Find area of blue shaded circle inscribed in a right triangle | Important Geometry skills explained

[@margarettee1380]

Amazing. Surprise was 3+r. Brilliant

[@PreMath]

Excellent! Thanks for your feedback! Cheers! You are awesome, Margaret. Keep it up 👍 Love and prayers from the USA! 😀

[@AnonimityAssured]

I think it's a good idea to remember a simple formula for the radius of the incircle of a right triangle. Spoiler alert. If a and b are the lengths of the orthogonal sides, and c is the length of the hypotenuse, then: r = (a + b − c) / 2 units. (There are other formulae, but this is perhaps the simplest.) The area, A, of the incircle is therefore: A = πr² square units. In this particular case: r = (9 + 12 − 15) / 2 = 3 units. A = π (3²) = π (9) = 9π square units. Edited to correct a small mistake.

[@harrymatabal8448]

Now Premat these kind of videos are worth watching. Makes you think and that is what maths os all about. Excellent

[@alster724]

Thanks for the refresher on the Two Tangent Theorem midway, the latter part became easier to solve

[@johnbutler4631]

I did this almost exactly the same way using the two tangent theorem. It turned out to be much easier than I expected.

[@jmlfa]

Big triangle + sum of the three smaller ones: 12x9/2 = sum r x 12/2 + + + hence r

[@sail2byzantium]

Wonderful! That was great! A memorable learning experience. Thank you, PreMath!

[@vidyadharjoshi5714]

R be the radius. 9 - R. Tangents. 12 - R = 15 - (9-R) = 6 + R. 2R = 6. R = 3 Area = 9pi = 28.27

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome, Joshi. Keep it up 👍 Love and prayers from the USA! 😀

[@Ramkabharosa]

If a, b, c are the sides of any triangle & s = semi-perimeter = (a+b+c)/2, then the radius r, of the in-circle, is given by r = √ {(s-a)(s-b)(s-c) / s} = √{(18-9).(18-12).(18-15) / 18} = √{(9).(6).(3) / 18} = √9 = 3 units. So, the area of the in-circle = π.r² = π.3² = 9π square units. The radius of the circum-circle is given by R = abc / 4√{(s-a)(s-b)(s-c).(s)} = 9.(12).(15) / 4√{(9).(6).(3).(18)} = 9.(3).(15) / (3).(18) = 15/2. So, area of the circum-circle = π.R² = π.(15/2)² = 225π/4 square units. .

[@MathBasics1]

Sir, I think we can do this shorter. Using the formula {r=Area/Semiperimeter} r=(1/2×12×9)/(9+15+12)/2 cm =3 cm Area of circle = πr² =π×3² cm² =9π cm² - Math Basics

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@margarettee1380]

Wow. So impressed. Semiperimeter.

[@murry141]

best method but what if the triangle is not right angled triangle

[@jjcadman]

Does that formula work for all triangles (or even possibly other polygons), or just right triangles?

[@murry141]

@@jjcadman i guess it this formula work for right triangles but the length of side must me give to find perimeter or the perimeter must be there to solve question with this formula, by the way this is best formula to solve this question i am really amazed by maths basics

[@normanc918]

@2:32 we have (9-r) + (12-r) = 15, after simplify it we have 2r= 6, so r=3. Area of circle is pi r square.

[@MrPaulc222]

Very similar method for me using the tangents. I had the hypotenuse as (12-r) + (9-r), so 21 - 2r = 15, giving r as 3. 9pi area.

[@HappyFamilyOnline]

Awesome explanation👍 Thanks for sharing😊

[@dimitarganchev3222]

We can also use two different formulas for area of the triangle: p.r = 9.12/2 or and so on.

[@jamesrocket5616]

Via Heron's formula: Area of triangle = radius of circle x semi-perimeter Semi-perimeter = (9+12+15)/2 = 18 Area of triangle = 0.5 (12x9) = 54 Radius of circle = 54/18 Radius of circle = 3 units Area of circle = π (3 units)^2 Area of circle = 9π sq.units

[@murdock5537]

Awesome and nice, many thanks, Sir! 15 - (9 - r) = 12 - r → r = 3 → πr^2 = 9π

[@PreMath]

Great job Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@tedn6855]

I did lengths on all 3 sides. 9-r , 12-(9-r),which equals 12-r. Therefore 2r = 12-6, r = 3

[@_HaRsH_AgArWaL_]

Actually there is a formula kind of something for these type of cases. Where the radius =sum of smallest sides - largest side ÷ 2 From this standard Smallest sides are 9 and 12 Largest side is 15 Then =12 + 9 - 15 /2 = 21- 15 /2 = 6 ÷ 2 = 3 Therefore , r = 3 And area = 9pi Though it can be derived too

[@RAJEET69]

The answer will have units² not only units¹...

