Question

The radius of circumcircle of triangle PRS with three sides \[2\sqrt{3},6\sqrt{2},10\] is
A. \[5\]
B. \[3\sqrt{3}\]
C. \[3\sqrt{2}\]
D. \[2\sqrt{3}\]

Answer 1

Hint: We have to know the cosine rule in a triangle that is \[\cos P=\dfrac{{{r}^{2}}+{{s}^{2}}-{{p}^{2}}}{2rs}\] and we will get \[\cos P\] and then find \[\sin P\]by trigonometric identity that is \[{{\sin }^{2}}P+{{\cos }^{2}}P=1\] and then we have to know the sine rule in a triangle that is \[\Delta PRS:\dfrac{p}{\sin P}=\dfrac{r}{\sin R}=\dfrac{s}{\sin S}=2R\]. By this way we will get the circumradius of the triangle.

Complete step-by-step solution -

Given, the sides of triangle PRS are \[2\sqrt{3},6\sqrt{2},10\]
Hence p= \[2\sqrt{3}\]. . . . . . . . . . . . . . . . . . . . . . . (1)
            r= \[6\sqrt{2}\]. . . . . . . . . . . . . . . . . . . . . . . .(2)
              s= \[10\]. . . . . . . . . . . . . . . . . . . . . . . . . (3)
we know that in a triangle according to cosine rule \[\cos P=\dfrac{{{r}^{2}}+{{s}^{2}}-{{p}^{2}}}{2rs}\]
\[\cos P=\dfrac{{{10}^{2}}+{{\left( 6\sqrt{2} \right)}^{2}}-{{\left( 2\sqrt{3} \right)}^{2}}}{2\times 10\times 6\sqrt{2}}\]
\[\cos P=\dfrac{100+72-12}{120\sqrt{2}}\]
\[\cos P=\dfrac{160}{120\sqrt{2}}=\dfrac{4}{3\sqrt{2}}\]
Rationalizing the both numerator and denominator by multiplying with \[\sqrt{2}\] we will get,
\[\cos P=\dfrac{4}{3\sqrt{2}}\times \dfrac{\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{3}\]. . . . . .. . . . . . . .(4)
We know the trigonometric identity that \[{{\sin }^{2}}P+{{\cos }^{2}}P=1\]
By substituting the value of \[\cos P\] we will get \[\sin P\]
\[\sin P=\sqrt{1-\dfrac{8}{9}}=\sqrt{\dfrac{1}{9}}\]
\[\sin P=\dfrac{1}{3}\]. . . . . . . . .. . . .(5)
According to sine rule we know that in a triangle, \[\Delta PRS:\dfrac{p}{\sin P}=\dfrac{r}{\sin R}=\dfrac{s}{\sin S}=2R\]
We can take the term \[\dfrac{p}{\sin P}=2R\]
By substituting the values of p and \[\sin P\] we will get the circumradius of triangle PRS as follows
\[\dfrac{2\sqrt{3}}{\dfrac{1}{3}}=2R\]
\[2R=6\sqrt{3}\]
\[R=3\sqrt{3}\]
So, the correct option is option (B).

Note: In the above we find \[\cos P\] by using cosine rule it is not mandatory to find only \[\cos P\] we can find \[\cos R\] or else \[\cos S\] and then using the trigonometric identity find \[\sin R\] and \[\sin S\], by applying the sine rule we can find the circumradius of triangle PRS.