Can you find the area of the circle? | Triangle inscribed in a circle | (Math skills) |

[@predator1702]

Very nice problem and very elegant explanation 👍, thank you teacher 🙏.

[@PreMath]

Glad you liked it ❤️ Many thanks 🌹

[@bitsavas]

I THINK MISTER PREMATH YOU MUST TELL THAT A

[@ybodoN]

@@bitsavas it can be deduced from the following rule: • when the circumcenter is inside the triangle, A < 90° • when the circumcenter is outside the triangle, A > 90° • when the circumcenter is on the hypotenuse, A = 90°

[@ybodoN]

The area of a triangle is ½ ab sin θ, in our case 81 = ½ (15) (18) sin θ ⇒ sin θ = ⅗ ⇒ cos θ = ⅘. By the law of cosines, we then have CB² = 15² + 18² − 2 (15) (18) (⅘) so CB = √117. The diameter of a triangle's circumcircle is abc / 2A, in our case (15) (18) (√117) / 162 = √325. The area of a circle is ¼ π d², in our case ¼ π 325 square units or 81.25 π square units.

[@bitsavas]

in my opinion to do this you must suppose Θ

[@ybodoN]

​@@bitsavas Since the center of the circumcircle is inside the triangle, the angle θ can only be less than 90°. If we didn't know where the center was, it is true that the area of the circle could also be ¼ π 2725 sq. units.

[@bitsavas]

thanks ,i am mathimatician in greece i dont rember to see this is school book but i think you are rigth(sorry for my english)

[@giuseppemalaguti435]

Tu hai usato S=1/2absinθ...io ho usato Erone,ma è la stessa cosa...magari i tuoi calcoli sono più semplici

[@hallo-g1r]

I did this as well, but i used the formula r = a/(2sin∅)

[@jimlocke9320]

Solution by two applications of the intersecting chords theorem. At 3:25, extend CD to construct a chord. Label the other intersection with the circle as point F. By intersecting chords theorem on AB and CF, (DF)(CD) = (AD)(BD) or (DF)(9) = (12)(6) and length DF = 8, so CF = 9 + 8 = 17. Construct perpendicular chord through midpoint of CF. It is 8.5 away from point C and from point F, so this chord and AB are 0.5 apart. This chord passes through O and O is 0.5 distant from AB. Construct chord perpendicular to AB and through O. It is a diameter, therefore has length 2r, where r is the radius, and chord AB divides it into (r + 0.5) and (r - 0.5). It divides AB into two equal segments, total length 18, so each length 9. By intersecting chords theorem, (r + 0.5)(r - 0.5) = (9)(9) and r² = 81.25. Area of circle = πr² = (81.25)π cm².

[@holyshit922]

Value of sine from formula for area Value of cosine from Pythagorean identity Missing side length from cosine law Radius from sine law

[@amarendrasingh7327]

Keep uploading these types of videos and the question is fantastic 🤩🤩😊

[@PreMath]

Thank you, I will ❤️ Keep rocking 👍

[@jeffeloso]

Knowing that the area of a circle is pi D squared over 4 and D squared=325, there was no point in taking the square root to calculate the area of the circle (having said that it was squared again later in your procedure but involved unnecessary steps).

[@MrPaulc222]

After finding the height (12) and BD = 6, I used intersecting chords with 12*6 = 9*8. I then moved CD across to the middle of AB, which split AB into two sets of 9, then extended CD down. AB becomes 9 and 9, and CD becomes (x+9)(x+8). (x+9)(x+8) = 81, and the radius can be calculated from there because CD has been moved to pass over the centre of the circle.

[@aljawad]

After obtaining the value of (h) I proceeded to find the X-Y coordinates of the points: A, B & C. From those and the equation of the circle I calculated the XY coordinate of the center of the circle and radius to reach the same value for the area (81.25*Pi).

[@santiagoarosam430]

Altura CD=h=2×81/18=9》AD =sqrt(AC^2 - h^2)=12》DA×DB=DC×DE》DE =12×6/9=8》(2r)^2 =CE^2+(2OD)^2 =(9+8)^2+[2(12-9)]^2》r^2=325/4》Área del círculo =325Pi/4 Gracias y un saludo cordial.

[@PreMath]

Thanks dear ❤️🌹

[@robertbourke7935]

Got it! Many thanks again.

