Відео

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)

This type of exercise should be do e without calculator…so dumbs downs to you

Fantastic aproach. I used the Pytagoras Theorem to find the radius, and the aproximately value to the circle area, was 5,5 . PI 👍👍

Let O be the center of the circle, and M and N be the points of tangency between circle O and AB and BC respectively. Let R be the radius of circle O. Draw radii OM and ON. As AB and BC are tangent to circle O at M and N respectively, ∠BNO = ∠OMB = 90°. As ∠MBN = 90° as well, then ∠NOM = 360°-3(90°) = 90°. As all four internal angles of OMBN are 90° and adjacent sides OM and ON are both of equal length r, OMBN is a square with side length r. Draw BD. As OMBN and DPQS are both squares and thus each have all sides of equal length, Q and O are collinear with BD. By Pythagoras, it can be shown that the diagonal of a square (let's use OMBN for an example) is equal to √2 times the side length: OM² + MB² = OB² r² + r² = OB² OB² = 2r² OB = √(2r²) = √2r Therefore as BD = 6√2 and QD = 2√2, QB = 6√2-2√2 = 4√2. QB also equals OB+OQ = √2r+r = r(√2+1) r(√2+1) = 4√2 r = 4√2/(√2+1) r = 4√2(√2-1)/(√2+1)(√2-1) r = 4√2(√2-1)/(2-1) = 4√2(√2-1) Circle O: Aₒ = πr² = π(4√2(√2-1))² Aₒ = π(32(2-2√2+1) Aₒ = 32π(3-2√2) ≈ 17.25 sq units

Let's extend line PQ to the intersection with the small circle. This is point L. If we connect points PT L O we get a trapezoid. We construct a new circle around this trapezoid. Lines OT and P L pass through the center Q . Then TQ * QO = LQ * QP. That is, TQ = R; QO = 16- R; PQ = 8+ R; Q L = R . We have R * (16- R) = (8+ R) * R . We get 8 = 2 R . That is, R = 4.

x/AB = AB/3x AB^2 = 3•x^2 AB = sqrt(3)•x tanC = sqrt(3)•x / 3x = 1/sqrt(3) Thus, C = 30degree

(2)^2.H/A/DpQSDino°(6)^2 H/A/MBONCoso° 4H/A/DPQS,Sino°+36H/A/MBON} ={40H/DPQS/Sino°Cos°MBON 360°/Tano°}= 90°H/A/DPQSSino°MBONCoso° 3^30 3^5^6 3^5^3^2 1^1^3^2 3^2 (H/A/DPQSSino°MBONCoso°Tano° ➖ 3H/A/BPQSSino°MBONCoso°Tano°+2)

Let the circle with diameter AB. (construction) Extend the line segment CD that intersects the circle at point P. ( construction) Now we must calculate the area of the circular sector with center O and arc PC , subtract the area of the triangle OPC and then divide by 2 . *We need only to estimate the angle < POC , but we can’t do it using Geometry !!!!!!* In Geometry , if we only have relations between straight segments then the requested angle will be 30°,36°,45°,60°, 72°, 90°,120°, 150° ....... finish

(6-2)√2=4√2=r+r√2=r(1+√2) → r=4√2/(1+√2)→ r²=32/(3+2√2)=32(3-2√2)→ Área del círculo =32π(3-2√2). Gracias y un saludo.

I understood the mechanics and thought process completely. Does that mean that I should be able to do this?

Nice! r(√2 + 1) + 2√2 = 6√2 → r = (4√2)(√2 - 1) → πr^2 = 32π(3 - 2√2)

S=32(3-2√2)π≈17,097

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R+R/√2+2=6...R(1+1/√2)=4...R=8-4√2...mah,troppo semplice?!?

Construct triangle ECF as symetrical image of ECB by axis EC Construct rectangle ABGH such that F lies on GH Angle BEF has 90° and so triangles ABE and EHB are congruent So |BG| = |AH| = |AE|+|EH| = 1+3 |FG| = |HG|-|HF| = |AB|-|AE| = 3-1 = 2 |BC| = |CF| = x Using Pythagoras theroem on triangle CFG: |CF|^2 = |CG|^2 + |FG|^2 x^2 = (4-x)^2 + 2^2 x^2 = 16 - 8x + x^2 +4 8x = 20 x = 5/2 S = 1/2*|BC|*|AB| = 1/2*5/2*3 = 15/4

X=3×(5^(1/2)).

I smelled this coming! No exact answer! Resort to numerical result!

