Can you find the radius of the inscribed circle? | (Heron's Formula) |

[@HarHarMahadev-931]

Nice sir Plss pin

[@PreMath]

Done 😀

[@PreMath]

😀

[@ybodoN]

The inradius _r_ of the incircle in a triangle with sides of length _a, b, c_ is given by the formula *_r = √[(s − a) (s − b) (s − c) / s]_* where *_s = ½ (a + b + c)_* is the semiperimeter This follows from Heron's formula. In our case, _r_ = ⅙ √(231) units. Thank you PreMath! 🙏

[@PreMath]

Excellent! You are very welcome! Thanks for sharing ❤️

[@Abby-hi4sf]

Great one! You are teaching a proof theory easily! Amazing teaching method! Thank you!

[@unknownidentity2846]

Let's find the radius: . .. ... .... ..... When the area A and the perimeter P of the triangle are known, the radius R of the inscribed circle can be calculated as follows: A = (1/2)*P*R ⇒ R = 2*A/P The perimeter P and the area A can be obtained from the given side lengths: P = a + b + c = 8 + 9 + 10 = 27 A = √[s*(s − a)*(s − b)*(s − c)] with s = (a + b + c)/2 = P/2 A = √[(27/2)*(27/2 − 8)*(27/2 − 9)*(27/2 − 10)] = √[(27/2)*(27/2 − 16/2)*(27/2 − 18/2)*(27/2 − 20/2)] = √[(27/2)*(11/2)*(9/2)*(7/2)] = (9/4)√(3*11*7) = (9/4)√231 Now we are able to calculate the radius: R = 2*A/P = 2*(9/4)√231/27 = √231/6 ≈ 2.53

[@PreMath]

Excellent! Thanks for sharing ❤️

[@robertlynch7520]

There's a VERY different alternative way to solve this using trigonometry. Basically, (using any orientation, but where one side is horizontal) the line functions of the half-angles of the two base angles intersect at some 𝒙. The height of the intersecting lines is the radius. So what's the slope of a θ/2 line? Well, it turns out (spectacular FAIL) that several online "trigonometry identities" references got this wrong. Doing it by hand from [1.1]  tan 2θ = 2 tan θ / (1 - tan² θ) … can be re-written as the half-angle formula [1.2]  tan θ = 2 tan θ/2 / (1 - tan² θ/2) … then rearranged to [1.3]  tan θ/2 = (-1 + √(1 + tan² θ)) / tan θ From some basic geometry of finding a height of an arbitrary triangle we get [2.1]  𝒒 = ( 𝒂² - 𝒃² + 𝒄² ) ÷ 2𝒄 … q is the distance 'in from the left' to find height [2.2]  𝒉 = √( 𝒂² - 𝒒² ) ... is height Where [𝒒] is the distance from (0, 0) along the horizontal axis where the [𝒉] height intersects the peak. This is NOT yet a solution. The tangent is [𝒉 / 𝒒]. So, that plugs into [1.3] readily enough to give the half-angle tangent. Good enough. This results in a 'half angle line'. Do the same thing for the right side, and you get the 'other half angle line' function. Cross them mathematically, and you get the intersection point; the height then is the mutual radius of the circle that fits inside the triangle. NOTE. If you do this … its good to test that your math is right by changing the lengths of the sides 8-9-10, 10-8-9 and so on. If you didn't make mistakes in the algebra or math, the SAME radius pops out at the end for all combinations. Here's the PERL solution, using this technique: (Note that I did NOT use the tan() function directly!) ________________________________________ !/usr/bin/perl; # on a MAC; my $a = 8; my $b = 10; my ¢ = 9; my $q = ($a••2 - $b••2 + ¢••2 ) / (2 × ¢); my $h = √( $a••2 - $q••2 ); my $eta = $q••2 / $h • (-1 + √(1 + ($h/$q)••2)); my $del = (¢ - $q)••2 / $h • (-1 + √(1 + ($h / (¢ - $q) )••2 ) ); sub dx { my $x = shift; return $del • (¢ - $x) / (¢ - $q ); } sub ex { my $x = shift; return $x • $eta/$q; } sub fx { my $x = shift; return ($h • $x) / ( $q); } sub gx { my $x = shift; return $h • (¢ - $x) / (¢ - $q); } my $dx = 0.1; # an iterative way to find 'where the lines cross'. my $x; # its not human-fast, but computer do it in my $y; # very short time. Thousandths of a second; for( $x = 0; $x < ¢; $x += $dx ) { ¥ = ex( $x ); my $yy = dx( $x ); next if $yy > $y; $x -= $dx; $dx /= 2.718281828; # can be anything between 2 and 3 last if $dx < 1e-11; } --------- OUTPUT --------------------------------------------------------------------- q = 2.5 h = 7.5993 eta = 1.8094 delta = 2.9937 x at crossover dx() ex() = 3.5 NOTE … not 'q'! Sweet… y at crossover dx() ex() = 2.5331 This is the RADIUS of the in-circle. And lookee there ... same answer as PreMath!!!

