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

A Very Nice Geometry Problem | You should be able to solve this!
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[@prossvay8744]

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.❤❤❤

[@amlanmukherjee7041]

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

[@Irtsak]

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

[@marioalb9726]

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 √ )

[@PrithwirajSen-nj6qq]

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

[@RealQinnMalloryu4]

(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)

[@giuseppemalaguti435]

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..

[@GabrieleIris-is7bg]

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

[@quigonkenny]

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

[@michaeldoerr5810]

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

[@ludmilaivanova1603]

we call sin inverse: arcsin.

[@bpark10001]

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

[@destruidor3003]

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

[@haiduy7627]

❤❤❤❤❤❤❤❤🎉🎉😊😊😊😊❤❤🎉🎉🎉🎉