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Find area of triangle with sides 4, 6, and 8 in simplest radical form

 Dec 18, 2018

Best Answer 

 #1
avatar+4296 
+2

Heron's: \(\sqrt{s(s-a)(s-b)(s-c)}\), where \(s\) is the semiperimeter and \(a,b,c\) are the sides. The semiperimeter can be found by \(\frac{A+B+C}{2}=\frac{4+6+8}{2}=\frac{18}{2}=9.\) Now, we have \(\sqrt{9(9-4)(9-6)(9-8)}=\sqrt{9*5*3*1}=\sqrt{135}=3\sqrt{15}.\)

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 Dec 18, 2018
 #1
avatar+4296 
+2
Best Answer

Heron's: \(\sqrt{s(s-a)(s-b)(s-c)}\), where \(s\) is the semiperimeter and \(a,b,c\) are the sides. The semiperimeter can be found by \(\frac{A+B+C}{2}=\frac{4+6+8}{2}=\frac{18}{2}=9.\) Now, we have \(\sqrt{9(9-4)(9-6)(9-8)}=\sqrt{9*5*3*1}=\sqrt{135}=3\sqrt{15}.\)

tertre Dec 18, 2018

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