Problem: In β³ABC,AC=24β²β²,BC=10β²β², and AB=26β²β². The radius of the inscribed circle is:
Answer Choices:
A. 26 in.
B. 4 in.
C. 13 in.
D. 8 in.
E. none of these answers
Solution:
For any right triangle we have aβr+bβr=c where r is the radius of the inscribed circle.
β΄2rβ΄rβ=a+bβc=24+10β26=8;=4.β