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Find the inradius of a triangle with 6, 25, and 29.

 Jun 30, 2020
 #1
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deleted

 
 Jun 30, 2020
edited by thelizzybeth  Jun 30, 2020
 #2
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Find the inradius of a triangle with 6, 25, and 29.

 

\(\text{Let $a=6$}\\ \text{Let $b=25$}\\ \text{Let $c=29$}\\ \text{Let the inradius of the triangle $= \rho$}\\ \text{Let $s=\dfrac{6+25+29}{2} = 30 $}\\ \text{Let the area of the triangle $A=\rho s $}\\ \text{Let the area of the triangle $A=\sqrt{s(s-a)(s-b)(s-c)} \quad (Heron) $}\)

 

 

\(\begin{array}{|rcll|} \hline A = \rho s &=& \sqrt{s(s-a)(s-b)(s-c)} \\\\ \rho s &=& \sqrt{s(s-a)(s-b)(s-c)} \\\\ \rho &=& \dfrac{ \sqrt{s(s-a)(s-b)(s-c)} } {s} \\\\ \rho &=& \sqrt{\dfrac{s(s-a)(s-b)(s-c)} {s^2} } \\\\ \mathbf{ \rho } &=& \mathbf{\sqrt{ \dfrac{ (s-a)(s-b)(s-c) } {s}}} \\\\ \rho &=& \sqrt{ \dfrac{ (30-6)(30-25)(30-29) } {30}} \\\\ \rho &=& \sqrt{ \dfrac{24*5*1 } {30}} \\\\ \rho &=& \sqrt{ \dfrac{120 } {30}} \\\\ \rho &=& \sqrt{4} \\\\ \mathbf{\rho} &=& \mathbf{2} \\ \hline \end{array}\)

 

The inradius of a triangle is \(\mathbf{2}\)

 

laugh

 
 Jun 30, 2020
edited by heureka  Jun 30, 2020
edited by heureka  Jun 30, 2020

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