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

  Рет қаралды 9,362

Math Booster

Math Booster

Ай бұрын

A Very Nice Geometry Problem | You should be able to solve this!
MY OTHER CHANNELS
••••••••••••••••••••••••••••••••
Calculus Booster : / @calculusbooster
Math Hunter : / @mathshunter
--------------------------------------------------------------------------------
Join the channel to become a member
/ @mathbooster

Пікірлер: 16
@amlanmukherjee7041
@amlanmukherjee7041 Ай бұрын
I thought some exciting things will happen for finding the shaded region but in the end sin inverse destroyrd all my excitement.
@marioalb9726
@marioalb9726 29 күн бұрын
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 √ )
@Irtsak
@Irtsak 29 күн бұрын
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
@michaeldoerr5810
@michaeldoerr5810 Ай бұрын
Given the title and my previous exposure to rats and angles, this seems easy.
@giuseppemalaguti435
@giuseppemalaguti435 Ай бұрын
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..
@quigonkenny
@quigonkenny Ай бұрын
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
@Grizzly01-vr4pn
@Grizzly01-vr4pn 25 күн бұрын
Shame the answer doesn't have a nice exact form.
@PrithwirajSen-nj6qq
@PrithwirajSen-nj6qq Ай бұрын
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
@RealQinnMalloryu4 29 күн бұрын
(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)
@prossvay8744
@prossvay8744 Ай бұрын
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.❤❤❤
@bpark10001
@bpark10001 29 күн бұрын
I smelled this coming! No exact answer! Resort to numerical result!
@GabrieleIris-is7bg
@GabrieleIris-is7bg 28 күн бұрын
AD:CD=CD:BD 2:4=4:8 8+2=10=AB 10/2=5 R
@destruidor3003
@destruidor3003 29 күн бұрын
This type of exercise should be do e without calculator…so dumbs downs to you
@ludmilaivanova1603
@ludmilaivanova1603 Ай бұрын
we call sin inverse: arcsin.
@haiduy7627
@haiduy7627 Ай бұрын
❤❤❤❤❤❤❤❤🎉🎉😊😊😊😊❤❤🎉🎉🎉🎉
Poland Math Olympiad | A Very Nice Geometry Problem
13:08
Math Booster
Рет қаралды 97 М.
Why Is He Unhappy…?
00:26
Alan Chikin Chow
Рет қаралды 58 МЛН
Fast and Furious: New Zealand 🚗
00:29
How Ridiculous
Рет қаралды 43 МЛН
Double Stacked Pizza @Lionfield @ChefRush
00:33
albert_cancook
Рет қаралды 112 МЛН
НЫСАНА КОНЦЕРТ 2024
2:26:34
Нысана театры
Рет қаралды 1,4 МЛН
Russia | Math Olympiad Question | You should know this trick!!
8:01
Is this even solvable? What is the radius?
12:21
MindYourDecisions
Рет қаралды 153 М.
Russian Math Olympiad | A Very Nice Geometry Problem
14:34
Math Booster
Рет қаралды 96 М.
Viral question from China
8:04
MindYourDecisions
Рет қаралды 693 М.
Why Is He Unhappy…?
00:26
Alan Chikin Chow
Рет қаралды 58 МЛН