## Chiomatn93 Group Title find the area of the triangle whose sides have the given lengths. a=9, b=12, c=15 2 years ago 2 years ago

1. shivam_bhalla Group Title

Do you know Heron's formula @Chiomatn93 ? Apply that and you will get your answer

2. shivam_bhalla Group Title
3. ipm1988 Group Title

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

4. Chiomatn93 Group Title

whats the answer and how do u solve?

5. ipm1988 Group Title

6. shivam_bhalla Group Title

@Chiomatn93 , Didn't you see the wiki link I gave above ?

7. Chiomatn93 Group Title

how do u solve it though? like the steps???

8. shivam_bhalla Group Title

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 Group Title

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 Group Title

damn calm down...jeez i will do but i just want an example

11. ipm1988 Group Title

see my reply solution is there

12. Chiomatn93 Group Title

13. ipm1988 Group Title

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 Group Title

I cannot put it more simply

15. Chiomatn93 Group Title

thank you! i understand it much more now!