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Chiomatn93

  • 2 years ago

find the area of the triangle whose sides have the given lengths. a=9, b=12, c=15

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  1. shivam_bhalla
    • 2 years ago
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    Do you know Heron's formula @Chiomatn93 ? Apply that and you will get your answer

  2. shivam_bhalla
    • 2 years ago
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    http://en.wikipedia.org/wiki/Heron%27s_formula

  3. ipm1988
    • 2 years ago
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    \[s= (9+12+15)/2 =18 now by heron's formula Area=\sqrt{18* (18-9) (18-12)(18-15)}\] \ [\sqrt{18*9*6*3} = \sqrt{2916} = 54 therefore your area is 54

  4. Chiomatn93
    • 2 years ago
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    whats the answer and how do u solve?

  5. ipm1988
    • 2 years ago
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    Answer is 54

  6. shivam_bhalla
    • 2 years ago
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    @Chiomatn93 , Didn't you see the wiki link I gave above ?

  7. Chiomatn93
    • 2 years ago
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    how do u solve it though? like the steps???

  8. shivam_bhalla
    • 2 years ago
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    Commo'n the formula is there, the quantities are given. Now can't you multiply and take the square root by yourself

  9. shivam_bhalla
    • 2 years ago
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    See the formula is \[Area = \sqrt{s(s-a)(s-b)(s-c)}\] where \[s= {a+b+c \over 2}\] where a , b, c are the length of the sides of the triangle. First find s. then substitute to get your Area ??

  10. Chiomatn93
    • 2 years ago
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    damn calm down...jeez i will do but i just want an example

  11. ipm1988
    • 2 years ago
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    see my reply solution is there

  12. Chiomatn93
    • 2 years ago
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    wheres the answer?

  13. ipm1988
    • 2 years ago
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    Ok first you would have to take out the s that is semi perimerter formula for s is S= (a +b +c)/2 here the S is 18 Area of triangle is \[A=\sqrt{s*(s-a)*(s-b)*(s-c)}\] So the area would be now \[A=\sqrt{18*(18-9)*(18-12)*(18-15)}\] \[A=\sqrt{18*5*6*3}\] \[A=\sqrt{2916}\] A=54 Therefore the area of triangle is A=54 which is your answer

  14. ipm1988
    • 2 years ago
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    I cannot put it more simply

  15. Chiomatn93
    • 2 years ago
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    thank you! i understand it much more now!

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