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

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

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Күн бұрын

A Very Nice Geometry Problem | You should be able to solve this | 2 Methods
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Пікірлер: 38

@fphenix 3 ай бұрын

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

@WahranRai 3 ай бұрын

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

@Mediterranean81 3 ай бұрын

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

@bpark10001 3 ай бұрын

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 3 ай бұрын

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

@brettgbarnes 3 ай бұрын

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 2 ай бұрын

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 2 ай бұрын

@@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 2 ай бұрын

@@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.

@brettgbarnes 2 ай бұрын

@@Grizzly01-vr4pn Yes, by definition, line AC only intersects with line BC at point C on the circumference, but that doesn't prove that line AC is tangent to the circumference at point C. Yes, "if it were not a tangent, but still intersected at point C, it would also intersect with the circumference at a second point." That's true, but only if angle ACB is less than 90 degrees and that would be an assumption that can't be proven from information on the starting diagram. If angle ACB is greater than 90 degrees, line AC would not "intersect with the circumference at a second point." The only case in which line AC could be tangent to the circumference at point C is if angle ACB is 90 degrees and, again, there is no way of knowing that from the information given at the start.

@Grizzly01-vr4pn 2 ай бұрын

@@brettgbarnes it certainly would intersect the circumference at a second point, that point being on the 'lower' unseen semicircle, that completes the full circle.

@machintruc8302 3 ай бұрын

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)

@Irtsak 2 ай бұрын

Let CD ( construction ) Obviously CD⊥AB. In right triangle ABC => BC²=AB·BD => 6²=AB·4 => AB=9 AD=AB-BD=9-4=5 Also in the same triangle : AC²=AB·AD =>x²=9·5 => x=3√5

@juanalfaro7522 2 ай бұрын

cos (B) = 4/6 = 6/AB -> AB=9 -> X^2 = 9^2 - 6^2 = 45 -> X = 3*sqrt (5). Another way: AB = 4+a = 9 in previous step. Now X^2 = a * (4+a) = (9-4) * 9 = 5*9 = 45 -> X = 3*sqrt (5). A 3rd way: cos (B) = 2/3 --> sin (B) = sqrt (5)/3 = X/AB where AB=9 determined in previous methods --> X = 9*sqrt (5)/3 = 3*sqrt (5).

@Captain_Thunder_22 3 ай бұрын

Wow ... Good video ❤

@marcgriselhubert3915 3 ай бұрын

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).

@JobBouwman 3 ай бұрын

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

@ВикторШеховцов 3 ай бұрын

Thanks for the beautiful ideas!!

@murdock5537 3 ай бұрын

∆ 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

@MarieAnne. 2 ай бұрын

When solving, you say that ∠ACB = 90° because AC is tangent to semi-circle (and radius and tangent are perpendicular), but when stating the problem, you don't mention that AC is tangent to semi-circle, and we can't assume it's tangent just because it looks tangent.

@MathTube-yazdani Ай бұрын

Yes. I think you are right.

@santiagoarosam430 3 ай бұрын

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.

@imetroangola4943 3 ай бұрын

Parabéns 💐👏🏻 são excelentes seus vídeos!

@quigonkenny 3 ай бұрын

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

@giuseppemalaguti435 3 ай бұрын

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

@sergeyvinns931 3 ай бұрын

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?

@yakupbuyankara5903 3 ай бұрын

X=3×(5^(1/2)).

@reolee924 2 ай бұрын

It was wrong solution. Hiw can say 90 degree of anggle BCD?

@michaeldoerr5810 3 ай бұрын

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!

@RealQinnMalloryu4 3 ай бұрын

(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)