[@KAvi_YA666]

Thanks for video.Good luck sir!!!!!!!!!!

[@theoyanto]

Super brilliant, you seem to make your solutions just suddenly jump off the page.. Your years of experience shine through, thank you 👍🏻

[@ludmilaivanova1603]

If we connect the vertices wirh the centre and get areas of these three triangles -ar/2, br/2 and cr/2- the sum of areas will be ab/2. we can find a r.

[@procash1968]

Super. I worked on different method but it was roundabout Good problem, well solved

[@johnbrennan3372]

The hypotenuse is divided into 9-r and 15-(9-r)=6+r.Therefore 6+r=12-r so r=3.

[@adgf1x]

18r=54=>r=3.so area of inscribed circle=9.pai sq.unit when pai=22/7

[@jjcadman]

Dang! I forgot the two-tangent theorem, and I know I've seen that here before. 🤔 Obviously I need to watch your channel more frequently! 😂 Great video. Thanks for posting!

[@yamunaravi7132]

Impressive explanation thanks aiot

[@mathbynisharsir5586]

Excellent presentation sir

[@PreMath]

Glad you think so! Thanks for your continued love and support! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@AmirgabYT2185]

S=9π≈28,28

[@adgf1x]

18R=54=>R=3=>Area of inscribed circle=9.pai where pi=22/7

[@adgf1x]

Radius of circle =r.then 18r=54=>r=3 unit.Area of circle=9pai.where pi=22/7.

[@tanaysingh7913]

Nice very innovative 💡 video keep it up

[@PreMath]

Thanks a lot 😊 Glad you think so! You are awesome, Tanay. Keep it up 👍 Love and prayers from the USA! 😀

[@almosawymehdi3416]

Hello, where do you found your exercice? It is very interesting

[@skipmars7979]

what figure is "100% true to scale"? Love your channel. always looking for the next one to see if i can quickly solve it.

[@soli9mana-soli4953]

I assumed that the two legs are the Cartesian axes and therefore knowing that the coordinates of the center of the circle are equal Xc=Yc=r (because they are located on the edge of the square built with the radii of the circle), I equaled the distances of the center from the abscissa axis and the center from the line passing through the points A(0,9) and B(12,0) which is -3x-4y+36=0 and I obtained the equation (-3r-4r-36)/sqr((-3)^2+(-4)^2)=r and so r=3 which is the radius of the circle... But Prof's solution is better.

[@adgf1x]

9×12/2=18R=>R=54/18=3=>pai.R^2(ar.of inscribed circle)=9.pai(when pai=22/7).ans

[@yuusufliibaan1380]

❤❤❤ thanks for the help lesson keep going 💕

[@rupendrakumar7857]

Finally I made it,... It took little bit more time 😊😊😊😊 thank you.... 9π.

[@PreMath]

Great job! Glad to hear that! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@williamwingo4740]

No peeking: Let x = the radius of the circle. Draw perpendiculars from the center of the circle to each of the sides of the triangle. Each of these has length x. Now draw a line from the center of the circle to the rightmost vertex of the triangle. This line is the hypotenuse of two identical right triangles, each with altitude x and having bases 12 -- x (lower) and 15 -- (9 -- x) = 6 + x (upper). Invoking Pythagoras, we have (6 + x)^2 + x^2 = (12 -- x)^2 + x^2; subtract x^2 from both sides: (6 + x)^2 = (12 -- x)^2; expand both sides: 36 + 12x + x^2 = 144 -- 24x + x^2; again subtract x^2 from both sides: 36 + 12x = 144 -- 24x; collect terms: 36x = 144 -- 36 = 108; divide both sides by 36: x = 108/36 = 3. So the area of the circle is pi r^2 = pi x^2 = pi (3^2) = 9 pi. Carpe Diem. 🤠 Added in edit: actually, we don't even need to call on Pythagoras. the bases of the two identical right triangles are 12 -- x and 15 -- (9 -- x) = 6 + x, so 12 -- x = 6 + x; 2x = x; and x = 3. Even simpler.😃

