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    Find the exact area of the triangle with side lengths of 4cm, 5cm and 6cm.
    I get the area to be 9.9cm^2.
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    Anyone?
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    (Original post by Ano9901whichone)
    Anyone?
    No angles?
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    http://www.mathopenref.com/heronsformula.html

    9.9 is right but it isn't exact
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    I was going to suggest that but Heron's doesn't involve trigonometry and he wants the "exact" answer, so 9.9 is not correct but rather rounded.
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    https://www.wolframalpha.com/input/?...F2-5)(15%2F2-6))

    \frac{15\sqrt{7}}{4}}

    1. Use Heron's formula
    2. Simplify
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    (Original post by Ano9901whichone)
    Find the exact area of the triangle with side lengths of 4cm, 5cm and 6cm.
    I get the area to be 9.9cm^2.
    Show your working, you don't necessarily need to use Heron's for this. I got \frac{15\sqrt7}{4}
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    Is Heron's formula GCSE?
    I've certainly never came across it, in either GCSE or A Level and had never heard of it before this thread.
    Did this by getting the cosine of an angle from the cosine rule, then working out sine of the same angle using SOHCAHTOA, and finally subbing the sine of the angle into the formula 1/2 ab sin C
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    (Original post by AlexS101)
    Is Heron's formula GCSE?
    I've certainly never came across it, in either GCSE or A Level and had never heard of it before this thread.
    Did this by getting the cosine of an angle from the cosine rule, then working out sine of the same angle using SOHCAHTOA, and finally subbing the sine of the angle into the formula 1/2 ab sin C
    No it's not. I just noticed this is in the GCSE section. Yes what you said is the best way for a GCSE student (I just couldn't be bothered to use the cosine rule and then the area formula)
 
 
 
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