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Math Help - Area of an Isosceles Triangle

  1. #1
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    Area of an Isosceles Triangle

    In grade 9 math i got homework about finding the area of an isosceles triangle and i don't know how to find the area.
    Thanks,
    Josh
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  2. #2
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    Quote Originally Posted by josh16 View Post
    In grade 9 math i got homework about finding the area of an isosceles triangle and i don't know how to find the area.
    Thanks,
    Josh
    The area of an isosceles triangle, and any triangle for that matter, is:

    (1/2)*b*h, where b is the base and h is the height. Further, for an ISOSCELES triangle, height can be found by the following formula:

    h = sqrt[(b^2 - (1/2)*a^2)]

    The area of an isosceles triangle can also be found using trigonometry, but I am assuming you are not quite up to that level yet.

    Another formula you can consider using is Heron's formula.

    Given a triangle with sides of length a, b, c, the area of the triangle is:

    A = sqrt(s*(s-a)*(s-b)*(s-c)),

    where s = (a + b + c)/2.
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  3. #3
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    Quote Originally Posted by Ideasman View Post
    The area of an isosceles triangle, and any triangle for that matter, is:

    (1/2)*b*h, where b is the base and h is the height. Further, for an ISOSCELES triangle, height can be found by the following formula:

    h = sqrt[(b^2 - (1/2)*a^2)]

    The area of an isosceles triangle can also be found using trigonometry, but I am assuming you are not quite up to that level yet.

    Another formula you can consider using is Heron's formula.

    Given a triangle with sides of length a, b, c, the area of the triangle is:

    A = sqrt(s*(s-a)*(s-b)*(s-c)),

    where s = (a + b + c)/2.
    Small correction:

    For: h = sqrt[(b^2 - (1/2)*a^2)], I think you mean:

    h = sqrt[(b^2 - (1/4)*a^2)]
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