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Math Help - Area of triangle.

  1. #1
    Newbie LaMandie's Avatar
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    Area of triangle.

    The question is as follows:
    The three sides of a triangle are 10m, 15m, and 20m. What is the area of the triangle?

    My diagram:
    Area of triangle.-problem.png
    (20 - x)^2 - x^2 = 15^2 - 10^2

    (20 -x)(20 - x) - x^2 = 125

    400 - 40x -2x^2 = 125

    -40x - 2x^2 = -275

    x - 2x^2 = 6.875

    -2x^2 = 6.875x

    x^2 = -3.4375x

    x = -3.4375

    Obviously it's not the right answer, but I have no idea how to get the right one. Answers are greatly appreciated!
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  2. #2
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    Quote Originally Posted by LaMandie View Post
    The question is as follows:
    The three sides of a triangle are 10m, 15m, and 20m. What is the area of the triangle?

    My diagram:
    Click image for larger version. 

Name:	problem.png 
Views:	38 
Size:	6.8 KB 
ID:	19805
    (20 - x)^2 - x^2 = 15^2 - 10^2

    (20 -x)(20 - x) - x^2 = 125

    400 - 40x -2x^2 = 125

    -40x - 2x^2 = -275

    x - 2x^2 = 6.875

    -2x^2 = 6.875x

    x^2 = -3.4375x

    x = -3.4375

    Obviously it's not the right answer, but I have no idea how to get the right one. Answers are greatly appreciated!
    Use Heron's formula: Heron's formula - Wikipedia, the free encyclopedia
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  3. #3
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    Quote Originally Posted by LaMandie View Post
    -40x - 2x^2 = -275
    x - 2x^2 = 6.875
    Why did you not also divide the -2x^2 by -40?
    There are other similar errors.
    Last edited by mr fantastic; November 22nd 2010 at 05:10 PM. Reason: Softened.
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  4. #4
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    Quote Originally Posted by LaMandie View Post
    The question is as follows:
    The three sides of a triangle are 10m, 15m, and 20m. What is the area of the triangle?

    My diagram:
    Click image for larger version. 

Name:	problem.png 
Views:	38 
Size:	6.8 KB 
ID:	19805
    (20 - x)^2 - x^2 = 15^2 - 10^2

    Using Pythagoras' theorem

    h^2=15^2-(20-x)^2=10^2-x^2

    225-(400-40x+x^2)=100-x^2

    -175+40x=100


    (20 -x)(20 - x) - x^2 = 125

    400 - 40x -2x^2 = 125 ......you should have 400-40x+x^2-x^2 here

    -40x - 2x^2 = -275

    x - 2x^2 = 6.875

    -2x^2 = 6.875x

    x^2 = -3.4375x

    x = -3.4375

    Obviously it's not the right answer, but I have no idea how to get the right one. Answers are greatly appreciated!
    You have correctly applied Pythagoras' theorem but should have cancelled the squares of x.
    Last edited by Archie Meade; November 22nd 2010 at 02:05 PM.
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  5. #5
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    Hello, LaMandie!

    The three sides of a triangle are 10m, 15m, and 20m.
    What is the area of the triangle?

    My diagram:
    Click image for larger version. 

Name:	problem.png 
Views:	38 
Size:	6.8 KB 
ID:	19805

    (20 - x)^2 - x^2 \:=\: 15^2 - 10^2 . Right1

    (20 -x)(20 - x) - x^2 \:=\: 125

    400 - 40x -2x^2 \:=\: 125 . . . . no

    (20-x)^2 - x^2 \:=\:125 \quad\Rightarrow\quad 400 - 40x + x^2 - x^2 \:=\:125

    . . 400 - 40x \:=\:125 \quad\Rightarrow\quad 40x \:=\:275 \quad\Rightarrow\quad x \:=\:\frac{275}{40} \:=\:\frac{55}{8}


    Then: . h^2 + x^2 \:=\:10^2 \quad\Rightarrow\quad h^2 + \left(\frac{55}{8}\right)^2 \:=\:100

    . . . . . h^2 \:=\:100 - \frac{3025}{64} \:=\:\frac{3375}{64}

    . . . . . h \:=\:\sqrt{\frac{3375}{64}} \:=\:\frac{15}{8}\sqrt{15}


    Therefore: . \text{Area} \;=\;\frac{1}{2}bh \;=\;\frac{1}{2}(20)\left(\frac{15}{8}\sqrt{15}\ri  ght) \;=\;\dfrac{75\sqrt{15}}{4}

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  6. #6
    Newbie sugarT's Avatar
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    hey there

    i have not used heron's formula before but you could do something else with some simple trig.

    1) find either angle using the Cosine Rule
    2) then find the area using A = BCsinA/2
    where A is the angle you found and B and C are the two sides that make the angle.
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  7. #7
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    Well, to each his own.
    I like this one, prefer it over Heron's:
    Area = cSQRT(a^2 - x^2) / 2 where x = (a^2 - b^2 + c^2) / (2c)

    Of course, a,b,c are triangle's sides: a =< b =< c
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