[@PreMath]

Thanks for putting heart and soul❤️ You are awesome. Keep smiling👍

[@prossvay8744]

triangle area=√s(s-a)(s-b)(s-c) s=(a+b+c)/2=(8+9+10)/2=13.5 Triangle area=√13.5(13.5-8)(13.5-9)(13.5-10)=9√231/4 square units 1/2(r)(8+9+10)=9√231/4 27r=9√231/2 3r=√231/2 so r=√231/6 =2.53 units. ❤❤❤ Thanks sir.

[@PreMath]

Great job You are very welcome! Thanks for sharing ❤️

[@uwelinzbauer3973]

Hello! I solved the question using trigonometry: c²=a²+b²-2ab cos(gamma) and a sin(beta)=b sin(alpha) Btw, which is the proper name: Rule of cosine, law, theorem, sentence, formula,...? - I can't remember learning Herons formula at school, either the teacher didn't tell it or perhaps I was ill that time. But it is a really interesting and useful method 👍 Thanks a lot for the interesting video, greetings and best wishes!

[@ludmilaivanova1603]

Sorry, this is not an alternative method because you use the area found earlier. It would be an alternative if you find the hight of a big triangle by using Pyphagorean theorem.

[@PreMath]

Thanks ❤️

[@dirklutz2818]

AB=10, so AD=5.85 and DB=4.15 (Pyth.th twice). Height triangle is DC=sqrt(46,7775)=9/20 * sqrt(231) Area triangle= 10/2 * 9/20* sqrt(231) = 9/4*sqrt(231)

[@LuisdeBritoCamacho]

1) H = Sqrt [13,5*(13,5-10)*(13,5-9)*(13,5-8))] ; H = Sqrt(1.169,4375) Square Units 2) H ~ 34.19704 Square Units 3) 8*R + 9*R + 10*R = 2 * 34.19704 ; 27*R = 68,35 ; R = 68,35/27 ; R ~ 2,533 Linear Units 4) My Best Answer is : The Radius is Approx. Equal to 2,533 Linear Units.

[@PreMath]

Excellent! Thanks for sharing ❤️

[@CloudBushyMath]

Great , Mind work-out

[@PreMath]

Yes indeed! Thanks ❤️

[@jamestalbott4499]

Thank you!

[@PreMath]

You are very welcome!🌹 Thanks ❤️

[@zdrastvutye]

the following code does not calculate bisectors of angles, it just uses the pythagoras and perpendicular formula ( the line between 2 of the triangle corners must be on the x axis): 10 print "innenkreis eines dreiecks ohne die winkelhalbierenden":dim x(5,2),y(5,2) 20 l1=8:l2=9:l3=10:n=l1^2+l2^2+l3^2:if l3>l2+l1 then else 40 30 print "ungueltige eingabe":end 40 lh=(l1^2-l2^2+l3^2)/2/l3:h=sqr(l1^2-lh^2):xb=l3:xc=lh:yc=h:xb=l3:yb=0 50 masx=1200/l3:masy=900/h:if masx0 then 110 120 xm=(xm1+xm2)/2:gosub 70:if dg1*dg>0 then xm1=xm else xm2=xm 130 if abs(dg)>1E-10 then 120 140 x1=l3:y1=0:x2=lh:y2=h:xp=xm:yp=ym:gosub 150:goto 170 150 dx=x2-x1:dy=y2-y1:zx=dx*(xp-x1):zy=dy*(yp-y1):k=(zx+zy)/(dx^2+dy^2) 160 dxk=dx*k:dyk=dy*k:xs=x1+dxk:ys=y1+dyk:return 170 xs2=xs:ys2=ys:x1=0:y1=0:gosub 150:xs3=xs:ys3=ys 180 x(0,0)=0:y(0,0)=0:x(0,1)=xm:y(0,1)=0:x(0,2)=xm:y(0,2)=ym 190 x(1,0)=xm:y(1,0)=0:x(1,1)=l3:y(1,1)=0:x(1,2)=xm:y(1,2)=ym 200 x(2,0)=l3:y(2,0)=0:x(2,1)=xs2:y(2,1)=ys2:x(2,2)=xm:y(2,2)=ym 210 x(3,0)=xs2:y(3,0)=ys2:x(3,1)=lh:y(3,1)=h:x(3,2)=xm:y(3,2)=ym 220 x(4,0)=lh:y(4,0)=h:x(4,1)=xs3:y(4,1)=ys3:x(4,2)=xm:y(4,2)=ym 230 x(5,0)=xs3:y(5,0)=ys3:x(5,1)=0:y(5,1)=0:x(5,2)=xm:y(5,2)=ym 240 print xm,"%",ym:r=ym:goto 260 250 xbu=x*mass:ybu=y*mass:return 260 for a=0 to 5:gcol 8+a:x=x(a,0):y=y(a,0):gosub 250:xba=xbu:yba=ybu 270 for b=1 to 3:ib=b:if ib=3 then ib=0 280 x=x(a,ib):y=y(a,ib):gosub 250:xbn=xbu:ybn=ybu:goto 300 290 line xba,yba,xbn,ybn:xba=xbn:yba=ybn:return 300 gosub 290:next b:next a:gcol 10:x=xm:y=ym:gosub 250:circle xbu,ybu,r*mass 310 innenkreis eines dreiecks ohne die winkelhalbierenden 4.5% 2.53311403 > run in bbc basic sdl and hit ctrl tab to copy from the results window

