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

@PreMath

6 ай бұрын

Glad you liked it ❤️ Many thanks 🌹

@bitsavas

6 ай бұрын

I THINK MISTER PREMATH YOU MUST TELL THAT A

@ybodoN

6 ай бұрын

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

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

6 ай бұрын

in my opinion to do this you must suppose Θ

@ybodoN

6 ай бұрын

​@@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

6 ай бұрын

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

6 ай бұрын

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

Got it! Many thanks again.

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

@PreMath

6 ай бұрын

Thank you, I will ❤️ Keep rocking 👍

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

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.

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

Thank you

Very well explained

@PreMath

6 ай бұрын

Thanks for liking ❤️🌹

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

6 ай бұрын

Thanks dear ❤️🌹

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

😉💯👍

@PreMath

6 ай бұрын

Thanks ❤️🌹

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

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

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.

Trusting that point O is inside the triangle, define it to have coordinates (0,0) Make AB a horizontal line, point O is therefore vertically above the centre of AB, by some distance z After we establish that CD=9 and AD=12 the coordinate of point B is (9,-z) while the coordinate of point C is (3,9-z) Points B and C lie on the circle, so (9)^2+(-z)^2=r^2=(3)^2+(9-z)^2 Expanding both sides and equating them yields 81+z^2=9+81-18z+z^2 Cancelling the identical terms from both sides reduces it to 0=9-18z hence z=0.5 Therefore r^2=(9)^2+(-0.5)^2=81.25 and the area of the circle is 81.25pi or 325pi/4

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

Day 1 of requesting premath to reveal his face

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!!

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. 😊

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).

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.

@PreMath

6 ай бұрын

Thanks❤️

@srirajan1933

6 ай бұрын

This was also my approach, but I have a question: should it not be the *double* angle at the center, since the angle at the center is *twice* the angle at the circumference when they subtend the same arc on the circle? if so, we have sin(2A) at the center = 3*sqrt(13)/r. By the double-angle trigonometric identity, sin(2A) = 2*sin(A)*cos(A) = 2*(3/5)*(4/5) = 3*sqrt(13)/r. But this leads to an incorrect answer under my assumption. Do you or @PreMath know where I went wrong? Thank you!

@misterenter-iz7rz

6 ай бұрын

@@srirajan1933 angle at center = 2×angle at circumference, so 1/2 angle is exactly equal to angle at circumference, angle at center is divided by two identical right angled triangles r×3sqrt(13)/2.....

@srirajan1933

6 ай бұрын

@misterenter-iz7rz Thank you for responding, and so quickly!! I agree completely with your derivation; i was trying to identify my mistake by using sin(2A) on center angle: if I use the full triangle spanned by triangle segment CB (and not 1/2 as you did by dividing into 2 equal right triangles by symmetry and perpendicular bisector theorem), then I must use the Law of Sines. Doing this way now correctly yields 3*sqrt(13)/sin(2A) = r/sin((180 - 2A)/2). This is because the 2 radii r are equal length and therefore is an isosceles triangle, OCB, leaving (180 - 2A) for the 2 triangle base angles, which are each divided equally to compute each angle. So: sin(2A) = 2*sin(A)*cos(A), and sin((180 - 2A)/2)) simplifies to sin(90 - A) for each base angle, which further simplifies to cos(A), and the cos(A) terms cancel each other. Then, simplifying the resulting equality produces exactly your answer for circle radius r and the circle area. Again, thank you for your response!

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

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.

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

6 ай бұрын

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

@user-wf3zv7ds3b

6 ай бұрын

Yes@@ybodoN

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

@PreMath

6 ай бұрын

Thanks a lot ❤️🌹

Area of circle =255.357 (maybe )

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

6 ай бұрын

Great❤️

@anthonycheng1765

6 ай бұрын

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

area of the circle=81π=254.47cm^2

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