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Mathematics: Post your doubts here!

Thought blocker

Thought blocker

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Alyjohn said:
PLEASE HELP!
Click to expand...
i am sure, a would be done by you, if not then temme..
here is b)
For the one triangle BXC :
The area of segment formula u found out in a(ii) use that to find the area of segment BC with r = 4 and alpha = pi/6 = 8pi/3 - 4sqrt(3)
The area of triangle BXC = {1/2 * 2 * 2sqrt(3)/3 } * 2 = 4sqrt(3)/3
The shaded region BXC = area of triangle BXC - Area of segment BC = 16sqrt(3)/3 - 8pi/3

The shaded area BXC = shaded area AXC = shaded area AXB

Therefore total shaded area = 3 * shahded region BXC = 16sqrt(3) - 8pi
 
A

Alyjohn

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Thought blocker said:
i am sure, a would be done by you, if not then temme..
here is b)
For the one triangle BXC :
The area of segment formula u found out in a(ii) use that to find the area of segment BC with r = 4 and alpha = pi/6 = 8pi/3 - 4sqrt(3)
The area of triangle BXC = {1/2 * 2 * 2sqrt(3)/3 } * 2 = 4sqrt(3)/3
The shaded region BXC = area of triangle BXC - Area of segment BC = 16sqrt(3)/3 - 8pi/3

The shaded area BXC = shaded area AXC = shaded area AXB

Therefore total shaded area = 3 * shahded region BXC = 16sqrt(3) - 8pi
Click to expand...
THANK YOU SO MUCH! (y):)
 
nehaoscar

nehaoscar

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upload_2016-10-5_1-11-45.png
Help please
I can't seem to find answer to part i and so I can't do the second part either :/
 
Fareez

Fareez

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Hey I have a problem for p1 maths 9709 may/june 2007
Question 5i) I dont understand why √3/3= AX/12
 
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