Russian Math Olympiad | A Nice Geometry Problem 👇👇👇👇

[@kshitishmahata8614]

∆=(1/2)×12×9=54,s=(1/2)×(9+12+15)=18,∆=rs,r=∆/s=54÷18=3. Circle area=πr^2=π×3^2=9π.

[@sureshbabuvelpula4560]

👌

[@ankurdhiman8068]

Gazab

[@GopalJakhar-h2q]

Gagan sir

[@AsanMondal-oq1kx]

r=(p+b-h)/2 so r=(9+12-15)/2=3 Area of circle= 9pie.

[@adgf1x]

ar. triangle=18R=>18R=54=>R=3.=>ar circle=9.pai when pai=22/7.ans

[@ManojkantSamal]

^= read as to the power *=read as square root #= read as area of the triangle @= read as triangle # of the right angle @ =(1/2)× 9×12=54 S=(9+12+15)/2=36/2=18 Radious of the circle (r)= (#@)/S =54/18 =3 Aera of the circle =π(r^2) =π(3^2)=π×9=9π

[@PARADOX01291]

We can also solve this question. In this way Area of traingle 1/2 b×h = 54 cm². r+9/2 + r+15/2 + r+12/2 = 54 cm r = 3 cm Area of circle πr² 3.14× 3×3 = 28.26 cm².

[@amitanand9126]

3,4,5 then r = 1 Multiply by 3 then r=3 Area = πr² = 9π✔️✔️

[@SudhirN.-ot5ze]

Step1 all 3 sides of triangle are given Find area of triangle using Herons formula Step2. Area of. Triangle = r x s r is radius, s is half perimeter Find r radius Step3. Area of circle = pie x r x r

[@rudeffx_1p]

No need to use heron's formula, we know the measure of bases and heights (which is r) so it will become easy to get r with that

[@baishnabroy9655]

In Base , two parts : r and 12-r , In Perpendicular , two parts : r and 9-r In Hypotenuses , two parts , (9-r ) and {15-(9-r)} Now {15-(9-r)} =12-r r= 3 Area = 9Π

[@ManojkantSamal]

9×(22/7)= 9 pie =28.26(approx) Let the circle intersects P, b, & h at the point m, n, q respectively In the triangle B=90 degree BC=base AB=perpendicular AC=hypotenuse BC=Bn+Cn AC=Aq+Cq AB=Am+Bm Let Aq=x=Am So Cq=Cn=15-x Bn=12-(15-x)=12-15+x=x-3 Bm=9-x The center of the circle =O So, Om=On=Bm=Bn 9-x=x-3 2x=12 X=6 Aq=Am=6 AC=Aq+Cq 15=6+Cq Cq=15-6=9 Cq=Cn=9 Bn=BC-Cn=12-9=3=Om So, R=3 Area =pie ×(r)^2 =pie × (3)^2=9 pie =9×(22/7)

[@michaeldoerr5810]

The answer is that the radius is 3. I learned this by comparing your video to another regarding circles in Pythagorean triplet triangles. And because this triangle is a 3-4-5 triangle scaled by a factor of 3 AND that the circle within 3-4-5 triangle is of a radius of 3, the radius of THIS CIRCLE is just 3. And that is just another theorem regarding primitive triples!!!

[@upsciaspcs2607]

radius r = (P+B-H)/2, now area of circle = TT x r^2

[@pradnyawantshakya9331]

108=r/2(9+12+15) 216/36=r r=3 Area of circle =πr. r=198/7=28.285...sq. units

[@shyamalanthoosath6652]

a+b-c=2r, 9+12-15=2r, r=6/2=3. Area is 9π

[@SakthiBliss]

Very calm explanation every seen. Since its math, i could understand even in hindi language. However please put it in title Hindi so others will unerstand it is not in english.

[@adgf1x]

ar.of circle=9.pai where pai=22/7 as radius of circle=3 unit.

[@IshwariPrasad-iv7tr]

Short method to calculate radius of inner circle always a+b-c/2. r=3 rest you can do by -πr2. =9π

[@srikantdixit9935]

πr² = 9π (how)?

[@rudeffx_1p]

​@@srikantdixit9935 Step1: construct 3 lines from the vertices of the triangle to centre O Step 2: label them as triangles [1,2 and 3] Step 3: now find the area of the large traingle which is Area[triangle ABC) = 1/2 ×b × h = 1/2 × 12 × 9 = 54 Step 4: find the areas of the smaller triangles and equate them to the larger triangle Area(1) = 1/2 × 9 × r =9/2 × r Area(2) = 1/2 × 15 × r =15/2 ×r Area(3) = 1/2 × 12 × r =6×r Step 5: equate the sum of all these areas to the area of larger triangles -> 9/2 ×r + 15/2 ×r + 12/2 ×r = 54 = 1/2 ×r (15 + 9 + 12) = 54 = r = 108/36 = r = 3 Step 6: to find area of circle, put the value πr² => π × 3 × 3 => 9π

[@somakoshi3681]

Area. .=1/2 p×r. 1/2×bh=1/2×12×9 = 54=1/2×(9+12+15)×r. 54=1/2×36×r. 54=18r. r=3

[@nabakumardasbairaggya948]

1/2×r×15+1/2×r×12+1/2×r×9=1/2×9×12 Or r(15+12+9)=12×9 Or 36×r=108 Or r=3 Hence required area= π×3×3 =9π

[@TheGautam13]

r = area / semi perimeter of triangle Then put circle formula= πr^2

[@ajitsm5657]

how do you find exact center of the circle ?

