AIEEE 2005 (Offline)Question 1
In a triangle $ABC$, let $\angle C=\frac{\pi}{2}$. If $r$ is the inradius and $R$ is the circumradius of the triangle $ABC$, then $2(r+R)$ equals :
Trigonometric Ratio & Identities
Solution
Concept. In a right triangle the hypotenuse is a diameter of the circumcircle, and the tangent length from the right-angle vertex is $r$. Formulas used. (1) $2R=c$ when $C=90^\circ$(2) $r=(s-c)\tan\frac C2=s-c$ when $C=90^\circ$Step 1 - Each radius. By (1) and (2): $2R=c$ and $r=\frac{a+b-c}{2}$. Step 2 - Add. $2(r+R)=(a+b-c)+c=a+b$. Answer: (B)