve it much easily without geometry using tangents.

[@xarlixe7565]

@@SkiroGaming How ?

[@SkiroGaming]

@@xarlixe7565 That is easy but a long method. So I can't explain in comments. But another shorter method is by using Heron's formula. Divide the circle into three triangles and the height of all those triangles would be equal to radius of circle. Equate the sum of area of these three small triangle equal to larger traingle whose area would be found by Heron's formula. Then you would get radius and then you can find area of circle. If you would know all of this, the answer can be found under 1 minute.

[@1959mikel]

Looked simple at beginning but ended up being more complicated than I expected, but as usual you did a great job simplifying it! How exciting!😊

[@bausHuck]

Man, I love these videos. And I love the sign off because so many people don't find math exciting, but I do (at least when I see a master do it).

[@all_the_moga]

These videos and this channel is the equivalent of watching beauty magazines. But now, instead of feeling ugly, I feel dumb AF.

[@Yucuuf]

@@all_the_moga lmao

[@croissantdx]

We can also take into account that the centre of the circle is the incentre of the triangle, which isn't just any triangle, but a very special one called Heron's Triangle, where its sides are in arithmethic progression (13, 14, 15), and the height relative to 14 is equal to 12. By tracing the bisectors, which all have as a common point the incentre 'I', we can see that the triangle is formed by three smaller triangles, where their bases are the triangle sides and the heights are all inradii (r). So we get this: A▵ = 13r/2 + 14r/2 + 15r/2 = bh/2 where r is inradius and b = 14, h = 12. hence, we get: 42 * r = 14 * 12, simpifying: r = 4 after finding the inradius, we can calculate the area of the circle using the formula: Aₒ = πr² Aₒ = π * 4² Aₒ = 16π sq. units //

[@PerKeltMusic]

YES!!! I tried it before watching the video and found a much faster solution: just find the average value of sides of the triangle (14) and calculate it as if the triangle had all sides 14 (it’s just a simple tan30=r/7) result is the same and I feel good about myself :)

[@gleydin]

The tangent thingy was a grear refresher! Thank you!

[@rishikeshmuppana1155]

Draw the height to the side with length 14, you get two right triangles, 9-12-15 and 5-12-13. So the area of the triangle is 14*12/2=84. Now draw the radii from the incenter to each vertex and to each tangency point. This creates 3 triangles, each with the radius as the height and one of thr sides of the triangle as the base. Let r be the inradius. Using bh/2 for the triangles gives 13r/2+14r/2+15r/2=21r. Thus 21r=84 and r=4. Area of green circle is pi(4)^2=16pi.

[@bismuthe]

I know a really quick approach to this problem. Everytime there is a circle inscribed in a triangle, the area of a triangle with a circle inscribed in it is: ((a+b+c)/2)*r with a,b,c being the sides of the circle and r being the radius of the inscribed circle. We also know that the area of the triangle can be computed using Heron's formula. Through substitution u get the radius. Hopefully this helps

[@JobBouwman]

The area is pi*(6*7*8)/(6+7+8) = 16*pi. Here 6, 7 and 8 are the three distances from the corners of the triangle to the tangent points of the circle.

[@kenhaley4]

What do you use to animate your solutions? You make things so clear!

[@syazry149]

I used to love Math very much since i was a kid, however stress of college life and other stuff make me forgot those feeling. Thanks for reminding me how fun math is to me. 😊

[@hemantrevankar3800]

a better method let one coordinate be (0,0) and one be (15,0) so applying distance formula we can get the 3rd coordinate as x^2 + y^2 = 169 (x-15)^2 + y^2 = 196 solving this we get 3rd coordinate as (33/5 , 56/5) now apply in-centre formula I =[(ax1 + bx2 + cx3)/(a+b+c) , (ay1 + by2 + cy3)/(a+b+c)] so coordinate of in-centre comes out to be I=(7,4) since we took one coordinate as (0,0) ie the origin and other as (15,0) the line joining this is the x axis the distance between the in-centre and x axis is the radius here that is = 4 (by simple observation) hence area = pi(4)^2 = 16pi

[@broderick2464]

That’s the same method I think of

[@1ReasonableGuy]

Very nice problem and solution. I like that you always go straight to the point instead of fluffing around repeating the same thing 4 or more times, as other people do here in KZbin. Heck, there are people who would had made a 1 hour long video for this problem!

[@cakes37]

i never got past algebra 2 but i really enjoy these videos.....its awesome to see someone who is really good at something

[@brendanbeardy9190]

At this point, doing the math must be like meditation for you. Must feel great finishing a question like that.

