SEE ATTACHMENT FOR QUESTION DETAILS
Call the horizontal leg of the triangle x-axis.
Call the vertical leg of the triangle y-axis.
On the x-axis draw a line perpendicular such that it will be tangent to the circle on the left side.
On the y-axis draw a line perpendicular such that it will be tangent to the circle on the top.
Already you know the measurements of the entire x-axis and y-axis, or legs of the triangle.
It should not be very difficult to measure the diameter of the circle from the box you just drew around it.
If that is not in the spirit of your text, look up TANGENT TO DIAMETER THEOREM:
If a chord in a plane of a circle is perpendicular to a tangent at the point of tangency, the chord is a diameter.
You have at least two points of tangency in your drawing.
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Here are instructions from a text:
At the point of tangency of each tangent line, draw a chord perpendicular to each tangent.
Because of the TANGENT TO DIAMETER THEOREM, these two chords will be diameters and will intersect each other at the center of the circle.
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That means you will also have two right triangles that point to nose-to-nose in the center.
The base of each triangle will be a radius.
You are no better off as you will still have to measure.
If you find a numeric solution, please let me know.
I've modified your sketch a little bit.
1. Use Tangens to calculate $\displaystyle \angle \alpha = 36.87^\circ$
2. The left right triangle will give you:
$\displaystyle x = (8-x) \cdot \tan\left(\frac{\alpha}{2}\right)$
3. Expand the bracket, collect like terms:
$\displaystyle x+ x \cdot \tan\left(\frac{\alpha}{2}\right) = 8 \cdot \tan\left(\frac{\alpha}{2}\right)$
and therefore:
$\displaystyle x = \frac{8 \cdot \tan\left( \frac{\alpha}{2}\right)}{1+\tan\left(\frac{\alpha} {2}\right)}~\implies~ x\approx 2.0$
With your problem you were looking for the length of the radius of the inscribed circle.
1. If you know the length of the three sides of a triangle you can calculate
- the interior angles (Cosine rule)
- the area of the triangle
2. The center of the incircle is the intersection of the angle bisectors of the interior angles.
3. Divide the triangle into 3 triangles so that the radius of the incircle is the height. $\displaystyle r_1 = r_2 = r_3 = r$
4. Let $\displaystyle A_{total}$ denote the area of the complete triangle. Then you know:
$\displaystyle A_{total} = \underbrace{\frac12 \cdot c \cdot r}_{red \ area} + \underbrace{\frac12 \cdot a \cdot r}_{blue \ area} + \underbrace{\frac12 \cdot b \cdot r}_{green \ area}$
$\displaystyle A_{total}=\frac12 \cdot (a+b+c) \cdot r$
$\displaystyle \boxed{r = \frac{2A_{total}}{a+b+c}}$
With your problem:
a = 8 cm, b = 6 cm, c = 10 cm
$\displaystyle A_{total} = 24 \ cm^2$
And therefore: $\displaystyle r = \frac{2 \cdot 24}{6+8+10} = 2$