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

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

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 √ )

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

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..

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

(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)

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

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

Shame the answer doesn't have a nice exact form.

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

AD:CD=CD:BD 2:4=4:8 8+2=10=AB 10/2=5 R

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.❤❤❤

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

we call sin inverse: arcsin.

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