# Core Mathematics C2 Solomon paper

Watch
Announcements
#1
Solomon paper L C2 can someone please solve the third question.

https://pmt.physicsandmathstutor.com...%20Edexcel.pdf
0
#2
https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

I am having trouble answering the third question. Could someone please show how to solve it
Last edited by Samuditha; 1 year ago
0
1 year ago
#3
Start by noticing OBC is an equilateral triangle, then think about how the desired area is different. What would be the correction term?
(Original post by Samuditha)
https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

I am having trouble answering the third question. Could someone please show how to solve it
0
#4
Well here's what I did
In sector OAB Area is A=0.5*r^2*theta where theta is pi/3 lets call it A1
also there is a segment in that sector its area is A=0.5r^2(theta -sin theta) lets call it A2
then I added the two areas( lets call this area A) and multiplied it by 2.
then I did (pi/2) minus 2A
but I still dont get the right answer
I got (r^2/6)*(-6square root 3 -5pi)
(Original post by mqb2766)
Start by noticing OBC is an equilateral triangle, then think about how the desired area is different. What would be the correction term?
0
1 year ago
#5
(Original post by Samuditha)
Solomon paper L C2 can someone please solve the third question.

https://pmt.physicsandmathstutor.com...%20Edexcel.pdf

Just consider this blue quarter of the circle. [see first image] If you can determine what the shaded area in that quarter is, then you just multiply it by 2 to get the final answer.

So how do we get this shaded bit? The simplest way is to begin with the full area of this quarter, then subtract the blank part.
So what's the area of the blank part? [see second image] You can find it by splitting it up into three sections: the green shaded equilateral triangle, and the two equal purple shaded parts. What is are the lengths of this equilateral triangle? You can now work these sections and out, and hence add them to get the blank shaded area.

0
1 year ago
#6
You've got the right idea.
The desired area is
* area of OBC - area of pi/3 segment
* segment = pi/3 sector - area of OBC
So the result should be
* twice area of OBC - pi/3 sector

(Original post by Samuditha)
Well here's what I did
In sector OAB Area is A=0.5*r^2*theta where theta is pi/3 lets call it A1
also there is a segment in that sector its area is A=0.5r^2(theta -sin theta) lets call it A2
then I added the two areas( lets call this area A) and multiplied it by 2.
then I did (pi/2) minus 2A
but I still dont get the right answer
I got (r^2/6)*(-6square root 3 -5pi)
0
#7
Thank you every much RDKGames and mqb2766!!! I finally got the right answer.
1
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (214)
74.83%
No (72)
25.17%