[@misterenter-iz7rz]

a+b=8, b+c=9, c+a=10, a+b+c=27/2, a=27/2-9=9/2, b=27/2-10=7/2, c=27/2-8=11/2, half of sum of angles in a triangle is 90, so 1=r^2×4/(7×9+9×11+11×7)=4r^2(1/63+1/77+1/99)=4r^2((11+9+7)/7×9×11)=r^2×108/77×9=r^2×12/77, r^2=77/12, r=sqrt(77/12).😊

[@PreMath]

Thanks for sharing ❤️

[@DanielNeedham2500]

I've lost you after you said half the angles in a triangle is 90 degrees

[@misterenter-iz7rz]

@@DanielNeedham2500 three angle bisectors intersecting at the incenter, so there are six angles in that sum of three is just 90°.

[@murdock5537]

Nice! φ = 30°; ∆ ABC → AB = 10 = AE + BE = k + (10 - k); BC = 9 = BF + CF = 10 - k + CF AC = 8 = AD + CD = k + 8 - k → CD = CF = 8 - k → BC = 9 = 8 - k + 10 - k → k = 9/2 → 10 - k = BE = 11/2 DO = EO = FO = r = ? → sin⁡(OEB) = sin⁡(ODA) = sin⁡(OFC) = sin⁡(3φ) = 1 EBO = OBF = δ; DAO = OAE = α; ∆ ABC → 64 = 100 + 81 - 2(10)(9)cos⁡(2δ) → cos⁡(2δ) = 13/20 → sin⁡(δ) = √((1/2)(1 - cos⁡(2δ))) = √70/20 → cos⁡(δ) =√(1 - sin^2(δ)) = √1320/40 → tan⁡(δ) = sin⁡(δ)/cos⁡(δ) = √231/33 = 2r/11 → r = √231/6

[@PreMath]

Thanks for sharing ❤️

[@michaelstahl1515]

Thanks for your video. Even you used the formula for calculating the radius for the inside circle as I did.

[@PreMath]

Great! You are very welcome! Thanks ❤️

[@AmirgabYT2185]

√231/6≈2,53

[@MrPaulc222]

I haven't watched the video yet, but I might split the large triangle into three smaller triangles that each have r as the height. This gives the large triangle's area as 5r + 4r + 4.5r = 13.5r. I would then calculate the area with Heron's Formula (can't recall it to mind but can look it up). I would then divide the area from HF by 13.5 to find r. Perhaps I would double the area and divide by 27, but it depends what number HF gives me. I see my way approximates your first method. Your second method looks much cleaner though.

[@laxmikatta1774]

Moral of the video :- whenever circle is inscribed in a triangle it's AREA OF Δ RADIUS= -------- S where s stands for semiperimeter😊❤😊

[@PreMath]

True! Thanks for sharing ❤️

[@MrPoornakumar]

Today I learnt another triangle formula.

[@wackojacko3962]

@ 10:40 this morning I am Cuckoo for Cocoa Puffs. 🙂

[@PreMath]

😀 Thanks for sharing ❤️

[@זאבגלברד]

Instead of Heron it is simpler to say that we find the intersection of two circles. One is centered at 0,0 redius 9 . The other is centered at 10,0 radius 8

[@PreMath]

Thanks for sharing ❤️

[@זאבגלברד]

@@PreMathDo you agree that there is no need for Heron ? A simple set of equations solve it. Giving 3 sides of a triangle, this is the best way, I think. Thank you for many excellent videos.

[@PreMath]

@@זאבגלברד Many approaches are possible to find the solution to this problem! Thanks for the feedback 🌹

[@sambhubaitharu2250]

Class 9th exampler question

[@misterenter-iz7rz]

No alternatives offered by premath? 😮 I just offer one alternative solutions,😅 I consider the sum of angles of three angle bisectors, that is just 90. 😊

[@PreMath]

Thanks ❤️

[@trazom446]

Permettez-moi de vous dire que vous conduisez assaz mal vos calculs. On ne supprime pas un dénominateur alors qu'on doit le refaire apparaitre quelques lignes plus loin. Un prof de math à la retraite.