This Geometry Challenge Took a While

Another Catriona Agg Problem. I hope you guys like it!

[@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

[@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.

[@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.

[@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.

[@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.

[@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

[@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.

[@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

[@mcjohngd3583]

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

[@michaellacaria910]

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).

[@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.

[@richoneplanet7561]

Your reasoning is just enjoyable - 😳 that is unbelievable!

[@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.

[@syazry1498]

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. 😊

[@kenhaley4]

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

[@saminko0791]

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

[@cakes37]

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

[@timothyshiu2263]

Very interesting! Do you have any recommendation for algebra, calculus, finite math, etc

[@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!

[@Sg190th]

2:02 I was expecting a "nice"

[@bebektoxic2136]

Yea me too 😂

[@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!

[@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 :)

[@brendanbeardy9190]

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

[@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.

[@BuenoSama-xk7re]

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

[@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

[@JaharNarishma]

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

[@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.

[@geoblk3000]

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

[@user-qj2ti1ro4o]

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

[@fatcatgaming695]

bro, wtf just happened. That was awesome work

[@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.

[@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).

[@gegessen159]

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

[@tommaniacal]

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

[@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.

[@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.

[@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) .

[@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

[@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

[@Mrqwerty2109]

This was the kind of math problems I'd see on the homework and my teacher would be like "why is the homework not done yet"

[@reyray7184]

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

[@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.

[@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.

[@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

[@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

[@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

[@someonespadre]

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

[@hans-joachimdreher2287]

For a triangle with side lengths a, a+1, a+2 the incircle radius r = 1/2*sqrt((a+3)(a-1)/3) Heron‘s Theorem is used for the proof.

[@user-iy1vu5vx1b]

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

[@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

[@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.

[@MuhammadAlam-ne9cf]

this one was a banger thanks andy

[@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

[@yepyepmusic]

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

[@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

[@bpark10001]

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

[@kinanradaideh5479]

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

[@filipeoliveira7001]

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

[@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

[@conatser]

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

[@mikelin8884]

This one blew my mind six ways to Sunday

[@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

[@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.

[@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

[@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...

[@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.)

[@eatmorebread2]

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

[@ivansabo8225]

Premjera igara iz Crnorečja je bila 1993.godine u SNP Novi Sad, u izvođenju ANIP "Sonja Marinković". Muzičku obradu je napisao prof. Ivan Sabo, a frulu i gajde je svirao Slobodan Trkulja. Koreografiju je uradila Desanka Đorđević

[@DanG1001]

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

[@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

[@Origen17]

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

[@stevez6460]

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

[@wizardtarun]

Or we can do by heron's formula , ∆=rs where r is radius of incircle, s = semi perimeter,∆= area of triangle

[@nabil4389]

How exciting!!

[@what.-_-._.118]

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

[@hridanshsurana2797]

A similar question is in the NCERT Book for class 10 maths. I was just scrolling through yt one day before my maths board exam and saw this diagram. I didn't instantly remember how to do it but still screamed 16π.

[@Brett_G_Barnes]

NO TRIG SOLUTION: Draw a perpendicular altitude line from the base of the triangle to the top vertex which splits the triangle into 2 additional triangles. h = the height of all 3 triangles drawn x = the distance from the left vertex to the perpendicular altitude line (15 - x) = the distance from the perpendicular altitude line to the right vertex h² + x² = 13² ; h² + (15 - x)² = 14² h² = 13² - x² ; h² = 14² - (15 - x)² 13² - x² = 14² - (15 - x)² 13² - x² = 14² - 15² + 30x - x² 13² = 14² - 15² + 30x 13² - 14² + 15² = 30x 169 - 196 + 225 = 30x 198 = 30x x = 198/30 x = 33/5 h² = 13² - x² h² = 13² - (33/5)² h² = 169 - (1089/25) h² = (169)(25/25) - (1089/25) h² = (4225/25) - (1089/25) h² = (4225 - 1089)/25 h² = 3136/25 h = √3136/√25 h = 56/5 area of the triangle = sum of the areas of the 3 inner triangles that can be constructed by drawing a line from the center of the incircle to each of the 3 vertices r = radius of the incircle (1/2)(15)(56/5) = (1/2)(13)(r) + (1/2)(14)(r) + (1/2)(15)(r) (1/2)(15)(56/5) = (1/2)(r)(13 + 14 + 15) (3)(56) = (r)(42) 168 = 42r r = 168/42 r = 4 A = area of the incircle A = πr² A = π4² A = 16π units²

[@jerryfields4837]

next time i have insomnia im gonna watch this one

[@akhipazham7270]

We can use the fact that the (Radius of Circle)*(Semi-perimeter of Triangle) = Area of Triangle We can find the Area using Heron's formula to be 84 cm^2, and the Semi-Perimeter is 21 cm So Radius of Circle is 4 cm, and Area is 16pi cm^2

[@aounelias]

loved it!

[@nfnawaf9154]

U can solve this in less than a minute just by taking half the base of the triangle and u will get an approximate number to the diameter of the circle wich will lead to the area (its not precise but efficient )

[@benjamingross3384]

You did this the hard way. Area of a triangle with Heron's formula. Then A=sr where r is the inradius and s is the semi perimeter. Then pi*r^2

[@aura_beast2678]

Guys listen just remember this and know how to prove it : radius of inscribed circle = area of triangle / semi perimeter

[@sasaritosa8055]

idk if this is legal but i gave angles the value of their opposing side as in 15k, 14k and 13k, then added them up and solved for k; 42k=180 and then i figured the true angle size, then i used the sin theorem and got the value of the outside radius R and then got the area of tge triangle with A=abc/4R, and at last i used the formula A=s×r with r being the inner radius and then i got the area of the inner circle wich came up as right! Please excuse me for my english and correct me if i did smt wrong!

[@greendruid33]

That was a tough one!

[@allrounderchannel4198]

Use heronsformula and circle center thorum in triangle (1:2)

[@sohansaha4693]

this one's a 10th grade problem for us over here in india (saw it in my textbook, and got surprised seeing it here)

[@ShiroiAkumaSama]

Is it possible to make out of the 14 a 13.5 and out of the 13 a 13.5? And then out of the 15 a 14 and the one we took away there can't I add 0.5 to the 13.5 of both sides now so that all sides are now 14?

[@meepy2739]

Crying while watching this because I'm torturing my mind.

[@karanjain166]

Bro please give JEE advanced you suits it 😊

[@kujmous]

I avoid trig whenever possible.

[@user-zn6bw8gp6g]

Dividing fractions don't know why, flip the second number and multiply.

[@TheTallRaver]

Cool!👍

[@mauriziobelluomini]

everything is much simpler using Heron's formula.

[@OrenLikes]

When you got to r/6=2/3, instead of cross multiplying and then dividing both sides by 3, you could have just multiplied both sides by 6.

[@shashanknayak1818]

Isnt there formula for in-radius abc/4R= area of triangls