[@danielsemendinger4540]

Here is my version of the kite approach. Drawing all three bisectors and the three radii tangent to the three sides, creates 6 triangles. Pair the triangles up and you get 3 kites. Calculate the lengths of the sides of the kites (or the 6 triangles) as Andy did. The area of the 13, 14, 15 triangle is 1/2*6*r + 1/2*6*r +1/2*7*r + 1/2*7*r + 1/2*8*r +1/2*8*r = 21r The area of the 13, 14, 15 triangle is also 1/2 * base * height Using side 15 as base, get the height using two pythagorian theorems. 13^2 + (15-w)^2 = 14^2 + w^2 w = 6.6 (14^2-6.6^2)^0.5=11.2 Area 13,14,15 triangle is 0.5*11.2*15=84 84=21r r=4 area=16π This was not my first solution. We will not go there.

[@danielsemendinger4540]

Sorry Should be 13^2 - (15-w)^2 = 14^2 - w^2 w = 8.4 (14^2-8.4^2)^0.5=11.2 This is why I could never be a mathematician

[@MiroslavBlagoev-t7t]

heron formula to find the area of the triangle gives Area = 84. Area is also equal to (half the perimeter) * r, from which we find r = 4, therefore Area of the circle is = 16 Pi

[@robertlynch7520]

There's another reasonably interesting way to do this, though a bit off-beat as far as KZbin geometers go. Strategy • find height of the triangle • use that to find θ (say on left corner) and φ (right corner); • tangent of ½θ is the slope of a line from corner to center of incircle… likewise × -tangent of ½φ is the 'other slope' of the incircle center to right corner • mathematically cross 'em, to find 𝒙 • and use that times the first slope to find 𝒓, the radius. Doing it: [1.1]  𝒉² = 𝒂² - 𝒔² … where 𝒔 is an unknown bit of 𝒃 [1.2]  𝒉² = 𝒄² - (𝒃 - 𝒔)² … the other half of 𝒃 Set those two to each other, and expanding, moving things around, solve [1.3]  𝒔 = (𝒂² - 𝒄² + 𝒃²) ÷ 2𝒃 the height then follows [2.1]  𝒉 = √(𝒂² - 𝒔²) Having that, we can now find θ [3.1]  θ = arctan( 𝒉 / 𝒔 ) And the slope of the incircle-center line from corner is [4.1]  𝒎₁ = tan( ½θ ) … for the line [4.2]  f(𝒙) = 𝒎₁𝒙 ⊕ 0 The exact same logic can be used on the right side finding φ and tan( ½φ ) to give a slope. [5.1]  𝒎₂ = -tan( ½φ ) [5.2]  𝒃₂ = -𝒎₂ • 𝒃 … intercept; [5.3]  g(𝒙) = 𝒎₂𝒙 + 𝒃₂ Now we have formulæ for lines that can be mathematically crossed [6.1]  𝒙 = 𝒃₂ / (𝒎₁ - 𝒎₂) and of course, the radius is [6.1] times 𝒎₁ ⊕ 0. In PERL: (just a convenient (for me) calculator) --------- CODE --------------------------------------------------------------------- my $a = 13; my $b = 15; my ¢ = 14; my $h; my $x; my $r; my $s; $s = ($a •• 2 - ¢ •• 2 + $b •• 2) / (2 × $b); $h = √($a •• 2 - $s •• 2); $x = $b • tan( ½ • atan2( $h, $b - $s ) ); $x /= ( tan( ½*atan2( $h, $s ) ) + tan( ½*atan2( $h, $b - $s ) ) ); $r = $x • tan( ½ • atan2( $h, $s ) ) ⊕ 0; --------- OUTPUT --------------------------------------------------------------------- a = 13 b = 15 c = 14 s = 6.6 h = 11.2 x = 7 r = 4

[@MrPaulc222]

There is a much simpler way, at least with these particular numbers: I used Heron's formula for the triangle: which gave (sqrt(21)(6)(7)(8)). Break down as 3*7*3*2*7*4*2 so sqrt(7056) = 84. Split into three triangles whose areas total 21r. 84/21=4, so r=4. Circle area 16pi un^2

[@TusharKanoi]

There is a much simpler way to do this with a bit of calculation First devide the traingle into 3 more parts such that each smaller triangle has the radius of the circle as it's height ( r is perpendicular to t). Then find the area of the triangle using herons formula. Find the area of each smaller triangle in terms of r (1/2base x height) and add them Substitute the values and find r

[@RAG981]

Heron's formula makes it a lot simpler to find the area of triangle, and use semi perimeter times r = area of triangle.

[@richoneplanet7561]

Your reasoning is just enjoyable - 😳 that is unbelievable!

