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    (Original post by Dalek1099)
    1/2base*height works for every triangle.For non right angled trigonometry the height isn't one of the side lengths, which can make it difficult to work out the height so you generally use 1/2absinc but the height and base of that triangle are easy to calculate in the 1/2absinc bsinc is the height.
    I see thanks
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    (Original post by Dalek1099)
    1/2base*height works for every triangle.For non right angled trigonometry the height isn't one of the side lengths, which can make it difficult to work out the height so you generally use 1/2absinc but the height and base of that triangle are easy to calculate in the 1/2absinc bsinc is the height.
    So we consider the perpendicular height, the height on this diagram cannot be identified, it is not a right angled triangle and the angle has not been given and no calculators are allowed for this question as I'm sure you're aware of.
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    (Original post by zed963)
    So you're doing C1,C2 and S1? I'm assuming.
    Yh in year 11 though
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    (Original post by zed963)
    The triangle I've got isn't a right angled triangle. So I can't apply that formula. It's times by 3.
    yes x 3, that's right
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    (Original post by zed963)
    So we consider the perpendicular height, the height on this diagram cannot be identified, it is not a right angled triangle and the angle has not been given and no calculators are allowed for this question as I'm sure you're aware of.
    How do you measure the height of a person in real life? top to bottom-you are making this harder than it is.
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    (Original post by Pride)
    yes x 3, that's right
    I don't understand how though, if you look at my working out you'll have a better idea.
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    (Original post by Dalek1099)
    How do you measure the height of a person in real life? top to bottom-you are making this harder than it is.
    No I'm not, If we were to calculate the triangle area wouldn't we split it up into two triangle and then calculate the area?

    So when we have this question how can we get away with 0.5ab?
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    (Original post by zed963)
    No I'm not, If we were to calculate the triangle area wouldn't we split it up into two triangle and then calculate the area?

    So when we have this question how can we get away with 0.5ab?
    we are not the height is not a side of the triangle(it is bsinc) the height is simply 3 the distance from the base to the bottom point(0-3) ignore the negative.
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    (Original post by zed963)
    So you're doing C1,C2 and S1? I'm assuming.
    Name:  2014-04-17 20.00.00.jpg
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    (Original post by Super199)
    Yh in year 11 though
    Check my last post mate.
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    (Original post by Galileo Galilei)
    Name:  2014-04-17 20.00.00.jpg
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    I see.
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    (Original post by Dalek1099)
    we are not the height is not a side of the triangle(it is bsinc) the height is simply 3 the distance from the base to the bottom point(0-3) ignore the negative.
    I see that its 3 away but I don't know how that actually works, like the theory behind it.
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    (Original post by zed963)
    I see.
    Should I explain it?
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    (Original post by Galileo Galilei)
    Should I explain it?
    Why not.
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    (Original post by zed963)
    Why not.
    What is it that you don't understand? The right angle?
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    (Original post by Galileo Galilei)
    What is it that you don't understand? The right angle?
    I see that there's a right angle at P,

    However we don't know the length of AP nor PB so how can we then use 7.5*3?
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    (Original post by zed963)
    I see that there's a right angle at P,

    However we don't know the length of AP nor PB so how can we then use 7.5*3?
    1/2 base x height is the formula I used. Base= 8-1/2= 7.5 or 15/2 and Height= 3 due to the fact the point P has a y coordinate of -3 and so is 3 units from the x axis and the x axis is our base of the triangle.
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    (Original post by zed963)
    I see that there's a right angle at P,

    However we don't know the length of AP nor PB so how can we then use 7.5*3?
    I've made this easier for you by flipping the triangle and showing the distances we know marked in red underlining. NB we know every single length of the triangle because we know every single point. But we don't need AP or PB.

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    1/2 x b x h is a basic rule for calculating the area of any triangle.

    If you don't get it, you need to go over the basics again.
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    (Original post by NikolaT)
    I've made this easier for you by flipping the triangle and showing the distances we know marked in red underlining. NB we know every single length of the triangle because we know every single point. But we don't need AP or PB.

    Name:  triangle.png
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    1/2 x b x h is a basic rule for calculating the area of any triangle.

    If you don't get it, you need to go over the basics again.
    Oh, that makes it clearer, I just wasn't looking at it like that.
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    (Original post by zed963)
    No I'm not, If we were to calculate the triangle area wouldn't we split it up into two triangle and then calculate the area?

    So when we have this question how can we get away with 0.5ab?
    Name:  alt.gif
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    Take a look at the above image.

    Let's find the area of each triangle individually:

    The area of the triangle on the left side of the diagram is \frac{1}{2}(h)(w)
    And the area of the triangle on the right is \frac{1}{2}(h)(c-w)

    Add them together to get the total area for the large triangle

    \frac{1}{2}(h)(w) + \frac{1}{2}(h)(c-w) = \frac{1}{2}(h)(w+(c-w)) = \frac{1}{2}(h)(c)

    So, as you can see, the area for the large triangle is indeed \frac{1}{2} \times base \times height.

    Image URL: http://www.maths.surrey.ac.uk/hosted...Pythag/alt.gif
 
 
 
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