What is the radius of the circle inscribed in triangle ABC if AB = 5, AC=6, BC=7? Express your answer in simplest radical form.
+14905 What is the radius of the circle inscribed in triangle ABC if AB = 5, AC=6, BC=7?
Hello friends !
The radius of the circle is r .
\(a=7\\ b=6\\ c=5 \)
\(c^2=a^2+b^2-2ab\cdot cos\gamma\\ \gamma=arccos\frac{a^2+b^2-c^2}{2ab}=arccos\frac{7^2+6^2-5^2}{2\cdot 7\cdot 6}=arccos 0.7142857\\ \gamma=44.4153°\)
\(b^2=a^2+c^2-2ac\ cos\beta\\ \beta=arccos\frac{a^2+c^2-b^2}{2ac}=arccos\frac{7^2+5^2-6^2}{2\cdot 7\cdot 5}=arccos\ 0.54285714\\ \beta=57.12165°\)
\(a=s+t\) (two sections on the side a,
on both sides of the contact with the circle)
\(tan\frac{\gamma}{2}=\frac{r}{s}\\ s=\frac{r}{tan \frac{\gamma}{2}}\) \(tan \frac{\beta}{2}=\frac{r}{t}\\ t= \frac{r}{tan \frac{\beta}{2}}\)
\(a=\frac{r}{tan \frac{\gamma}{2}}+\frac{r}{tan \frac{\beta}{2}}=7\)
\(\frac{r}{tan \frac{44.4153°}{2}}+\frac{r}{tan \frac{57.12165°}{2}}=7\)
\(\frac{r}{0.408248}+\frac{r}{0.544331}=7\)
\(0.544331r+0.408248r=7\cdot 0.408248\cdot 0.544331\\ r=\frac{7\ \cdot\ 0.408248\ \cdot\ 0.544331}{0.544331+0.408248}\)
\(r=1.63299\)
\(The\ radius\ of\ the\ circle\ inscribed\ in\ triangle\ ABC\ is\ 1.63299.\)
!
+349 First, we illustrate the problem:
I - the incenter (center of inscribed circle)
Then, draw lines like this:
Notice that the triangle has been split into 3 smaller ones, each with a height of the radius, and with a base of the sides of the large triangle. Which means that we can say:
\(a\triangle ABC=a\triangle AIB+a\triangle BIC+a\triangle CIA\)
\(a\triangle ABC={5r\over 2}+{6r\over 2}+{7r\over 2}\)
\(a\triangle ABC =9r\)
Now, we also know that \(a\triangle ABC=\sqrt{s(s-AB)(s-AC)(s-BC)}\) wherein \(s={AB+AC+BC\over2}\). This is called Heron's Formula. Substituting the values, we get:
\(\sqrt{9*4*3*2}=9r\)
\(6\sqrt{6}=9r\)
\(r={2\sqrt{6}\over3}\)
