In a triangle ABC if a,b,c be length of sides AB,BC,AC then then radius of ciircumcircle is given by \[\frac{abc}{4 \Delta }\]
where \[ \Delta \]\[=\sqrt{s(s-a)(s-b)(s-c)}\] area of triangle
and \[s=\frac{a+b+c}{2}\] s is called semiperimeter of the triangle.
Question. In a triangke ABC ,a:b:c=4:5:6. then find the ratio of the radius of semicircle to the incircle is
solution.
let \[ \frac{a}{4}=\frac{b}{5}=\frac{c}{6}=k\]
then radius of circumcircle is given by \[ R=\frac{abc}{4 \Delta }\]
and radius of incircle is given by \[ \frac{ \Delta }{s}\]
\[ \frac{R}{r}=\frac{abc}{4 \Delta }\times \frac{s}{ \Delta }=\frac{sabc}{4 \Delta ^2}\]
=\[ \frac{sabc}{4s(s-a)(s-b)(s-c)}\]
=\[ \frac{4k\times 5k\times 6k}{4(\frac{15}{2}k-4k)(\frac{15}{2}k-5k)(\frac{15}{2}k-6k)}\]
=\[ \frac{5\times 6\times 8}{7\times 5\times 3}\]=\[ \frac{16}{7}\]
