the sides of a triangle are 80 cm, 100cm, and 140 cm. determine the radius of the inscribed circle

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- Nov 3rd 2010, 11:10 PMaeroflixinscribed circle
the sides of a triangle are 80 cm, 100cm, and 140 cm. determine the radius of the inscribed circle

- Nov 4th 2010, 05:16 AMearboth
1. Draw a sketch.

2. Calculate the area of the triangle: Use Cosine rule to determine one interior angle (i.e. $\displaystyle \alpha$) of the triangle, for instance the angle opposite the side a. Then the area is:

$\displaystyle A = \frac12 \cdot c \cdot b \cdot \sin(\alpha)$

3. The center of the inscribed circle is the common vertex of three interior triangles whose area is:

$\displaystyle \frac12 \cdot a \cdot r + \frac12 \cdot b \cdot r + \frac12 \cdot c \cdot r = A$

$\displaystyle \frac12 \cdot (a+b+c) \cdot r = A~\implies~\boxed{r = \dfrac{2A}{a+b+c}}$ - Nov 5th 2010, 11:35 PMearboth
Here comes a different approach to solve this question:

1. Use Cosine rule to determine the interior angles of the triangle.

2. Let denote $\displaystyle \alpha$ the angle at A and $\displaystyle \beta$ the angle at B. Then you know:

$\displaystyle \underbrace{\dfrac r{\tan\left(\frac{\alpha}{2} \right)}}_{x} + \underbrace{\dfrac r{\tan\left(\frac{\beta}{2} \right)}}_{y} = c$

3. Solve for r:

$\displaystyle r = c \cdot \dfrac{\tan\left(\frac{\alpha}{2} \right) \cdot \tan\left(\frac{\beta}{2} \right)}{\tan\left(\frac{\alpha}{2} \right) + \tan\left(\frac{\beta}{2} \right)}$