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

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Do you know Heron's formula @Chiomatn93 ? Apply that and you will get your answer
\[s= (9+12+15)/2 =18 now by heron's formula Area=\sqrt{18* (18-9) (18-12)(18-15)}\] \ [\sqrt{18*9*6*3} = \sqrt{2916} = 54 therefore your area is 54

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whats the answer and how do u solve?
Answer is 54
@Chiomatn93 , Didn't you see the wiki link I gave above ?
how do u solve it though? like the steps???
Commo'n the formula is there, the quantities are given. Now can't you multiply and take the square root by yourself
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 ??
damn calm down...jeez i will do but i just want an example
see my reply solution is there
wheres the answer?
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
I cannot put it more simply
thank you! i understand it much more now!

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