# Numerical solutions of eq

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#1
I dont understand how to do question 9(a)
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1 month ago
#2
(Original post by Shas72)
I dont understand how to do question 9(a)
What do you understand?
Its a relatively simple area question with a bit of working required on the radius of the sector?

If you work back from the answer, can you spot any structure?
Last edited by mqb2766; 1 month ago
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#3
(Original post by mqb2766)
What do you understand?
Its a relatively simple area question with a bit of working required on the radius of the sector?

If you work back from the answer, can you spot any structure?
Is it that i have to calculate the area of triangle ABC?
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1 month ago
#4
(Original post by Shas72)
Is it that i have to calculate the area of triangle ABC?
You could do but you'd have to do the additional segment as well. Isn't the sector easier?
Try and write down some equations and upload when you get stuck.

Did you try working backwards?
Last edited by mqb2766; 1 month ago
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#5
(Original post by mqb2766)
You could do but you'd have to do the additional segment as well. Isn't the sector easier?
Try and write down some equations and upload when you get stuck.

Did you try working backwards?
So I did area of shaded region= area of circle - area of sector abc.
Area of shaded region = 3/8× area of circle
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#6
(Original post by mqb2766)
You could do but you'd have to do the additional segment as well. Isn't the sector easier?
Try and write down some equations and upload when you get stuck.

Did you try working backwards?
Area of sec acb = 1/2 r^2× 2 theta
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1 month ago
#7
(Original post by Shas72)
So I did area of shaded region= area of circle - area of sector abc.
Area of shaded region = 3/8× area of circle
isn't the shaded region the sector?
What variables, formulae, ..
Did you try working backwards...
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1 month ago
#8
(Original post by Shas72)
Area of sec acb = 1/2 r^2× 2 theta
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#9
(Original post by mqb2766)
isn't the shaded region the sector?
What variables, formulae, ..
Did you try working backwards...
You mean I use the expression theta=cos^-1sqroot

(Original post by mqb2766)
oh yeah right
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#10
(Original post by mqb2766)
(Original post by mqb2766)
isn't the shaded region the sector?
What variables, formulae, ..
Did you try working backwards...
You mean i use the expression given in question a and work out backwards?
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1 month ago
#11
(Original post by Shas72)
You mean I use the expression theta=cos^-1sqroot

oh yeah right
A couple of things
* Step away from the reply button for a time and draw/think about the problem
* Work back from the answer, can you get an "area = 3/8 area" type expression. Does that give you any hints?

Upload your thoughts/working after a bit of thinking. There is nothing beyond a hard gcse question here. Working backwards from the desired expression can, in some cases, give you hints about how the analysis is done.
Last edited by mqb2766; 1 month ago
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#12
(Original post by mqb2766)
A couple of things
* Step away from the reply button for a time and draw/think about the problem
* Work back from the answer, can you get an "area = 3/8 area" type expression. Does that give you any hints?

Upload your thoughts/working after a bit of thinking. There is nothing beyond a hard gcse question here. Working backwards from the desired expression can, in some cases, give you hints about how the analysis is done.
Ok sure. I will try. OA and OB = r
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#13
(Original post by mqb2766)
A couple of things
* Step away from the reply button for a time and draw/think about the problem
* Work back from the answer, can you get an "area = 3/8 area" type expression. Does that give you any hints?

Upload your thoughts/working after a bit of thinking. There is nothing beyond a hard gcse question here. Working backwards from the desired expression can, in some cases, give you hints about how the analysis is done.
Iam not understanding how to approach this question. Can you pls explain in a simpler way if possible?
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1 month ago
#14
(Original post by Shas72)
Iam not understanding how to approach this question. Can you pls explain in a simpler way if possible?
What picture have you drawn? What equations are you using for the area of the circle / sector. Working backwards steps would be appreciated. Can you p!s upload what you've done.
Please give me something to work with.
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1 month ago
#15
(Original post by Shas72)
Iam not understanding how to approach this question. Can you pls explain in a simpler way if possible?
So, I assume you're happy that if we knew the radius R of the sector, then the area of the shaded region would be R^2 theta?

So what you really need to do at this point is find out what R is. This is basically GCSE geometry, but it seems to be something a lot of A-level students struggle with.

You know two sides of the triangle AOC = r. So, that triangle is isoceles. Does that help?

Spoiler:
Show
If AOC is isoceles, then if you set M to be the midpoint of AC, drawing the line OM divides the triangle into 2 smaller right angled triangles. From there it's easy to find AM (or CM).
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#16
(Original post by mqb2766)
What picture have you drawn? What equations are you using for the area of the circle / sector. Working backwards steps would be appreciated. Can you p!s upload what you've done.
Please give me something to work wit
(Original post by DFranklin)
So, I assume you're happy that if we knew the radius R of the sector, then the area of the shaded region would be R^2 theta?

So what you really need to do at this point is find out what R is. This is basically GCSE geometry, but it seems to be something a lot of A-level students struggle with.

You know two sides of the triangle AOC = r. So, that triangle is isoceles. Does that help?

Spoiler:
Show
If AOC is isoceles, then if you set M to be the midpoint of AC, drawing the line OM divides the triangle into 2 smaller right angled triangles. From there it's easy to find AM (or CM).
so Ac = sqroot(2) ×r
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1 month ago
#17
(Original post by Shas72)
so Ac = sqroot(2) ×r
Why (and no)?
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#18
(Original post by mqb2766)
Why (and no)?
Since oc is perpendicular to oa and ob
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1 month ago
#19
(Original post by Shas72)
Since oc is perpendicular to oa and ob
Where does it say that in the question?
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#20
(Original post by mqb2766)
Where does it say that in the question?
Yeah it doesn't say that.
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