[@dyslexic_corn]

im literally one of the worst students in my school and this guy actually makes me excited to learn geometry

[@medoostuf]

You don't have to use any trig formulas or annoying variables, just use herons formula to find the area which provides A = 84 Then split up the triangle into three seperate triangles along the center of the circle. Notice that the radius of the circle is the altitude of all three triangles Then set up the equation 1/2(13+14+15)r=84 where r is the radius and the altitude of all three mini triangles. Divide 42 from both sides(13+14+15) and multiply both sides by two. r = 4 so the radius is 16pi.

[@davidhovey4645]

Interestingly, this problem can be solved in a pretty clean manner. You can actually break a 13 14 15 triangle into a 9 12 15 right triangle and a 5 12 13 right triangle. Then, you can find that the area of the triangle is 84 since bh/2 = (12)(14)/2 = 84. Since inradius * semiperimeter = Area of the triangle, (21)r=84 and r=4. Therefore, the area of the circle is 16(pi).

[@ventsislavminev]

If you have all 3 sides of the triangle just use Herons formula to find the area of the triangle. Then the radius of the inscribed circle is 2S/(a+b+c). No need for all the trigonometry. Heron did it for us 2000 years ago.

[@arihantpatel8043]

as a school going class 10th Indian student... this was a piece of cake. (I solved it using heron's formula and the basic formula for area of triangle (1/2*b*h)... no need for all that trigonometry)

[@Sumaia915]

I solved it in a totally different way but had the same result.. what i did is I divided the shape into smaller squares that had smaller quarter of circles and then solved it

[@CliffSedge-nu5fv]

Long, complicated way to do it. I broke it up into 3 pairs of congruent right triangles and set the sum of those areas equal to the total area of the triangle. From that, I solved for the radius of the circle. r/2(2a + 2b + 2c) = r(a + b + c) = sqrt[s(s-2a)(s-2b)(s-2c)] = rs: s = a + b + c. Square both sides, divide by s and multiply by pi. Done.

[@ryannarby4519]

Please make a video for 2:53! Again one of the best channels on KZbin. One of the few completely ethical avenues of entertainment in my life.

[@JaharNarishma]

This is the first video I've seen where my way of doing it is the same as Andy's. How exciting!

[@RDM1776]

I recently discovered your videos and love it. Can you give us a sense of what grade (if any) each problem rises too? I have a couple elementary age kids so I don't want to scare them too soon!

[@ahmedfahadkhan4558]

Found an easeir way. First find the included angle of side 13 and 15 which is 59.48.... Using the laws of cosine then usinf the trignometic ratio of sin find the shortest distance of triangle (the height) which is sin59.48 x 13 and the shortest distance i.e the perpendicular will be 11.2. Then find the included angle of 13 and 14 which would be 67.38... Then divide this by two as we know from cirlce properties (as told by andy) that tangents from same points have every angle and side same thus the angle wil be 33.69... Then as seen in the picture we can subract the radius,r, from from 11.2 i.e 11.2 -r to get the hypotnease of the triangle of the congurent triangles and then using the trignometic ratio of sin on the congurent triangles we can do sin33.69..=r/11.2-r which will give us the radius 3.99 rounded of to 4 and usinf the area formula we will get the answer 16pi

[@robertyoung9611]

Don't always follow the route to the solutions, but always find it satisfying.

[@knotwilg3596]

It depend on what you know oc. If you know the formula for the radius of the incircle in terms of the sides, it's just a computation :). If you know Heron's formula, then you can see how the incircle helps cutting the triangle into 3 pieces with height R and base = the sides. If you don't know Heron's formula but you can solve this problem, then you're on par with the ancient Greek mathematicians.

[@fatcatgaming695]

bro, wtf just happened. That was awesome work

[@BuenoSama-xk7re]

Holy shit dude that's probably the hardest math question I've seen until now

[@nabil4389]

How exciting!!

[@fuglbird]

The radius of a circle inscribed in a triangle is two times the area of the triangle divided by the perimeter of the triangle. In your case r = 2 x 84 / 42 = 4.

[@geoblk3000]

You are a better teacher than all of my teachers from middle school to college combined.

[@yoshi27661]

my journey in solving this problem is even longer. i first figured out what the two lengths of a side are considering the three altitudes intersecting them, then i found the measure of the angles and the lengths of the altitudes, then I figured out the lengths of the angle bisectors, then i converted the angle bisectors into equations considering the lower left corner as (0,0), then i found the intersection, then i plugged the y value into the area of a circle. this took me roughly 2 hours. didnt know heron's formula, didnt remember law of cosines or double angle formula

[@I_Am_Milnek]

I see i'm not the only one who found the solution with heron's formula. I actually didn't know about that formula, and had to look it up. I also wouldn't have found the solution without the first note about tangent lines in the beginning.

