Superb!!!

@PreMath

Ай бұрын

Excellent! Glad to hear that! ❤️Thanks a lot

The second method is out of this world!

For a fast solution, let's note that B could be the midpoint of ED. In this case, EB² = DB² = 9² − 6² = 45 and AB = 45 / 6 = 7.5 units. Pythagoras theorem and the intersecting chords theorem did it! But this only relies on a special case 🤔 Thank you PreMath! 🙏

@soli9mana-soli4953

Ай бұрын

In my solution with Stewart theorem after a semplification I had: EB*DB = 9^2 - 6^2=45 The semplification was the common factor (m+n) of the formula

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

Great lesson as usual!

Awesome!

The second method can be simplified by connecting point C with midpoint M of side DE and apply the Pythagorean theorem twice on the right triangles with hypotenuse 9 and 6

Yes, @ 0:58 With or Without Trigonometry. 🤔 Without Trigonometry human beings would never have been able to travel to the moon. 🙂

@PreMath

Ай бұрын

Great point! Thanks for sharing ❤️

Fine!

@PreMath

Ай бұрын

Thanks ❤️

Applying Stewart theorem you get EB*DB, then with intersecting chord theorem AB=15/2

@PreMath

Ай бұрын

Thanks for sharing ❤️

@ybodoN

Ай бұрын

Very nice solution! And Stewart's theorem can be proved as an application of the law of cosines, as PreMath shows with the second method😉

Thank you!

@PreMath

Ай бұрын

You are very welcome! Thanks ❤️

Let angle ECD be 2t. Let the midpoint of ED be F. Then CF = 9 * cos t. Hence cos(BCF) = (9 * cos t)/6 = 3/2 * cos t. Let O be the centre of the circle. Since C, F and O are collinear, angle OCB = angle FCB, and cos (OCB) = cos(FCB) = 3/2 * cos t. Let r be the radius of the circle, we know 9 = 2*r*cos t. Hence r = 9/(2 * cos t). AC = 2*r*cos(OCB) = 2*9/(2 * cos t) * (3/2) * cos t = 13.5 The AB = AC - BC = 13.5 - 6 = 7.5

Here I am with my Resolution Proposal. 1) Observe that Triangle [CDE] is not a Right Triangle. It's an Isosceles Triangle with Sides : 9 - 9 - ED. 2) Triangle [BCE] is also an Isosceles Triangle, with Sides : 9 - 9 - 6. 3) The height of Triangle BCE is equal to : h^2 = 9^2 - 3^2 ; h^2 = 81 - 9 ; h^2 = 72 ; h = sqrt(72) ; h = 6*sqrt(2) ; h ~ 8,5 4) The Area of Triangle [BCE] is equal to : A = 3 * (6*sqrt(2)) ; A = 18*sqrt(2) ; A ~ 25,5 5) Draw a Line from Point C to intercept the Middle Point between DE; call it C' 6) The Distance between C and C' is equal to : CC' = 36*sqrt(2) / 9 ; CC' = 4*sqrt(2) ; CC' ~ 5,7 Now, 7) The Distance Between Point B and Point C' (let's name it X) is equal to : 8) h^2 = 36 - X^2 and h^2 = 81 - (9 - X) ^2 9) 36 - X^2 = 81 - 81 + 18X + X^2 ; 18X = 36 ; X = 36/18 ; X = 2 Now we can conclude that the Line ED = 14 ; EB = 9 ; BD = 5 10) 6 * AB = (9 * 5) ; AB = 45 / 6 ; AB = 7,5 lin un 11) Answer: The Line AB is equal to 7,5 Linear Units.

@PreMath

Ай бұрын

Thanks for sharing ❤️

similar triangles' approach🤔, joining AD and AE, 6/9=2/3=BD/DA, 6/9=2/3=BE/EA, 😮😮😮

@PreMath

Ай бұрын

Excellent! Thanks for sharing ❤️

asnwer=5.6cm

Triangle ACE is similar to triangle ECB not triangle BCE

@PreMath

Ай бұрын

Thanks for sharing ❤️

First comment Plss pin

PreMath, This is great! Let's be friends and have fun together!

@PreMath

Ай бұрын

Thanks ❤️🌹

PreMath, You're awesome! Let's be friends and play together!

@PreMath

Ай бұрын

Thanks ❤️