SOLUTION: Find the area of region ABCD in cm^2 if the radius of the circle is 2 cm and both AB and CD are perpendicular to EF. {{{drawing(400,400,-5,5,-5,5, circle(0,0,4), line(-2,3

Algebra ->  Customizable Word Problem Solvers  -> Geometry -> SOLUTION: Find the area of region ABCD in cm^2 if the radius of the circle is 2 cm and both AB and CD are perpendicular to EF. {{{drawing(400,400,-5,5,-5,5, circle(0,0,4), line(-2,3      Log On

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Question 1199182: Find the area of region ABCD in cm^2 if the radius of the circle is 2 cm and both AB and CD are perpendicular to EF.


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Found 4 solutions by Edwin McCravy, ikleyn, greenestamps, math_tutor2020:
Answer by Edwin McCravy(20054) About Me  (Show Source):
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Draw the two diameters AC and BD (in red) 



OA is a radius so OA = 2. And by the marks along the horizontal diameter,
OE is half a radius or half of 2 which is 1.  So OE = 1

So triangle AOE is a 30-60-90 right triangle, so AE=sqrt%283%29

Triangle AOE has area = expr%281%2F2%29%2Abase%2Aheight=expr%281%2F2%29%281%29sqrt%283%29%22%22=%22%22sqrt%283%29%2F2%29

The four triangles AOE, BOE, COE, DOE are congruent. 
So the four of them have area 4%28expr%28sqrt%283%29%2F2%29%29=2sqrt%283%29 

Next we find the area of the sector AOD.
   
Angle AOD is 60o because angles AOE and DOE are both 60o,
so angle AOD = 180o-60o-60o = 60o.

The area of a sector is expr%281%2F2%29%28theta%2F360%29pi%2Aradius%5E2 
and substituting the values, the area of sector AOD is

%281%2F2%29%2860%2F360%29%28pi%29%282%5E2%29%22%22=%22%22pi%2F3

The sector BOC is congruent to the sector AOD

Adding the 4 triangles and the two sectors,

Area of ABCD = %282sqrt%283%29%2Bpi%2F3%29cm%5E2

Edwin

Answer by ikleyn(52778) About Me  (Show Source):
You can put this solution on YOUR website!
.

I will borrow the plot from the post by Edwin.


    



The area of the figure  ABCD  is the sum of 


      the area of triangle AOB   + the area of triangle DOC + 
    + the area of the sector AOD + the area of the sector BOC.


The area of triangle AOB is the same as the area of equilateral triangle with the side of 2 units,
so it is 2%5E2%2A%28sqrt%283%29%2F4%29 = sqrt%283%29 square units.


The area of triangle AOB + the area of triangle DOC = 2%2Asqrt%283%29 square units.


Next, the area of the sector AOD is  1%2F6  of the area of the circle with the radius of 2;
so, the area of the sector AOD is  %281%2F6%29%2Api%2A2%5E2 = %284%2F6%29%2Api = %282%2F3%29%2Api square units.


The area of the sector AOD + the area of the sector BOC is twice that value, i.e.  %284%2F3%29%2Api  square units.


So, the  highlight%28highlight%28ANSWER%29%29  is  2%2Asqrt%283%29+%2B+%284%2F3%29%2Api  square units = 3.4641 + 4.1867 = 7.651 square units approximately.

Solved.



Answer by greenestamps(13198) About Me  (Show Source):
You can put this solution on YOUR website!


Also using Edwin's figure....



OD is the radius, and OF is half the radius; that makes each of the four triangles in the figure 30-60-90 right triangles. That means all six central angles are 60 degrees, so

(1) The two circular sectors AOD and BOC are each one-sixth of the circle; together their areas are one-third the area of the circle, which is %284%2F3%29pi; and
(2) each of the two triangular regions AOB and COD is a triangle with base 2sqrt%283%29 and height 1; together their areas are 2sqrt%283%29.

So the area of region ABCD is

ANSWER: %284%2F3%29pi%2B2sqrt%283%29

Answer by math_tutor2020(3816) About Me  (Show Source):
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Consider this shaded area in red

The angle DOF is 60 degrees as the other tutors have pointed out.
This means the red shaded pizza slice area is 60/360 = 1/6 of a full circle's area of radius 2.

full circle area = pi%2Ar%5E2+=+pi%2A2%5E2+=+4pi
1/6 of that full area = %281%2F6%29%2A4pi+=+2pi%2F3

The area of the red shaded region shown above is 2pi%2F3 square cm.

Then subtract off the area of triangle DOF
%282pi%29%2F3+-+sqrt%283%29%2F2
This represents the blue shaded region below


Quadruple this result since there are 4 identical symmetric (i.e. mirrored) such regions
4%2A%28+%282pi%29%2F3+-+sqrt%283%29%2F2+%29

8pi%2F3+-+2%2Asqrt%283%29

Then subtract this from the full circle area
4pi+-+%288pi%2F3+-+2%2Asqrt%283%29%29

4pi+-+8pi%2F3+%2B+2%2Asqrt%283%29

12pi%2F3+-+8pi%2F3+%2B+2%2Asqrt%283%29

4pi%2F3+%2B+2%2Asqrt%283%29
That is the area between the vertical lines.
It is the area of region ABCD.