The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

C2

https://b940eb267c31463d4d23ccfd3105c3b78e5616c6.googledrive.com/host/0B1ZiqBksUHNYUnQzSHRkZ3JoNGs/for-Edexcel/Solomon%20L%20QP%20-%20C2%20Edexcel.pdf

Question 3 help pls
Reply 1
Avatar for nerak99
nerak99
13
This was seen recently. The key is realising where euilateral triangles lie. Also, identifying the area of the segments cut off by chords.

One way of finding the required area is half the circle minus two equilateral tringales - 4 identical segments.

Here is the previous thread
Reply 2
Avatar for thethinone
thethinone
OP
Original post by nerak99
This was seen recently. The key is realising where euilateral triangles lie. Also, identifying the area of the segments cut off by chords.

One way of finding the required area is half the circle minus two equilateral tringales - 4 identical segments.

Here is the previous thread


thanks a ton
Original post by thethinone


right okay here's what you have to do....

replicate the diagram first and draw in the other circles, then draw two circle sectors where the arcs of each are OC and OB
Then, draw a hexagon inside the centre circle (the first one you drew)
Split the hexagon into six equilateral triangles (every angle is 60 degrees) and you should see that two of the triangles fall exactly into those sectors, the sides are length r (one side of each will line up with the radius already drawn in).
use Area = 1/2 absin(c) to work out the area of one of these triangles and you should get Area = 1/2 r x r sin(30) = 1/2 r^2 x (3^1/2)/2
Look at the diagram again and you'll see that because of the hexagon you can tell that the two circle segments are a sixth of the circle each as each of them fits inside one of the six triangles (within the hexagon) therefore the area of ONE of the sectors is (pi x r^r)/6
If you do the area of the segment - the are of the triangle you get this small area where the arc is and if you look at the triangle that the shaded region is inside you'll see that TWO of these areas (the one you just found) cross into that triangle and the rest is the shaded region that you want.
Therefore, double the area of that you just found (should be ((r)^2(3)^1/2)/4
Then do the area of one of the triangles made by the hexagon minus this area (which you just doubled) and then simplify by taking 1/6 and r^2 outside the bracket and multiplying the value with (3)^1/2 by 2 and you're sorted :smile:

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you