c2 past paper question-trapezium rule and area under curve?
Hello everybody, I'm a bit stuck on a particular part of this maths question, I was just wondering if anybody could help me out. It's question 6 of the January 2011 paper, I can do 6a and 6b pretty easily(the table and then trapezium equation) but I'm not sure on how you work out part C?
I know you have to do 1/2xBxH then you get the figure from b(0.3175) then minus it from the area of the triangle to find the area "S" The mark scheme got the height as 0.2, though I don't know how?
I know the base is "1" because (3-2) but I don't know how to work out the height, any assistance would be deeply appreciated! Thanks
Here is the link to the mark scheme as well it's q6.C
http://www.kingsleyschoolbideford.co.uk/Images-(1)/SeniorSchool/Maths-PDFS/C2rms_20110309.aspx
I know you have to do 1/2xBxH then you get the figure from b(0.3175) then minus it from the area of the triangle to find the area "S" The mark scheme got the height as 0.2, though I don't know how?
I know the base is "1" because (3-2) but I don't know how to work out the height, any assistance would be deeply appreciated! Thanks
Here is the link to the mark scheme as well it's q6.C
http://www.kingsleyschoolbideford.co.uk/Images-(1)/SeniorSchool/Maths-PDFS/C2rms_20110309.aspx
Simple, if you take 1 to be the base, then the heigh is the y coordinate of B (3,y).
Sub in x=3 into your equation.
Sub in x=3 into your equation.
(edited 12 years ago)
Original post by vc94
The height of the triangle is the y-value corresponding to x=3...so look in the given table!
Original post by In One Ear
Simple, if you take 1 to be the base, then the heigh is the y coordinate of B (3,y).
Sub in x=3 into your equation.
Sub in x=3 into your equation.
Thank you both, feel a bit silly now that I couldn't find it before hand, but again thank you
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