A Very Nice Geometry Problem | You should be able to solve this | 2 Methods

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A Very Nice Geometry Problem | You should be able to solve this | 2 Methods
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2 июл 2024

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Комментарии : 26

@WahranRai 6 дней назад

9:11 The theorem is the *Power of a point* relative to a circle

@fphenix 7 дней назад

I started like Method 1, but once I had the 2√5 length, I used the tan() in B.=> tan(DBC) = 2√5 / 4 = √5/2 = tan(ABC) = x / 6. Hence 2.x = 6.√5, so x = 3√5

@Mediterranean81 6 дней назад

Use the tangent secant theorem x is the tangent BD is the secant So AD=x^2/4 AB= 4+x^2/4 AB^2= 16+2x^2+x^4/16 (1) ABC is a right triangle So AB^2=BC^2+AC^2 AB^2 = 36+x^2 (2) Let x^2= u Compare (1) and (2) 16+2u+u/16=36+u 256+32u+u=16u+576 32u+u-16u=576-256 15u=320 3u=64 u=64/3 x=8/sqrt 3

@Captain_Thunder_22 7 дней назад

Wow ... Good video ❤

@bpark10001 6 дней назад

In 1st method, you need not solve for DC. It is simpler to solve for AB. Using same triangle similarity, BD/BC = BC/AD. From this you get BD = 9. Then you can use Pythagoras to get AC squared = 81 - 36 = 45.

@rabotaakk-nw9nm 6 дней назад

👍, but BD/BC=BC/AB => AB=9 Or, AD=AB-BD=9-4=5 AC²=AD•AB=5•9=45 😉

@santiagoarosam430 6 дней назад

Llamamos E a la proyección ortogonal de D sobre BC → 6²-4²=DC²→ DC=2√5 → BC*DE=BD*DC→ DE=4√5/3 → BD²-DE²=BE²→ BE=8/3 → DE/BE=AC/BC→AC=X=3√5 =6,7082.... Gracias y un saludo cordial.

@machintruc8302 4 дня назад

Cos(dbc) = cos(abc) = BD/BC = BC/BA So BA=BC²/BD=6²/4=9 Then AB²=BC²+AC² 9²=6²+x² x²=81-36=45 x=v(45)=3v(5)

@marcgriselhubert3915 7 дней назад

O the center of the circle, and t = angleCBA, In triangle OBD: OD^2 = BO^2 + BD^2 -2.BO.BD.cos(t) So, 9 = 9 + 16 - 2.3.4.cos(t), and cos(t) = 2/3 Then 1 + (tan(t))^2 = 1/(cos(t))^2 = 9/4, then (tan(t))^2 = 5/5 and tan(t) = sqrt(5)/2 In triangle CBA: X = CA = BC.tan(t) = 6.(sqrt(5)/2) Finally: X = 3.sqrt(5).

@user-ll5wl9gq4x 7 дней назад

Thanks for the beautiful ideas!!

@JobBouwman 5 дней назад

Thales tells DC = sqrt(6^2 - 4^2)=2sqrt5. Congruency tells that AC = 6/4*2sqrt5. So AC = 3sqrt(5)

@brettgbarnes 7 дней назад

Based on the starting diagram and information, it cannot be deduced that AC is tangent to the circle at point C. That assumption was not given until after the solution began.

@Grizzly01-vr4pn День назад

That's just not true. If BC is the diameter of the semicircle, and also a side of the triangle, then AC cannot be anything other than a tangent at point C. If it was a tangent at any point on the semicircle other than point C, then the horizontal base of the triangle would be longer than the semicircle diameter. If the horizontal base of the triangle was equal to the semicircle diameter, but AC wasn't a tangent, then it would be a secant to either the semicircle as shown (intersecting the arc above BC), or the reflected semicircle below BC.

@brettgbarnes 16 часов назад

@@Grizzly01-vr4pn "If BC is the diameter of the semicircle, and also a side of the triangle, then AC cannot be anything other than a tangent at point C." Line AC can intersect with diameter BC at any angle in-between (but not including) 0 and 180 degrees. However, the only angle at which line AC could intersect with diameter BC, tangent to the circle at point C, is 90 degrees. There's no indication on the diagram that angle ACB is 90 degrees and there's no way to deduce that it is or that AC is tangent to the circle at point C.

@Grizzly01-vr4pn 10 часов назад

@@brettgbarnes Please re-read my second paragraph in my above comment. The fact that AC _only_ intersects with BC at point C means it _must_ be a tangent. If it were not a tangent, but still intersected at point C, it would also intersect with the circumference at a second point. It would be a secant line.

@murdock5537 7 дней назад

∆ ABC → BC = 6; AB = AD + BD = y + 4; CD = z; AC = x; ABC = δ sin⁡(BCA) = sin⁡(CDB) = 1 → z = 2√5 → tan⁡(δ) = z/4 = √5/2 = x/6 → x = 3√5 → y = 5

@imetroangola4943 7 дней назад

Parabéns 💐👏🏻 são excelentes seus vídeos!

@giuseppemalaguti435 7 дней назад

DC=√20...4:√20=√20:AD..AD=5..x^2=(4+5)^2-6^2=45

@quigonkenny 7 дней назад

Let O be the center of the semicircle. Draw radius OD. OD = OB = BC/2 = 6/2 = 3. As OB = OD, ∆BOD is an isosceles triangle and ∠DBO = ∠ODB = θ. Triangle ∆BOD: cos(θ) = (OB²+DB²-OD²)/(2(OB)DB) cos(θ) = (3²+4²-3²)/(2(3)4) cos(θ) = (9+16-9)/24 cos(θ) = = 16/24 = 2/3 sin(θ) = √(1-cos²(θ)) sin(θ) = √(1-(2/3)²) sin(θ) = √(1-4/9) sin(θ) = √(5/9) = √5/3 tan(θ) = sin(θ)/cos(θ) tan(θ) = (√5/3)/(2/3) = √5/2 CA/BC = √5/2 √5BC = 2CA 6√5 = 2x x = 6√5/2 = 3√5 units

@sergeyvinns931 6 дней назад

Нарисовано коряво! Но DC^2=36-16=20, DC=2\/5, DC/x=4/6, 2\/5/х=2/3, 2х=6\/5, х=3\/5, найдём гипотенузу АВ, AB^2=[^2+BC^2=45+36=81, AB=9. А нарисовано так, что АС=ВС?

@haiduy7627 7 дней назад

🎉🎉🎉🎉🎉🎉🎉

@yakupbuyankara5903 5 дней назад

X=3×(5^(1/2)).

@michaeldoerr5810 6 дней назад

Based on one of the liked comments from Math Booster, there might actually be THREE methods. Just like the last video. I shall have to practice all three methods then!

@Tmwyl 6 дней назад

I got this one!

@haiduy7627 7 дней назад

❤❤❤❤❤❤❤🎉🎉😊😊😊❤❤❤❤🎉🎉🎉🎉

@RealQinnMalloryu4 7 дней назад

(4)^2H/A/BDASino°=16H/A/DBASino° (6)^2 A/H/BCoso°= 36A/H/BCCoso° {16H/A/BDASino°+36A/H/BCCoso°}= 52HA/AH/BDASino°BCCoso° {180°/52HA/AH/BDASino°BCCoso°} =3 .24 HA/AH/BDASino°BCCoso° 3 .4^6 3.2^23^2: 1.1^1 3^2 3^2: (HA/AHSino°BDABCCoso° ➖ 3HA/AH/BDASino°BCCoso°+2)