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    The question is 11v on this exam paper link:

    http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

    What distances do I need to find for the base & height?
    I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
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    area=(0.5)*(9*square_root(10)/10)*(α)square_root(10)

    combine the solutions of questions (iii) and (iv)
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    (Original post by Magenta96)
    The question is 11v on this exam paper link:

    http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

    What distances do I need to find for the base & height?
    I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
    Remember the area of a triangle is a half multiplied by the base multiplied by the height perpendicular to the base.
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    (Original post by Magenta96)
    The question is 11v on this exam paper link:

    http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

    What distances do I need to find for the base & height?
    I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
    remember that the area is given by A=0.5 * h * b

    So to find the hight you would need to draw a line from point B to the Y axis and call it X so the distance between XB is 6 ( which is my hight) using the formula and I know the base is from OA which is 3.

    => A = 0.5 * 6 * 3 = 9


    is this the answer?
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    (Original post by Magenta96)
    The question is 11v on this exam paper link:

    http://www.mei.org.uk/files/papers/c109ju_kusp.pdf

    What distances do I need to find for the base & height?
    I thought it'd be 1/2(distance OA x distance OB) but that doesn't give an area of 9 like it should
    It does work...

    Point O (0,0)
    Point A (0,3)
    Point B (6,1)

    Distance of OA is the Height - therefore is 3
    Distance of OB is the Length - therefore 6

    6x3 = 18. Thinking of the rule 0.5xhxb
    So 18 x 0.5 = 9!
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    (Original post by Sitrix)
    It does work...

    Point O (0,0)
    Point A (0,3)
    Point B (6,1)

    Distance of OA is the Height - therefore is 3
    Distance of OB is the Length - therefore 6

    6x3 = 18. Thinking of the rule 0.5xhxb
    So 18 x 0.5 = 9!
    I keep getting (root)37 as the distance of OB because O(0,0) and B(6,1)
    OB = (root)(0-6)^2 + (0-1)^2
    OB = (root)(-6)^2 + (-1)^2
    OB = (root)36 + 1
    OB = (root)37
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    (Original post by Magenta96)
    I keep getting (root)37 as the distance of OB because O(0,0) and B(6,1)
    OB = (root)(0-6)^2 + (0-1)^2
    OB = (root)(-6)^2 + (1)^2
    OB = (root)36 + 1
    OB = (root)37
    Do a sketch.

    View the line from (0,0) to (0,3) as the base of your triangle.

    The height is the distance from the y axis to the point (6,1).

    base = ?

    height = ?

    Area = ?
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    (Original post by BabyMaths)
    Do a sketch.

    View the line from (0,0) to (0,3) as the base of your triangle.

    The height is the distance from the y axis to the point (6,1).

    base = ?

    height = ?

    Area = ?
    Yes I've got one length of 3 but for the distance of OB, I'm making some kind of error in the distance formula as I keep getting (root)37 rather than (root)6 which would give me 6
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    (Original post by steve2005)
    Diagram ( Can you see the dots ?)

    Yes, I can tell where the base and height are now and have one distance of 3. I can't seem to get 6 for the distance of OB though, I'm making some error in the distance formula even though I've checked my working loads
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    (Original post by Magenta96)
    Yes I've got one length of 3 but for the distance of OB, I'm making some kind of error in the distance formula as I keep getting (root)37 rather than (root)6 which would give me 6
    Why are you finding OB surely you can see that the HEIGHT of the triangle in that diagram is 6
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    (Original post by Magenta96)
    Yes, I can tell where the base and height are now and have one distance of 3. I can't seem to get 6 for the distance of OB though, I'm making some error in the distance formula even though I've checked my working loads
    http://www.purplemath.com/modules/distform.htm Is the distance formula, I can't type it out cause I suck at latex.
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    (Original post by robbie242)
    http://www.purplemath.com/modules/distform.htm is the distance formula, i can't type it out cause i suck at latex.
    not needed
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    you can split it into two right angle triangles

    Then use 1/2 (base*height) for each triangle and add them together
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    (Original post by TenOfThem)
    Why are you finding OB surely you can see that the HEIGHT of the triangle in that diagram is 6
    but the diagram that Steve2005 has given isn't on the exam paper so we're not supposed to see it's 6 without working it out first, what do you type in the distance formula for the distance of OB because even when substituting the coordinates for O and B in, I get (root)37, I subsituted:

    OB = (root)(6-0)^2+(1-0)^2
    OB = (root)(6)^2 + (1)^2
    OB = (root)(36)+(1)
    OB = (root)37
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    (Original post by Magenta96)
    but the diagram that Steve2005 has given isn't on the exam paper so we're not supposed to see it's 6 without working it out first
    You are given that B = (1,6)

    What working out do you think is needed
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    (Original post by TenOfThem)
    not needed
    Yup I know, since the y coordinate of the height is 6, giving you 6, just for reference in the future.
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    (Original post by Robbie242)
    Yup I know, since the y coordinate of the height is 6, giving you 6, just for reference in the future.
    But he had found OB correctly just unnecessarily
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    (Original post by TenOfThem)
    You are given that B = (0,6)

    What working out do you think is needed
    Actually B is (6,1).

    I don't think Steve's diagram is very helpful. [His diagrams are usually good!]
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    (Original post by Magenta96)
    but the diagram that Steve2005 has given isn't on the exam paper so we're not supposed to see it's 6 without working it out first, what do you type in the distance formula for the distance of OB because even when substituting the coordinates for O and B in, I get (root)37, I subsituted:

    OB = (root)(6-0)^2+(1-0)^2
    OB = (root)(6)^2 + (1)^2
    OB = (root)(36)+(1)
    OB = (root)37
    1/2 x OA x 1
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    (Original post by BabyMaths)
    Actually B is (6,1).

    I don't think Steve's diagram is very helpful. [His diagrams are usually good!]
    Poop

    Never the less ... I am sure that the OP can sketch a diagram

    Something I have suggested on other threads he has asked on this topic
 
 
 
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