[@swaaaarts4]

I will never look at a calculator the same way ever again after watching this video.

[@DugRut]

i created a spread sheet after watching your videos. I just need to enter any triangle data and it gives me all the other information. Triangle 13, 14, 15 = inside circle area: 50.265

[@andrewkoziel7470]

Another way to solve it: The area of the triangle A = 13*r/2 + 14*r/2 + 15*r/2 = 21r; Also, A = SQRT(s*(s-a)*(s-b)*s-c)). This is Heron's formula. Where a, b, c are the triangle sides and s = (a+b+c)/2; Than s = (13+14+15)/2 = 21; A = SQRT(s*(s-a)*(s-b)*s-c)) = SQRT(21*(21-13)*(21-14)*21-15)) = SQRT(21*8*7*6) = 84; Than 21r=84 than r = 4; The circle area = PI*4*4 = 16*PI

[@vireaknou8835]

To be honest I don’t think this is a hard problem but indeed interesting. I just use heron formula and this formula I don’t really know what it call “s = (r*(a + b + c))/2” where r is the radius of incircle of a triangle, s is the surface area of the triangle that has the incircle that we are dealing with, a b and c are the length of each side of the triangle. So the problem ask us to find the radius of incircle of the triangle which we can use the second formula but first we have to rearrange it so s = (r*(a + b + c))/2 therefore r = 2*s/(a + b + c) but we don’t know the surface area of the triangle yet but in the problem they tell the length of each side of the triangle so we can use heron formula to calculate which s = squarerootof((semi_perimeter_of_triangle)*(semi_perimeter_of_triangle-a)*(semi_perimeter_of_triangle-b)*(semi_perimeter_of_triangle-c)) so s = squarerootof(((13+14+15)/2)*(((13+14+15)/2)-13)*(((13+14+15)/2)-14)*(((13+14+15)/2)-15)) so s = 84 unit square so plug numbers into the second formula we got r = 2*(84)/(13+14+15) so r is 4 unit length. Edit: My bad this problem actually ask for the surface area of the incircle but since we already figured out the radius of the circle we just need to use formula circle_surface_area = pi*(radius’s_length)^2 where pi is approximately 3.14 so circle_surface_area (approximately) = 3.14*(4)^2 = 3.14*16 = 50.24 square unit.

[@Origen17]

At 5:23 I nearly thought we had the answer, and wondered why you kept going - then I realized it was a ratio...

[@classicdrivingdevelopment8165]

"Just did a math problem I will never be able to do" -how exciting-

[@fahmidnaiefchowdhury]

But I have done that math in just 10 seconds with this formula. The radius of that inscribed circle is, r = Area of that Triangle / Semi-perimeter of that triangle. By evaluating r, we can now determine the area of that circle.

[@aounelias]

loved it!

[@reyray7184]

"How exciting!" Andy Math Idk about you but marh gives me a big ol rubbery one. 😂

[@nayeem7359]

Step 1: Find the area of the triangle (A) using Heron's formula: A=√(s(s-a)(s-b)(s-c)) where s=(a+b+c)/2 Step 2: use A=sr to find r, which is the radius of the inner circle Step 3: area of the inner circle =πr²

[@angeloschena1630]

Ok, this is a long but nice solution. I solve this with Heron’s formula and perimeter to find apothem (radius of circle inscribed in to a triangle) .

[@devgupta2490]

Take areas of AOB, BOC, AOC, add them. You get area of ABC, which you know, and you have a linear equation for r.

[@Sg190th]

2:02 I was expecting a "nice"

[@bebektoxic2136]

Yea me too 😂

[@MrAuswest]

Is it purely accidental that the numeric values of the sides and area of the inscribed circle are 13,14,15,16? Or is this repeated in other triangles? Sure i could do the arithmetic and trig myself (proof might be harder to do though) but it's late here and I'm tired! ;-) (it clearly doesn't for the 3-4-5 triangle btw: 3, 4, 5, Pi.)

[@AnkhArcRod]

If half angle theta(t) at vertex with sides of length 13 and 15 and half angle alpha(a) at vertex with sides of length 15 and 14, tan(t) = r/7; tan(a) = r/8. Here, r is radius of incircle. Also, tan(90-t-a) = r/6. Thus, 6 = r*tan*t+a) = r * (tan(t) +tan(a))/(1-tan(t)tan(a)). This leads to 6 = (15*r^2)/(56-r^2). This leads to r = 4. Thus area is 16pi.

