A Very Nice Geometry Problem | 3 Different Methods
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- Опубликовано: 28 сен 2024
- A Very Nice Geometry Problem | 3 Different Methods
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Intersecting chords theorem:
3x = 2²
x = 4/3 cm
2R = x+3 = 4/3 + 3 = 13/3
R = 2,1666 cm = 13/6 cm
Shaded area:
A = A2 - A1
A = ½πR² - ½b.h
A = ½π(13/6)² - ½3.2
A = 7,374 - 3
A = 4,374 cm² ( Solved √ )
A variation on method #3: After constructing AC, we find that ΔABC is similar to ΔCBD by angle -angle (common angle
*Outra solução:*
Seja OB=OC=r. Por Pitágoras no ∆ABC, temos:
BC=√13 e Cos B= 3/√13
Usando a lei dos cossenos no ∆COB, temos:
r^2= r^2 + 13 - 2r√13 × 3/√13
0= 13 - 6r=> *r=13/6.*
O resto da linha segue da mesma forma do vídeo apresentado.
YOU ARE NOT COHERENT !
you named radius gamma and the angles alpha and beta : respect your audience the grec letters are for angles ( INTERNATIONAL STANDARD FOR NAMING ANGLES IN GEOMETRY) and radius R or r
S=(169π-216)/72≈4,37
cb= sqroot 13. In right triangle acb, ac^2= 4(r^2)-13. In triangle acd, 4(r)^2-13=(2r-3)^2 +4. So r =13/6 etc
4th Method:
Centre of circle _O_ lies along _AB._ Draw _OC._
Let _∠OBC = θ_
By Pythagoras in _ΔBDC |BC| = √13_
∴ in _ΔBDC sinθ = 2/√13_ and _cosθ = 3/√13_
Since _|OB| = |OC| = r_ (the radius of the circle) _ΔBOC_ is isosceles.
∴ _∠OCB = ∠OBC = θ_
⇒ _∠DOC = 2θ (external angle to ΔBOC)_
∴ in _ΔDOC sin2θ = 2/r_
∴ _2/r = 2sinθcosθ = 2(2/√13)(3/√13)_
⇒ _1/r = 6/13_
⇒ *_r = 13/6_*
In a fuller circle:
2r = x+3
3x = 4
x = 4/3
2r = 3 + 4/3 = 13/3
r = 13/6
r^2 = 169/36
Full circle area is (169/39)pi, so semicircle area = (169/72)pi
Triangle area is (3*2)/2 = 3
Shaded areas total (169/72)pi - 3 approximates to 4.374 un^2
3X = 2*2 _> X = 4/3 -> 2R = 3+4/3 = 13/3 -> R = 13/6. Area = Pi * R^2/2 - 3*2/2 = 169*Pi/72 - 3 = 7.374 - 3 = 4.374
I used a third method similar to the second. Consider that any triangle with a base equal to the diameter inscribed in the semi-circumference is a right-angled triangle. Thus, I applied Pythagoras' theorem to the smaller right-angled triangle on the left, equalizing the equations, obtaining the value of "x" and, consequently, that of "R".
AD *BD =4
AD=4/BD=4/3(as 2 is the GM of AD & BD)
Area of the triangle =1/2*2*3=3
sq unit
Radius =(4/3+3)/2=13/6
Area of semicircle
= π*169/72
Area of coloured area =169π/72-3
=4.37 sq units (approx.)
Comment please.
Is it the 4th method?
a^2+2^2=r^2..a+r=3...r=13/6...
Looks like this gives a new definition of easier than it looks. There are three methods because there are three ways to compute the radius and consequently there three equally relevant constructions.
The first construction makes use of a circle theorem, the second construction makes use of a relationship between pairs of chords and the third construction makes use of alternating congruent angles that result in HL similarity.
No entiendo inglés. Pero cómo se justifica (en el segundo Método) que el Centro está en BD y nó en AD, pues no se conoce todavía el valor de AD. Si BD>AD entonces sí sería correcto
Let O be the center of the semicircle. Draw radius OC. As DB = 3 and OB = r, OD = 3-r.
Triangle ∆CDO:
OD² + CD² = OC²
(3-r)² + 2² = r²
9 - 6r + r² + 4 = r²
6r = 13
r = 13/6
Purple shaded area:
A = πr²/2 - bh/2
A = π(13/6)²/2 - 3(2)/2
A = 169π/72 - 3 ≈ 4.374 sq units
This action is wrong.
Who's still watching this at 1:41 am? #nostalgia
Tan B = 2/3 AC°D = DB°C
Tan C = 2/3 and CD is 2 so AD = 4/3
AB = 3 + 4/3 or 13/3
The radius is 13/6
The area of the half circle is 169/36 * 2 * pi = 169 pi/72
The are of the triangle is 3
Thus the answer is 169pi/72 - 216/72 = (169pi-216)/72
4.375
Draw a straight line from C to A to form a triangle ACD and ABC. Triangle ABC is a right triangle (Thales Theorem
Triangle ACD is similar to triangle ABC,
Hence, to find AD, 3/2 = 2/AD =
3 AD = 4
AD = 4/3
AD = 1.33333
Hence, the diameter of the semi-circle = 3+ 1.3333 = 4.3333 , Hence, the radius = 2.1665
hence, the area of the circle= 14.75
Hence, the area of semi-circle = 7.375
The area of the BCD = 3 (2*3*1/2)
Area of shaded = 7.375- 3
= 4.375 Answer
10:50-13:05 Right Triangle Altitude Theorem:
CD²=AD•BD => 2²=x•3 => x=4/3 😁
asnwer=2+/3
asnwer=13/6 isit
The last operation you did was wrong.....(h)*2 H squared=m×n...