We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
+1
137
5
avatar+208 

This is a mes.. I dont even know where to start...

 Jun 23, 2019

Best Answer 

 #2
avatar+8720 
+4

 

area of shaded region  =  area of sector ACD - area of blue segment - area of △ACD

 

----------

 

AB  and  DB  are both radiuses of circle B so they are the same length.

 

AD  =  DB  =  AB  =  20

 

So △ABD  is an equilateral triangle.

 

m∠ABD  =  60°

 

m∠ACD  =  2( m∠ABD )   =   2( 60° )  =  120°

 

----------

 

m∠GCD  =  (1/2)(m∠ACD)  =  (1/2)(120°)  =  60°

 

△CGD is a 30-60-90 triangle.

 

GD  =  (1/2)(AD)  =  (1/2)(20)  =  10

 

CG  =  10 / sqrt(3)

 

CD =   20 / sqrt(3)

 

----------

 

area of sector ABD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot(20)^2\cdot\big(\frac{60^\circ}{360^\circ}\big)\ =\ \frac{400\pi}{6}\)

 

area of △ABD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(10\sqrt3)\ =\ 100\sqrt3\)

 

----------

 

area of blue segment  =  area of sector ABD - area of △ABD =  \(\frac{400\pi}{6}-100\sqrt3\)

 

area of sector ACD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot\Big(\frac{20}{\sqrt3}\Big)^2\cdot\Big(\frac{120^\circ}{360^\circ}\Big)\ =\ \frac{400\pi}{9}\)

 

area of △ACD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(\frac{10}{\sqrt3})\ =\ \frac{100}{\sqrt3}\)

 

----------

 

area of shaded region  =  \(\frac{400\pi}{9}-(\frac{400\pi}{6}-100\sqrt3)-\frac{100}{\sqrt3}\ \approx\ 46\) square units

 Jun 23, 2019
edited by hectictar  Jun 23, 2019
edited by hectictar  Jun 23, 2019
edited by hectictar  Jun 23, 2019
edited by hectictar  Jun 24, 2019
 #1
avatar
-2

👛👛

 Jun 23, 2019
edited by Guest  Jun 23, 2019
 #2
avatar+8720 
+4
Best Answer

 

area of shaded region  =  area of sector ACD - area of blue segment - area of △ACD

 

----------

 

AB  and  DB  are both radiuses of circle B so they are the same length.

 

AD  =  DB  =  AB  =  20

 

So △ABD  is an equilateral triangle.

 

m∠ABD  =  60°

 

m∠ACD  =  2( m∠ABD )   =   2( 60° )  =  120°

 

----------

 

m∠GCD  =  (1/2)(m∠ACD)  =  (1/2)(120°)  =  60°

 

△CGD is a 30-60-90 triangle.

 

GD  =  (1/2)(AD)  =  (1/2)(20)  =  10

 

CG  =  10 / sqrt(3)

 

CD =   20 / sqrt(3)

 

----------

 

area of sector ABD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot(20)^2\cdot\big(\frac{60^\circ}{360^\circ}\big)\ =\ \frac{400\pi}{6}\)

 

area of △ABD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(10\sqrt3)\ =\ 100\sqrt3\)

 

----------

 

area of blue segment  =  area of sector ABD - area of △ABD =  \(\frac{400\pi}{6}-100\sqrt3\)

 

area of sector ACD  =  \(\pi\cdot(\text{radius})^2\cdot\big(\frac{\text{angle}}{360^\circ}\big)\ =\ \pi\cdot\Big(\frac{20}{\sqrt3}\Big)^2\cdot\Big(\frac{120^\circ}{360^\circ}\Big)\ =\ \frac{400\pi}{9}\)

 

area of △ACD  =  \(\frac12\cdot(\text{base})\cdot(\text{height})\ =\ \frac12\cdot(20)\cdot(\frac{10}{\sqrt3})\ =\ \frac{100}{\sqrt3}\)

 

----------

 

area of shaded region  =  \(\frac{400\pi}{9}-(\frac{400\pi}{6}-100\sqrt3)-\frac{100}{\sqrt3}\ \approx\ 46\) square units

hectictar Jun 23, 2019
edited by hectictar  Jun 23, 2019
edited by hectictar  Jun 23, 2019
edited by hectictar  Jun 23, 2019
edited by hectictar  Jun 24, 2019
 #3
avatar+26 
-2

?...

CharmingTeachers  Jun 23, 2019
 #4
avatar+103069 
+3

 

It appears that the chord AD is bisected by  BE

So angle BED is right   and ED  =  1/2 AD  =  10

And we can find angle EBD using the Law of Sines

sin BED          sin (EBD)

_______  =  __________

  BD                 ED 

 

sin 90              sinEBD

______  =       ______

  20                     10

 

sinEBD  = (10/20) sin 90

sin EBD  = (1/2) (1)

sin EBD  = 1/2

So....EBD must be  30° 

So.....by symmetry....angle ABD  must be twice this  = 60°  

So....now....we can compute the area of sector AEBD =

pi* BD^2(60/360)  = pi*  (20)^2 (1/6)  = pi * 400/6  = 200/3 pi  units^2

 

 

And since angle ACD is twice angle ABD.....then   (1/2)ACD = ABD  =  60° = angle ECD

 

Then....since DEC is right, then triangle ECD is a 30-60-90 right triangle

Then  

CD / sin DEC  =   ED / sin ECD

CD / 1  =  10/  [sqrt(3)/2 ]

CD  =  20 /sqrt (3)

 

And the area of sector   AFDC =

pi (CD^2) (120/360)  = pi * (400/3)(1/3)  = (400/9) pi units^2      (1)

 

Now....we need to caclculate the area of triangle ABD

(1/2)(AB)(BD) sin (60°)   =

(1/2)(20)(20)sqrt (3)/2)  =  100sqrt(3) units^2

 

So.....the area between chord AD and arc AED  =

 

Area of  AEBD  - area of triangle ABD  =

 

  ( [200/3] pi  - 100sqrt(3)  )    (2)  

 

Lastly....we need to calculate the area of triangle ACD  [note CD = AC ]  =

(1/2)(CD)(AC)sin(120°)  =  (1/2)(20/sqrt3)^2 sqrt(3)/2  =   (1/2)(400/3)sqrt(3)/2 = 100/sqrt(3) = 100sqrt(3)/3   (3)

 

So....the area we are looking for is : 

Area of sector AFDC  - area between chord AD and arc AED  - area of triangle ACD  =

 

(1)   - (2)  - (3)  =

 

[400/9] pi  - [ 200/3 ]pi + 100sqrt(3) - 100sqrt(3)/3=

 

[ 400/9]pi - [600/9]pi  + (2/3)100sqrt(3)  =

 

(2/3)100sqrt(3) - (200/9)pi  ≈  45.66 units^2  =  46 units^2

 

[If I didn't make any errors....LOL!!!  ]

 

cool cool cool

 Jun 23, 2019
 #5
avatar+103069 
+1

Hectictar and I agree.....that might be dangerous...... the Sign of the Apocalypse.!!!

 

cool cool cool

 Jun 23, 2019

41 Online Users

avatar
avatar