Question
The radius $r$ of a circle inscribed in any triangle whose sides are $a, b$, and $c$ is given by$$r=[(s-a)(s-b)(s-c) / s]^{1 / 2}$$where $s$ is an abbreviation for $(a+b+c) / 2 .$ Check this formula for dimensional consistency.
Step 1
The formula is $$ r=[(s-a)(s-b)(s-c) / s]^{1 / 2} $$ where $r$ is the radius of the circle, $a$, $b$, and $c$ are the sides of the triangle, and $s$ is the semi-perimeter of the triangle given by $(a+b+c) / 2$. Show more…
Show all steps
Your feedback will help us improve your experience
Ravindra Yadav and 68 other Physics 101 Mechanics educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
The radius $r$ of a circle inscribed in any triangle whose sides are $a, b$, and $c$ is given by $$ r=[(s-a)(s-b)(s-c) / s]^{1 / 2} $$ where $s$ is an abbreviation for $(a+b+c) / 2 .$ Check this formula for dimensional consistency.
The radius of a circle inscribed in any triangle whose sides are $a, b,$ and $c$ is given by the following equation, in which $s$ is an abbreviation for $(a+b+c) \div 2$ Check this formula for dimensional consistency. $$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
The radius of a circle inscribed in any triangle whose sides are $a, b,$ and $c$ is given by the following equation, in which s is an abbreviation for $(a+b+c) \div 2$. Check this formula for dimensional consistency. $$r=\sqrt{\frac{(s-a)(s-b)(s-c)}{s}}$$
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD