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AQA Maths C1 circle question help
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#1
I don't know how I can get the answer for 6c(ii) I have attached the question
![Image]()
my working for c(i) is:
C=(5,8) M=(7,12)
CM^2=(7-5)^2+(12-8)^2
CM^2=20
CM=square root of 20
CM = 2 x squareroot of 5
where n = 2
I am unsure of how to find the area of the triangle PCQ, using 1/2(bxh) i could only get as far as MQ square root 5 or MP square root 5 as the area. Please could someone help me out, thanks
my working for c(i) is:
C=(5,8) M=(7,12)
CM^2=(7-5)^2+(12-8)^2
CM^2=20
CM=square root of 20
CM = 2 x squareroot of 5
where n = 2
I am unsure of how to find the area of the triangle PCQ, using 1/2(bxh) i could only get as far as MQ square root 5 or MP square root 5 as the area. Please could someone help me out, thanks
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AmarPatel98
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#2
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#2
Use the length of CM. (This is the height of the triangle CPQ)
Work out the length PQ (this is the base of the triangle)
Area = bh/2
Hope this helps
Work out the length PQ (this is the base of the triangle)
Area = bh/2
Hope this helps
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#3
(Original post by AmarPatel98)
Use the length of CM. (This is the height of the triangle CPQ)
Work out the length PQ (this is the base of the triangle)
Area = bh/2
Hope this helps
Use the length of CM. (This is the height of the triangle CPQ)
Work out the length PQ (this is the base of the triangle)
Area = bh/2
Hope this helps
but i have no idea how to work out the length of PQ, i know the mid point of PQ is (7,12) thats all the information I have.
Would you be able to tell me the method of working out the lengh of CM
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Rabadon
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#4
Draw it out. Because P and Q lie on the circle, the lengths CP and CQ are the radius. then you have CM so you can use pythagoras and areas of triangles etc to figure it out.
To find the length CM you know the co-ords of M and the co-ords of C so it's just the length equation which is pythagoras too.
To find the length CM you know the co-ords of M and the co-ords of C so it's just the length equation which is pythagoras too.
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AmarPatel98
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#5
ohhh i see your problem now.
If you draw triangle CPM, you have CM and CP (radius). Use this to work out the side length PM by pythagoras.
You can double the length PM to find the length PQ.
To work out CM, use the co-ords of C and M (both are given). Do the following to find its length:![Image]()
Then proceed by using PQ as the base and CM as the height of the triangle to work out area by 'bh/2'.
If you draw triangle CPM, you have CM and CP (radius). Use this to work out the side length PM by pythagoras.
You can double the length PM to find the length PQ.
To work out CM, use the co-ords of C and M (both are given). Do the following to find its length:
Then proceed by using PQ as the base and CM as the height of the triangle to work out area by 'bh/2'.
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#6
Okay i see now i got:
5^2-20=pm^2
25-20 = 5
pm = square root of 5
pm = mq therefore base = 2 x square root of 5
(2 x square root of 5)x(2 x square root of 5) all divided by 2 using (bxh)/2
= 20/2 = 10
therefore area of PCQ = 10
Is that right?
5^2-20=pm^2
25-20 = 5
pm = square root of 5
pm = mq therefore base = 2 x square root of 5
(2 x square root of 5)x(2 x square root of 5) all divided by 2 using (bxh)/2
= 20/2 = 10
therefore area of PCQ = 10
Is that right?
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