[@Smartsols12]

Measuring exact distance between two opposite points of circle edges and cutting it in half... I think this is the way

[@nayakuniyadaiah4856]

Good question Ans: draw two chords and bisect them perpendicularly .it's intersecting poin is center of circle !

[@ajitsm5657]

Thank you 😊​@@nayakuniyadaiah4856

[@mr.roy4852]

wow nice theory 🎉🎉 helpful for ssc cgl❤❤❤❤

[@adgf1x]

18r=54=>r=3 and ar.red circle=3^2.pai=9. pai sq unit when pai=22/7.ans

[@hasinabanu3227]

Good. நன்று

[@biplobbipu7668]

r = 2× area of Triangle / perimeter of Triangle

[@muhammadashraf-mn2si]

Right triangle or Right angle triangle.......

[@piyushrvyas]

Radius of circle "r" It cannot be same as one side

[@lakhwindersingh3409]

Good method

[@AbrarHasan-g7e]

tan90= 9-r / 12-r 1/0= 9-r / 12-r 12-r =0 r=12 9-12=-3 πr^2 π(-3)^2 =28.3

[@srikantdixit9935]

Nice one 👍

[@MrWongndeso]

3, 4, 5 ==> r = 1 -------------------------x3 9,12,15 ==> r = 3 Area red = 3² π = 9π 🎉😊

[@santiagoarosam430]

Áreadel triángulo de la figura =9*12/2=54=r²+r(9-r)+r(12-r)→ r=3→ Área del círculo rojo =9π. Gracias y saludos.

[@SanatGhosh-03m]

1/2×r×(9+12+15)=1/2×12×9. ,r =3 and then 22/7×3×3

[@ramsanehisagar9912]

इसी त्रिभुज में तीन वृत एक दूसरे को स्पर्श करते तीनों वृत की त्रिज्या कैसे ज्ञात करेंगे इसका वीडियो बनाकर डाले

[@luiscostacarlos]

Triângulo ( 3,4 5) inscrito o raio é 1. Triângulo (9,12,15) inscrito raio é 3. Logo a área do círculo é: πr² = 9π

[@abdurrouf2996]

Good

[@satyamsah5624]

R=(9+12-15)/2=3 Area=9π

[@ajeethpandey]

(1/2)(r.a+r.b+r.c)=∆ r(a+b+c)=2.∆ r.2s=2.∆ r=∆/s r=(1/2)9.12/(9+12+15)/2 r=3 Area of Circle=π.3.3=28.27 sq unit

[@Hareshnayak16]

R=P+B-H/2

[@adgf1x]

area 9.pi when pi=22/7

[@Antonioff5dsffsgv]

Muito bom.

[@mastertutor6250]

At first you should find the centre of incircle instead of assuming it.

[@mastertutor6250]

The centre of incircle is the intersecting point of angle bisectors of the triangle.

[@massihmoghadam4353]

Brrrrrrrrrriliant.

[@kahashmi100]

9-r+12-r=15 21-2r=15 21-15=2r 6=2r r=3

[@PS-mh8ts]

are you the same as math booster? 🙂

[@stephenvarghese3657]

😂

[@ShivamChaudhary-pb3lq]

Antah vritt ki trijya, r,3,hi hoga

[@SajalSarkar-z1x]

Radius=3

[@SantoshVerma-xx2qe]

9π

[@annamoon20]

9 pie or 28.26

[@mahendrasaroj1913]

Class 10th

[@SajalSarkar-z1x]

198/7

[@gyanprakashraj4062]

😂😂😂😂SAMJHAWA😂😂😂

[@次野先生]

{(9+12-15)/2}²ㅠ

[@ramsanehisagar9912]

महोदय वृत पर खीजी गई स्पर्श रेखा बराबर होती है तो ,9-r=9-r ही होगा न कि 9-r=3+r हमारे विचार से आपका वीडियो सही नहीं है

[@KhushbooMaurya-jh6gr]

Mahadev aap two tangent theorem padhe pahle yese

[@abhishekchouhan3522]

कुछ बोलने से पहले सोच लिया करो। 9 -r , 3+r के बराबर क्यों नहीं हो सकता। अगर r=3 हो तब तो हो जाएगा ना बराबर बराबर बेटा😅 मेरे भोलू राम😂

[@abhishekchouhan3522]

वीडियो गलत नहीं ,तेरा समय गलत है। तेरी खोपड़ी के अभी पूरे तार सही सेकाम नहीं कर रहरहे

[@kalyanbasak6494]

Answer sharing 9π sq units

[@Misha-g3b]

9п un.²

[@satyamengineers1738]

🫡

[@ratikantapatra9874]

9×22/7

[@sakshikausal]

r = = (9*12)/(9+12+15) = 108/36 = 3 Area of circle = 9pi

[@wasimahmad-t6c]

1minat ka math bhasadh lambachaoda

[@himadrikhanra7463]

9 pi ?

[@ramsanehisagar9912]

इसके अलावा और कोई तरीका है

[@ShaileshKumar-px8di]

Aapka solution galat hai

[@ยี่สิบเก้าพฤศจิกา]

15-(12-r)=9-r ; 15-12+r=9-r ; 2r=6; r=3

[@wasimahmad-t6c]

2.4×3=7.2÷2.4142135624=2.9823×2.9823=8.89433×3.14169268=27.95