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Find the area of a triangle with side lengths frac43, 2, and frac83

waffles  Aug 10, 2017

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 #1
avatar+18712 
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Find the area of a triangle with side lengths frac43, 2, and frac83

 

Let \(a = \frac43\)

Let \(b = 2\)

Let \(c = \frac83\)

 

\(\begin{array}{rcll} \mathbf{Formula:} \\ \hline \text{Area of a triangle with Heron} \\ \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \\ s &=& \frac{a+b+c}{2} \\ \end{array} \)

 

 

\(\begin{array}{lcll} \mathbf{s = \ ? } \\ \begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \\ &=& \frac{\frac43+2+\frac83}{2} \\ &=& \frac23+1+\frac43 \\ &=& \frac63+1 \\ &=& 2+1 \\ &\mathbf{=}& \mathbf{3} \\ \hline \end{array} \\ \end{array} \)

 

\(\begin{array}{lcll} \mathbf{A = \ ? } \\ \begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \\ &=&\sqrt{3(3-\frac43)(3-2)(3-\frac83)} \\ &=&\sqrt{3(\frac{9}{3}-\frac43)(1)(\frac{9}{3}-\frac83)} \\ &=&\sqrt{3(\frac{9-4}{3})(\frac{9-8}{3})} \\ &=&\sqrt{3(\frac{5}{3})(\frac{1}{3})} \\ &=&\sqrt{5(\frac{1}{3})} \\ &\mathbf{=}&\mathbf{\sqrt{\frac53}} \\ \hline \end{array} \\ \end{array}\)

 

The area is \( \mathbf{\sqrt{\frac53} = 1.29}\)

 

laugh

heureka  Aug 10, 2017
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1+0 Answers

 #1
avatar+18712 
+2
Best Answer

Find the area of a triangle with side lengths frac43, 2, and frac83

 

Let \(a = \frac43\)

Let \(b = 2\)

Let \(c = \frac83\)

 

\(\begin{array}{rcll} \mathbf{Formula:} \\ \hline \text{Area of a triangle with Heron} \\ \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \\ s &=& \frac{a+b+c}{2} \\ \end{array} \)

 

 

\(\begin{array}{lcll} \mathbf{s = \ ? } \\ \begin{array}{|rcll|} \hline s &=& \frac{a+b+c}{2} \\ &=& \frac{\frac43+2+\frac83}{2} \\ &=& \frac23+1+\frac43 \\ &=& \frac63+1 \\ &=& 2+1 \\ &\mathbf{=}& \mathbf{3} \\ \hline \end{array} \\ \end{array} \)

 

\(\begin{array}{lcll} \mathbf{A = \ ? } \\ \begin{array}{|rcll|} \hline A &=& \sqrt{s(s-a)(s-b)(s-c)} \\ &=&\sqrt{3(3-\frac43)(3-2)(3-\frac83)} \\ &=&\sqrt{3(\frac{9}{3}-\frac43)(1)(\frac{9}{3}-\frac83)} \\ &=&\sqrt{3(\frac{9-4}{3})(\frac{9-8}{3})} \\ &=&\sqrt{3(\frac{5}{3})(\frac{1}{3})} \\ &=&\sqrt{5(\frac{1}{3})} \\ &\mathbf{=}&\mathbf{\sqrt{\frac53}} \\ \hline \end{array} \\ \end{array}\)

 

The area is \( \mathbf{\sqrt{\frac53} = 1.29}\)

 

laugh

heureka  Aug 10, 2017

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