[@massimogranzotto5979]

Stupendo video professore. Grazie

[@PreMath]

You are very welcome! Thanks for your continued love and support! You are awesome, Massimo. Keep it up 👍 Love and prayers from the USA! 😀

[@haric5948]

(a+b-c)/2 (12+9-15)/2 6/2=3 r=3 Area of circle 9π

[@gulfamsaifi4114]

28.29 sqcm (Apprx)

[@muttleycrew]

I got 9𝛑 as the area from a simple system of three simultaneous equations. E1... (a+b) = 15 E2... (a+r) = 12 E3... (b+r) = 9 E1 - E2 - E3 = (a+b) - (a+r) - (b+r) = -2r = 15 - 12 - 9 = -6. So, the radius, r = 3 and 𝛑r^2 = 𝛑9

[@santiagoarosam430]

Area triángulo =9×12/2=54 =r^2 +(9-r)r +(12-r)r 》r^2 -21r +54=0 》r=3 》Área azul =9Pi Gracias y saludos cordiales.

[@adlerbrietzke]

Area of the circle inscribed in a right triangle: r = (a + b - c)/2 Where a and b are base and height ou height and base, and c is the hypotenuse

[@michaelpolishchukCAInc]

S=p*r, where "p" - is a semiperimeter. After knowing radius - find the area

[@PreMath]

Thanks for your feedback! Cheers! You are awesome, Michael. Keep it up 👍 Love and prayers from the USA! 😀

[@IanNewYashaTheFinalAct]

Hello (multiple of the) 3-4-5 right triangle, my old friend

[@beeruawana6662]

Thanks sir

[@sorourhashemi3249]

The other way is the area =54, so we divided the triangle into 3 triangle in which I Their areas are 9r/2+12r/2+15r/2=54===>r=3==>area of circle= 9π

[@Copernicusfreud]

Yay, I solved it.

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@geraldillo]

15=(9-r)+(12-r) 15=21-2r 2r=6 r=3 area circle=πr^2 =π*3^2=9π

[@nicholascage3699]

The tutor even explain where radius and tangent is ...

[@gideonlapidus8996]

Love it

[@PreMath]

Glad to hear that! Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@weggquiz]

brilliant

[@mirjanamilosavljevic4261]

Apparently I had great math teacher…….got it just by looking at the dimensions of the triangle And knew the formula 40 years ago a had finished the school Hopefully I still don’t have dementia 😅

[@ioannismichalopoulos6936]

This is a 3:4:5 right triangle. No need to give all three side lengths or the fact that it is right.

[@PreMath]

Thanks for your feedback! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@yesterwilliamquebradaholgu4210]

Y+R=9 and R+X=12 and Y+X=15

[@缘投囝仔]

(9-r)+(12-r)=15. R=3

[@sytrostormlord3275]

9*PI

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@wesleysuen4140]

inradius=(9+12-15)/2=3

[@PreMath]

Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@rudychan8792]

"And once again, i'm thinking about, taking an easy way out" 🎵 Westlife^ A = (1/2)•12•9 = 54 s =( 9+12+15)/2 = 18 r = A/s !! = 54/18 = 3 O = π•r" = 9π = 28,27 units That is Phytagoras Famous Triangle* (3-4-5) 😉

[@PreMath]

Thanks for your feedback! Cheers! You are awesome, Rudy. Keep it up 👍 Love and prayers from the USA! 😀

[@narayaneevairamanipandian7441]

28.26

[@PreMath]

Excellent! Thanks for sharing! Cheers! You are awesome. Keep it up 👍 Love and prayers from the USA! 😀

[@giuseppemalaguti435]

6+r=12-r....r=3

[@reforma715]

👍👍👍

[@TheDHemant]

3

[@daredevil5508]

Why do so many calculation to find r man ● 9+12=21 ● 21-15÷2 = r ● 6÷2=3 Therefore r is 3 cm😎😎😎

[@_HaRsH_AgArWaL_]

exactly the same method i used 🔥

[@_HaRsH_AgArWaL_]

Actually there is a formula kind of something for these type of cases. Where the radius =sum of smallest sides - largest side ÷ 2 From this standard Smallest sides are 9 and 12 Largest side is 15 Then =12 + 9 - 15 /2 = 21- 15 /2 = 6 ÷ 2 = 3 Therefore , r = 3 And area = 9pi