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Math Help - How to measure the area of a plot of land consisting of the following 3 triangles:

  1. #1
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    How to measure the area of a plot of land consisting of the following 3 triangles:

    How to measure the area of a plot of land ABCDE consisting of the following 3 triangles(without using Heron's formula for the measure of areas of triangles where all sides are known):

    [top triangle 1 ]BCD -side BC = 124 m
    - side CD = 47 m
    - side BD = unknown length [side BD is the top side of middle triangle 2 aswell]



    [middle triangle 2]ABD - side BD = unknown length
    - side AD = 82 m [side DA is the top side of lower triangle 3 aswell]
    - side AB = 54 m
    - angle BAD is a right angle



    [lower triangle 3]ADE - side AD = 82m
    - side DE = 28 m
    - side AE = 96 m

    Below is a useful visual representation of the situation. I've been stuck on this question for quite a while and would appreciate it a lot if someone could help me out with this one.
    Attached Thumbnails Attached Thumbnails How to measure the area of a plot of land consisting of the following 3 triangles:-trigo.gif  
    Last edited by Trigo; August 9th 2005 at 10:08 AM.
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  2. #2
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    Why can't we not use the Heron's formula? That is the only way to solve your question here.

    Another way would have been if you gave us more angles but you gave only one---the rihgt angle. That is not enough to solve the question if we are not to usae the Heron's formula.
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  3. #3
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    I agree that not using Heron's formula is a pointless restriction, but you can get round it. Suppose you know the sides a,b,c of a triangle. The cosine formula tells you that angle A satisfies a^2 = b^2 + c^2 - 2bc.cos(A). Since you know a,b,c then you know cos(A) and hence A. But the height of a triangle from vertex C to edge AB is b sin(A) and so the area of triangle ABC is half base times height = (1/2) bc sin(A).

    Since bc sin(A) = sqrt{(bc)^2 - (bc cos(A))^2} = sqrt{b^2c^2 - (1/4)(b^2+c^2-a^2)^2}, a little manipulation will show you that you're using Heron's formula 'in disguise'.
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  4. #4
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    I mentioned not to use Heron's formula because this formula as it is in its 'undisguised form' is not referred to in my book from which I am learning and therefore it seemed obvious to me that it was either a mistake to put such question in the book or that they simply expected one to make use of the methods and formulas which are available in the book. I did not think about the option of simply using te cosine rule to find out an angle of a triangle where all that is known about the triangle are the length of all its sides (many thanks for your showing me this possibility), and also didn't know that by combining the cosine rule and the formula for an area of a triangle using sine, (or atleast that's the way it looked like to me, if I am not right please tell me so), we can deduce the formula called Heron's formula. I'd like to thank you for showing me this aswell.
    So in summary what i've learned is: there is a way to measure the area of a triangle without using Heron's formula which is by using the cosine rule and the formula to measure the area of a triangle using sin ( which are the two methods learned in my book) but which in its turn give the same answer as Heron's formula because Heron's formula is simply an incorporation of these two methods (which can be called the disguised or not compact form of Heron's formula) .
    If any of the above is uncorrect or unclear I would be very pleased with more corrections so I could use these to improve my logical thinking.
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  5. #5
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    I made the mistake of misreading the question and over looked the right angle BAD.

    Is it possible to solve the problem without that knowledge? The cosine rule helps with the angles in one triangle (with all known sides) but is it possible to proceed any further? I spent a while trying and gave up so I'm just curious.

    Has anyone tried the crossword puzzle that someone posted? There were 16 unkowns and 8 equations so I'm guessing the solution isn't unique. I've tried fixing the values of some squares (like the ones before and after the divisions)in order to work out what the remaining squares must be but it ends in a muddle. Is there another strategie, other than guessing?
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  6. #6
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    The right angle BAD is the key to the solution of the problem. With that, we can get the length of hypotenuse BD. Then all the sides of the three triangles are known.
    Without that, then the quadrilateral BADC is a problem. With no known measure of any of its four interior angles, the area of BADC cannot be solved for its exact area---eventhough all its four sides are known.
    Having said that, I say we can still solve for the exact area of quadrilateral BADC even with just its 4 sides known. We can use another formula similar to the Heron's, but it is for quadrilaterals. It is called Brahmagupta's Formula. This formula has two versions: one for cyclic quadrilaterals, and one for non-cyclic quadrilaterals. With non-cyclics, though, we need to know at least two interior angles of the quadrilateral---they are to be opposite angles.

    Also, if there is no known interior angle in quadrilateral BADC, then graphical method, by using scales, can be used to get an approximate length for BD or AC, but that is not the purpose of this question.
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  7. #7
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    I've never come across Brahmaguptas formula before, I'll have to look that one up. Thanks.
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