well my question is how to calculate the area of the circle when its in a triangle and i only have the units for the triangle and not the circle.
here is the actual problem: all the sides of the triangle is 8 units each. then there is a circle perfectly placed inside. and i need to find the area of the circle.

so far i think i need to find the area of a triangle first, which i did and i just need to get going...

2. Originally Posted by svetka
well my question is how to calculate the area of the circle when its in a triangle and i only have the units for the triangle and not the circle.
here is the actual problem: all the sides of the triangle is 8 units each. then there is a circle perfectly placed inside. and i need to find the area of the circle.

so far i think i need to find the area of a triangle first, which i did and i just need to get going...
In an equilateral triangle of sidelength $a$ the length of the median is $\frac{\sqrt{3}}{2} \cdot a$

The radius of the circle is 1/3 times the length of the median because the centre of the circle is the centroid of the triangle and hence divides the median in the ratio 2:1.

Therefore the radius of the circle is $\frac{a \sqrt{3}}{6}$.

In your question $a = 8$.

3. Hello, svetka!

Another approach . . .

A circle is inscribed in an equilateral triangle of side 8.
Find the area of the circle.
Code:
                        *
/ \
/   \
/     \
/       \
/         \
/   * * *   \
/*           *\
*               *
*  \           /  *
/    r\       /r    \
/*       \   /       *\
/ *         *         * \
/  *         |         *  \
/             |             \
/     *        |r       *     \
/       *       |       *       \
/          *     |     *          \
*-------------*-*-*-----------------*
: - - - - - - - - 8 - - - - - - - - :

The area of an equilateral triangle of side $x$ is: . $A \:=\:\frac{\sqrt{3}}{4}x^2$
We have $x = 8$, so: . $A \:=\:\frac{\sqrt{3}}{4}(8^2) \:=\:16\sqrt{3}$

Formula: .The area of a triangle is: . $A \:=\:\tfrac{1}{2}pr$
. . . . . . . where $p$ is the perimeter, and $r$ is the radius of the inscribed circle.

The perimeter of this triangle is 24, so we have: . $\tfrac{1}{2}(24)r \:=\:16\sqrt{3}$

Hence: . $12r \:=\:16\sqrt{3} \quad\Rightarrow\quad r \:=\:\frac{4\sqrt{3}}{3}$

Now you can find the area of the circle . . .