+0  
 
0
483
5
avatar+195 

What is the area of a triangle with side lengths $13$, $14$, and $15$?

 Feb 28, 2021
 #1
avatar+201 
+1

Herons formula is \(\sqrt{s(s-a)(s-b)(s-c)}\)\(s\) is the perimiter divided by 2, and a, b, and c are the side lengths. We sum 13, 14, and 15 to get 42, then we plug in the lengths. \(\sqrt{42(42-13)(42-14)(42-15)}\)\(=\boxed{126\sqrt{58}}\)

 Feb 28, 2021
 #2
avatar+195 
0

that is wrong

QuestionMachine  Feb 28, 2021
 #3
avatar+201 
+1

oh my bad I made a mistake

 

The answer is \(\sqrt{7056}=84\)

jxc516  Feb 28, 2021
 #5
avatar+37153 
0

QM....if you ACTUALLY looked at jcx516' s  origianl solution....YOU should have been able to spot the minor error made in the calculation ....   then you would not have to have been so rude/curt and unappreciative  :     "that is wrong"

ElectricPavlov  Feb 28, 2021
 #4
avatar+1223 
+1

You can split a 13-14-15 triangle into a 9-12-15 triangle and a 5-12-13 triangle, so the area is \(\frac{9 \cdot 12}{2} + \frac{5 \cdot 12}{2} = 84\).

 Feb 28, 2021

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