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    Hello,

    I was just doing a question from the edexcel C2 book. Question 7 in the mixed exercise chapter 6 on Radians

    This is part of the question so you will kinda know which one i am talking about.

    The diagrams show the cross sections of two drawer handles....etc


    It states that the areas of both shapes x and y are equal.

    For shape x :

    A = 2d^2+1/2(pi2d^2)

    Now shape Y is a sector of a circle so I have used the area equation for a sector to give:

    A = 1/2 (2d^2) theta

    In the answer book that half disappears to give:

    A= 2d^2 theta

    Why does that half disappear? Then you have to equate them both together to make theta the subject. Which I can do....

    But I just dont know why that 1/2 disappears....

    The fundamentals escape me again...

    Does anyone have any suggestions to get better at that - when it comes to rearranging formula and solving etc I have noticed I never do them in the right order!
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    (Original post by christinajane)
    x
    Pictures of the diagram would help to show that you've got the correct values.

    For your shape y, is A = 1/2 * (2d)^2 * theta?
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    Hi yeah sorry - not sure how to add diagrams on here.

    This is the link though to the answers

    https://0e90d90be05fe2865012661b731d...apter%2006.pdf

    Its page 55/69 on my computer.
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    (Original post by christinajane)
    A = 2d^2+1/2(pi2d^2)
    This should be just pi d^2.
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    (Original post by christinajane)
    Hi yeah sorry - not sure how to add diagrams on here.

    This is the link though to the answers

    https://0e90d90be05fe2865012661b731d...apter%2006.pdf

    Its page 55/69 on my computer.
    (2d)^2 = (2x2)dxd =4D^2 therefore 1/2 x 4d^2 is 2d^2
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    (Original post by christinajane)

    Now shape Y is a sector of a circle so I have used the area equation for a sector to give:

    A = 1/2 (2d^2) theta
    It's not A = \frac{1}{2} (2d^2), it's A = \frac{1}{2}(2d)^2 = \frac{1}{2} (2^2 \times d^2) = \frac{1}{2} (4 \times d^2) = 2d^2
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    (Original post by Zacken)
    It's not A = \frac{1}{2} (2d^2), it's A = \frac{1}{2}(2d)^2 = \frac{1}{2} (2^2 \times d^2) = \frac{1}{2} (4 \times d^2) = 2d^2
    Yours is prettier than mine
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    Ahhh ok!

    Thanks guys - its simple basic things like that - that catch me out ALL the time!!

    Its so frustrating!

    I threw myself off track by not writing the square outside the brackets - duh!
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    (Original post by christinajane)
    Ahhh ok!

    Thanks guys - its simple basic things like that - that catch me out ALL the time!!

    Its so frustrating!
    Remember (ab)^2 = a^2b^2.
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    Yeah thanks alot! You have saved me from hours of frustration trying to work out something simple like that!

    things like that and the order of operations when solving an equation or anything always get me - I know the methods its those silly mistakes that are gonna cost me on the exam.
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    (Original post by christinajane)
    Yeah thanks alot! You have saved me from hours of frustration trying to work out something simple like that!

    things like that and the order of operations when solving an equation or anything always get me - I know the methods its those silly mistakes that are gonna cost me on the exam.
    It comes with practice. Oh, and quick tip, if you're ever not sure about an order of operation, try plugging numbers in. So if you're not sure whether (ab)^2 = a^2b^2 or (ab)^2 = ab^2 then try doing something like:

    (3\times 2)^2 = 6^2 = 36 = 3^2 \times 2^2 = 9 \times 4 = 36 but (3\times 2)^2 = 36 \neq 3\times 2^2 = 3 \times 4 = 12
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    Yeah thats a good tip thanks! Its amazing how different an answer can be if you do the orders wrong.

    There is too much to learn.

    Hope the exams are not as hard or long winded as the questions in this text book!
 
 
 
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