(2)^A/O2/Coso° =4A/O/Tano° (4)^2A/O/Tano° =16A/O/Tano° {4A/OCoso°+16A/O/Tano°} =20A/O/Coso°Tano° 180°/20A/O/Coso°Tano°=9 A/O/Coso°Tano° {A/O/Coso°Tano° ➖ 9A/O/Coso°Tano°+9)

(6)^2H/0/Sino° = 36 (3)^2H/O/Tano°=9H/O/Tano° {36H/O/Sino+9H/O/Tano°}=45H/O/Sino°Tano° {45°D+30°A+90°C }= 165°DAC {165°DAC+15°B}= 180°DACB/Coso° {45H/O/Sino°Tano°÷180°DACB/coso°}= 4DACBH/O/Sino°Tano°Coso° (H/O/Sino°Tano°Coso°DACB ➖ 4H/O/Sino°Tano°Coso°DACB+4)

Intersecting chords theorem: (2R-2).2 = 4² 4R - 4 = 16 R = 5 cm tan α = 4/(R-2) = 4/3 α = 53,13° A = ¼ R² (2α - sin 2α) A = ¼ 5² (106,26°- sin 106,26°) A = 5,59 cm² ( Solved √ )

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

Triangle BOE is similar to BCA, so sqrt(15)/sqrt(10) = 2*sqrt(10)/x. Thus x=2*10/sqrt(15). Simplifying: x=2*10*sqrt(15)/15=4/3*sqrt(15)

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?

OD = x R is the radius Now (R-x )*(R +x )=16 Geometric mean theorem R-x =2 R+x=8 R=5 It is for derivation of radius

I thought some exciting things will happen for finding the shaded region but in the end sin inverse destroyrd all my excitement.

Given the title and my previous exposure to rats and angles, this seems easy.

Draw radius OC. As OC = OA = r and DA = 2, OD = r-2. Triangle ∆CDO: OD² + CD² = OC² (r-2)² + 4² = r² r² - 4r + 4 + 16 = r² 4r = 20 r = 5 The red shaded area is equal to the area of the sector subtended by minor arc AC minus the area of ∆CDO. Let ∠AOC = θ. θ = sin⁻¹(4/5) ≈ 53.13°. Red shaded area: Aᵣ = (θ/360°)πr² - OD(CD)/2 Aᵣ = (sin⁻¹(4/5)/360°)π(5)² - (5-2)(4)/2 Aᵣ = (sin⁻¹(4/5)/360°)π(25) - 3(2) Aᵣ = 5sin⁻¹(4/5)π/72 - 6 ≈ 5.59 sq units

we call sin inverse: arcsin.

4^2+(R-2)^2=R^2...R=5...πR^2:2π=As:(arctg3/4+π/2)...As=(arctg3/4+π/2)25/2..Ablue=(25π)/2-As-4*3/2=5,5911..

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2(2R-2)=(4)(4) 4R-4=16 4R=20 So R=5 OD=5-2=3; OC=5 tan(DOC)=4/3=53.13° π(5^2)(53.13°/360°=11.6 Shaded area=11.6-1/2(4)(3)=5.60.❤❤❤

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

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Thank you for your work

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

a= 1, BC = 5 + 1 = 6, AC = 3, phytagoras X² = 45, X = 3 sqrt 5

This just seems to be a much more useful explanation than the comments that got hearted.

I got this one!

Small circle's diameter is semicircle's radius, 16/2 =8 is circle's diameter, 8/2=4 is its radius

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.

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.

Unnecessarily complicated solution (first one, eapecially). From first look, you can say that ABD is similar to CBA (by AA, 90 degrees and theta common). Therefore, x/AB = AB/3x => AB = root 3. x Now, in ABD, as tan(theta) = x/root3.x = 1/root 3, Thus, theta = 30 degrees. Solved in 2 mins.

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!

Nice! φ = 30° → sin⁡(3φ) = 1; ∆ ABC → AB = c; BC = BD + CD = 6 + 5; AC = 10; CAD = θ DAB = 2θ; AD = x = ? BD = BE + DE = (6 - a) + a; DAE = EAB = θ → 10/5 = 2 = AE/a → AE = 2a; AQ = AD + DQ; AP = AE + PE = 2a + (10 - 2a) → PQ = CQ = z → sin⁡(AQC) = 1; DQ = y → QCD = CAQ = θ → y/5 = z/10 → y = z/2 → 25 = 5z^2/4 → z = 2√5 → y = √5 → cos⁡(θ) = z/5 = 2√5/5 = (x + y)/10 = (x + √5)/10 = 2√5/5 → x = 3√5 sin⁡(θ) = √5/5 → sin⁡(2θ) = 2sin⁡(θ)cos⁡(θ) = 2(√5/5)(2√5/5) = 4/5 = (a + 5)/10 → 40 = 5(a + 5) = 5a + 25 → 5a = 15 → a = 3 → (6 - a) = 3 → sin⁡(BEA) = 1 fast lane: AB = AD = x → ∆ AEC = pyth. triple: 2(3 - 4 - 5) → tan(θ) = 1/2 → x = 3√5

Using angle bisector theorem we may write 6/AC= x /5 ACx =30 AC^2* x^2=900 [AC ^2= 36+25+10x +x^2 =x^2+10x +61] x ^4 +10x^3+61x^2-900=0 Solving this we get two real and two imaginary solutions. One of the real solutions is negative The other is 3 Hence x =3 Comment please

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).

4 2/3 cannot be = to 8/3. Should be 14/3.

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

(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)