+1223 The inradius can be found using the formula A = rs, where A is the triangle's area, r is the inradius, and s is the triangle's semiperimeter.
\(s= \frac{29+29+40}{2} = 49\)
\(A = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(49)(20)(20)(9)} = 420\)
\(\rightarrow r = \frac{A}{s} = \frac{420}{49} = \boxed{\frac{60}{7}}\)
+14995 Find the inradius of triangle ABC if it's sides lengths are 29, 29, 40.
Hello Guest!
a = b = 29 c = 40
\(cos(\alpha)=\frac{c}{2\cdot b}=\frac{40}{2\cdot 29 }\\ \alpha =46.397^o\\ f(x)=tan( \frac{\alpha}{2})x\\ x=\frac{c}{2}=20\\ r=tan( \frac{\alpha}{2})\cdot x=tan(\frac{46.397^o}{2})\cdot 20 \)
\(r=8.\overline{571428}=\large \frac{60}{7}\)
!
