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Find Area of an isosceles triangle

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    • Thread Starter

    I want to find out the area of an isosceles triangle which has only a base of 12 cm and the lower two corners both have an angle of 54°

    Here you go. I am pretty sure this is how you do it but anyone else feel free to correct me.

    1. Cut the triangle in half so we have a right angled triangle where the Opposite side is representative of the height.

    2. Now we can do good old trig because we have a right angled triangle. We have the Adjacent side and we want to know the Opposite side so we use tan.(SOH CAH TOA)

    3. Substitute the values in with x being the height
    so x=6tan54 and that is 8.26 to 3sf

    4. Use the area of a triangle which is 1/2 bh. We now know the height is 8.26 so it would be 1/2 12x8.26 which is 49.56
    So the area of the triangle is 49.56cm² .

    To get a more accurate answer use 6tan54 all the way until the end.

    Messy paint diagram below (:
    Name:  yeah maths.png
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    You could use two methods for this:

    Method 1
    a) Since the two base angles are 54, the top angle is 180-(54*2) which is 72 degrees.
    b) You now know 3 angles and a side, so you can use the sine rule to find another side.
    c) Label the triangle with a,b,c, A,B, and C. Capital letters for angles, and corresponding letters (so side a is opposite angle A)
    d) Use the sine rule (which you can invert to make it easier to find what you want). so a/sin(a) = b/sin(b)
    e) Therefore 12/sin(72) = sin(54)/b
    f) Therefore b = (12/sin(72)) * sin(54)
    g) Therefore b = 10.2078097
    h) Now you can find the area by using the formula A = 0.5 x a x b x sin(C)
    g) So A = 0.5 x 12 x 10.2078097 x sin(54)
    f) Therefore the area is equal to 49.54974914cm^2, which is 49.5cm^2 to 3 significant figures.

    Method 2
    a) Split the isosceles triangle into half, creating two right angled triangles. Use one of them.
    b) The height of the right-angled triangle will be the height of the full triangle as well.
    c) The base is now 6, the top angle is 36 degrees (72/2) and the third angle is obviously 90 degrees
    d) The height is C. a/sin A = c/sin C
    e) Therefore 6/sin(36) * sin(54) = c, which is 8.258291523
    f) Area = 0.5 x base x height = 0.5 x 6 x 8.258291523
    g) Area = 24.77487457
    h) Since this was half the total triangle, the total area is double the previous number
    f) Therefore the area is equal to 49.54974914cm^2, which is 49.5cm^2 to 3 significant figures.

    Hope I helped!

    Area= \frac{1}{2} ab\sin C . Angle C would be 180-2(54) degrees. a=b = 12\text{cm} .
    • Thread Starter

    Thank you very much

    (Original post by Awesome Brother)
    Thank you very much
    To make it clear, the method suggested by Ano123, i.e. use of the formula 1/2*ab*sin(C), is by far the best, as it is much quicker than splitting into right-angled triangles etc.
    • Thread Starter

    Thanks to you i have got full 5 marks for my maths exam in this question. Thanks guys
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