[@kinanradaideh5479]

Is there a general way to find the area of a circle inscribed into a triangle, given triangle side lengths?

[@aleksadjordjevic8286]

There is. Use Heron's formula first to find out the area of the triangle (since we have all 3 sides) A=sqrt(S*(S-a)*(S-b)*(S-c)); a,b,c are the sides of the triangle and S is half of the perimeter S=(a+b+c)/2 Now that you have the total area, use the formula A=S*r to find out the radius of the inscribed circle Then it's simple, use the formula A○=pi*r^2 to get the area of the inscribed circle This is true for any triangle and a general way to find out the area of an inscribed circle (if you have all 3 sides)

[@judekirkcruz7177]

I love this channel

[@JoeDubs432]

Since the 345 triangle is involved in squaring the Circle and since that uses the golden ratio, the ratio of the circumference to diameter Hass to be congruent with the golden number. Pi is 3.144... or 4/phi squared

[@someonespadre]

I did this using my typical brute force math via Herron’s formula but the video is way more beautiful.

[@conatser]

I... feel like you're a wizard.

[@IsaacTounsi]

He's not? 😂

[@izaamshafeer3575]

We can find the radius of the circle by applying herons formula to the sides of the triangle and also finding the area of the triangle by using 1/2.b.h formula to all the three triangles inside it and finding the total area of the triangle and then find r. how exciting...

[@yepyepmusic]

That's exactly what I learn in school but it's really fun

[@eatmorebread2]

Is it just coincidence that the sides of the triangle are 13, 14, and 15, and the area of the circle is 16?

[@koliasic7456]

yes but also there is a formula S=pr, where s is the area of the whole triangle, p is half perimeter, r is radius

[@jesusvillarruel7476]

Ni siquiera hablo el idioma, pero todo quedó clarísimo, fue increíble.

[@c49tejasjha7]

You could take radius perpendicular to Side and dividing the whole triangle into 3 triangle then by herons formula, you could solve the problem It would take comparatively less time

[@terranosuchus]

It's so satisfying that the radius is what it is!

[@DaManCave123]

we dont need to do this much work, we can use the formula:- inradius r = Area of tri/s(semiperimeter), So here it would be := 84/21 = 4 now area = pi*r*r := 16*pi. easy peasy learnt this through my NTSE prep

[@abhi111182]

What about A=rs? It's way easier if you use it

[@Every2Days...]

You could EASILY solve this in only two steps. • Find the area of the triangle • Use the following formula: S (area of the triangle) = P (½ of a perimeter) × r (radius of the circle inscribed within a triangle) Easy

[@filipeoliveira7001]

Or you could just use Heron’s formula😭 but great video!

[@Eishan17]

We had this question in class 10 its easy not to prolong but you can just the use the formula area of the triangle = perimeter * radius of circle /2 to find the radius and then find area

[@FruitZeus]

I wasn’t even going to attempt this one. I knew it was too hard lol.

[@gegessen159]

It always amazes me if such challenges start AND end with even numbers. I expected root or at least some fraction as result

[@DDD111-n5v]

I’ve watched a surprising number of these in a row.

[@chimken7666]

I'm not even gonna pretend to understand but cool vid

[@bpark10001]

Heron's formula followed by area/s = radius is WAY simpler. Area = 84, s = 21, radius = 4 & area of circle = 16 pi.

[@mikelin8884]

This one blew my mind six ways to Sunday

[@ez258]

heron equation + area of triangle derived from radius and perimeter

[@oliveroadrigo]

this couldve been done way faster, since we know theres a formula for the radius of an inscribed circle (2 x area / perimeter) and the area was easy to solve, using herons formula (Area= sqrt(s x (s-a)x(s-b)x(s-c), where s is semi perimeter) and the area, in this case would be 84, and the perimeter 42. 168 / 42 = 4

[@oliveroadrigo]

and then you find the area of the circle whatever

[@kovako6723]

You don't need any law of cosinus. If you do the second part, calculating the length of parts divided by r in each side (6,7,8), and also calculate the area of the whole trangle by adding up the area of the small triangles 6r+7r+8r=21r, then apply the general rule for area of a triangle, here: 15 x height of the triangle/2. The height of the triangle = r+ square root of( 36+r2). Then you have two eguation: 21r= r+ square root of( 36+r2) x 15/2. Which also gives r=4

[@jerryfields4837]

next time i have insomnia im gonna watch this one

[@stevez6460]

I definitely feel like there is a way to solve it faster but I like this method.

[@TheTallRaver]

Cool!👍