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

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

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[predator]

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

[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.

[jimlocke]

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.

[giuseppemalaguti]

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..

[amarendrasingh]

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

[santiagoarosam]

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 del círculo =325Pi/4
Gracias y un saludo cordial.

[himo]

18h/2=81 h=9
15²-9²=144=12² 18-12=6
12*6=9*x x=8
(9+8)/2=17/2


area of the circle : r²π=325/4*π=81.25πcm²

[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).

[davidcung]

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

[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).

[MrPaulc]

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.

[vaggelissmyrniotis]

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!!

[misterenter-izrz]

Alternative solution, first we still compute sin A=3/5 and BC=3sqrt(13), consider the half angle at center from BC, just equal to A, then sin A=3/5=(3sqrt(13)/2)/r, r=(3/2)×(5/3) sqrt(13)=(5/2)sqrt(13), therefore the answer is (25/4)×13 pi=325/4 pi.

[misterenter-izrz]

Sin A=0.6=81×2/(15×18), then cosA=0.8, BC=sqrt(15^2+18^2-2×15×18×0.8)=sqrt(117)=3sqrt(13), thus the radius is 15×18×3sqrt(13)/sqrt((15+18+3sqrt(13))(15+18-3sqrt(13))(15+3sqrt(13)-18)(18+3sqrt(13)-15))=5sqrt(13)/2, thus the area is pi times its square=325/4 pi=255.254403 approximately. 😊

[holyshit]

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

[nenetstree]

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

[ai]

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

[ВасильМигович-шп]

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

[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.

[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.