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Triangle's sides are a=5 b=12 and c=9. What is the area of the triangle?

Guest Jul 25, 2014

Best Answer 

 #1
avatar+26971 
+5

Heron's formula can be used when you just know the length of each side. This says

Area = √[s*(s-a)*(s-b)*(s-c)]  where a, b and c are the lengths of the sides and s = (a+b+c)/2.

Here: s = (5+12+9)/2 = 13, so:

$${\mathtt{Area}} = {\sqrt{{\mathtt{13}}{\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{5}}\right){\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{12}}\right){\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} \Rightarrow {\mathtt{Area}} = {\mathtt{20.396\: \!078\: \!054\: \!371\: \!139\: \!3}}$$

Area ≈ 20.396

Alan  Jul 25, 2014
 #1
avatar+26971 
+5
Best Answer

Heron's formula can be used when you just know the length of each side. This says

Area = √[s*(s-a)*(s-b)*(s-c)]  where a, b and c are the lengths of the sides and s = (a+b+c)/2.

Here: s = (5+12+9)/2 = 13, so:

$${\mathtt{Area}} = {\sqrt{{\mathtt{13}}{\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{5}}\right){\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{12}}\right){\mathtt{\,\times\,}}\left({\mathtt{13}}{\mathtt{\,-\,}}{\mathtt{9}}\right)}} \Rightarrow {\mathtt{Area}} = {\mathtt{20.396\: \!078\: \!054\: \!371\: \!139\: \!3}}$$

Area ≈ 20.396

Alan  Jul 25, 2014

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