[@davidcung7585]

I did it by making a guide line extending CD straight to the circle... and calling it point E... until two intersecting chords are formed..dial point D with AD=12 CD=9 BD=6 and get DE=8... then we find the relationship between the radius of the circle and the two intersecting chords 4R²=A²+b²+C²+D², then R²=325/4

[@ramanivenkata3161]

Very well explained

[@PreMath]

Thanks for liking ❤️🌹

[@vaggelissmyrniotis2194]

My way of solving it was to take OK and OL perpendicular to AC and AB and used trigonometry on AOK and AOL triangles to find cosφ1 and cosφ2 where φ1+φ2=angleCAB=φ.Then i used sum of cosines where cos(φ1+φ2)=cosφ=4/5 and found the radius^2 and therefore the area of the circle.It was a pain to solve it like that though...Your solution was a lot smarter!!

[@giuseppemalaguti435]

Con Erone calcolo il 3'lato...a me viene c=32,99...per cui il raggio R della circonferenza circoscritta è R=15*18*32,99/4*81=10*32,99/36... però ho ovviamente dei dubbi sul calcolo... I calcoli sono errati, ma la procedura mi sembra corretta.. 1)calcolo del 3 lato con erone.. 2)R=abc/4S... ho rifatto i calcoli c=sqrt117.. R=15*18*sqrt117/4*81=(5/6)sqrt117...Ac=pi*117*25/36=325/4pi

[@himo3485]

18h/2=81 h=9 15²-9²=144=12² 18-12=6 12*6=9*x x=8 (9+8)/2=17/2 r²=(12-9)²+(17/2)²=9+289/4=36/4+289/4=325/4 area of the circle : r²π=325/4*π=81.25πcm²

[@raya.pawley3563]

Thank you

[@nenetstree914]

We can use the length of CB to get the same answer too. 😀

[@ai2657]

Sir we can use. Herons formula and find 3rd side and find cirumradius by using R=abc/4∆

[@ВасильМигович-ш5п]

sin(A)=CD/AC=9/15=3/5; cos(A)=4/5 or cos(A)=-4/5; 1) cos(A)=4/5; BC^2=18^2+15^2-2*18*15*4/5=117; 2R=BC/sin(A), 4R^2=117/(3/5)^2=325, S=325*pi/4. 2) cos(A)=-4/5; BC^2=18^2+15^2+2*18*15*4/5=981; 2R=BC/sin(A), 4R^2=981/(3/5)^2=2725, S=2725*pi/4.

[@ybodoN]

In the 1st case, A < 90° ⇒ the center of the circumcircle is inside the triangle. In the 2nd case, A > 90° ⇒ the center of the circumcircle is outside the triangle.

[@ВасильМигович-ш5п]

Yes@@ybodoN

[@Irishfan]

They always said to simplify our work. This I not simplified in my opinion until the multiplication of 81.25 and pi are carried out. The answer is 255.25 sq cm.

[@georgexomeritakis2793]

The diameter of the circle is a/sin A = b/sin B = c/ sin C

[@wackojacko3962]

😉💯👍

[@PreMath]

Thanks ❤️🌹

[@laxmikatta1774]

Day 1 of requesting premath to reveal his face

[@Copernicusfreud]

Yay! I solved the problem. Area = (1/2)(18)(15)sin Θ. Sin Θ = 0.6. Θ = 36.8698976458. Drawing a line OC = r. Drawing another line OB = r. The angle of COB is 2Θ 2Θ =73.7397952917. Cos Θ = 0.8. Cos 2Θ = 0.28. (BC)^2 = (18)^2 + (15)^2 - 2(15)(18)(cos Θ). Simplified, BC = 3* sq rt 13. For triangle OBC, c = 3*sq rt 13, a = r, b = r, 2Θ = 73.7397952917. 117 = r^2 + r^2 - 2*r*r*cos 2Θ. Simplified, r^2 = 325/4. The area of the circle = (325/4)*pi.

[@PreMath]

Great❤️

[@anthonycheng1765]

yes, need not compute 2*theta, use double angle formula

[@manojkantsamal4945]

Area of circle =255.357 (maybe )

[@prossvay8744]

area of the circle=81π=254.47cm^2

[@arnavkange1487]

1st view and comment today ........

[@PreMath]

Thanks a lot ❤️🌹

[@tx1nv264]

There are two different answers for this problem. You should either delete this post or correct it