To solve this problem, we will first clarify the setup of the triangle and the circles involved. The question involves an isosceles triangle, which is inscribed in a circle (the circumcircle), and has a circle inscribed within it (the incircle).

Given:

• The base of the isosceles triangle is $10 \, \text{cm}$.
• The length of each of the equal sides (slopes) of the isosceles triangle is $13 \, \text{cm}$.

We need to find the radii of both circles:

1. Circumradius (radius of the circumcircle): The radius of the circle that passes through all three vertices of the triangle.
2. Inradius (radius of the incircle): The radius of the circle that is tangent to all three sides of the triangle.

Step 1: Calculate the height of the isosceles triangle

First, we can find the height $h$ of the triangle by splitting the isosceles triangle into two right-angled triangles. The base of the isosceles triangle is $10 \, \text{cm}$, so each of the right-angled triangles has a base of $5 \, \text{cm}$ (half of the base of the isosceles triangle).

Using the Pythagorean theorem:

$h^2 + 5^2 = 13^2$ $h^2 + 25 = 169$ $h^2 = 169 - 25 = 144$ $h = \sqrt{144} = 12 \, \text{cm}$

So, the height of the triangle is $12 \, \text{cm}$.

Step 2: Circumradius (radius of the circumcircle)

The formula for the circumradius $R$ of a triangle is:

$R = \frac{abc}{4A}$ Where:

• $a$, $b$, and $c$ are the sides of the triangle,
• $A$ is the area of the triangle.

For our isosceles triangle:

• The base $a = 10 \, \text{cm}$,
• The two equal sides $b = c = 13 \, \text{cm}$,
• The height $h = 12 \, \text{cm}$.

The area $A$ of the triangle is:

$A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 10 \times 12 = 60 \, \text{cm}^2$

Now, applying the circumradius formula:

$R = \frac{10 \times 13 \times 13}{4 \times 60} = \frac{1690}{240} \approx 7.04 \, \text{cm}$

Thus, the circumradius is approximately $7.04 \, \text{cm}$.

Step 3: Inradius (radius of the incircle)

The formula for the inradius $r$ of a triangle is:

$r = \frac{A}{s}$ Where $s$ is the semi-perimeter of the triangle, and $A$ is the area.

The semi-perimeter $s$ is:

$s = \frac{a + b + c}{2} = \frac{10 + 13 + 13}{2} = 18 \, \text{cm}$

Now, applying the inradius formula:

$r = \frac{60}{18} = \frac{10}{3} \approx 3.33 \, \text{cm}$

Thus, the inradius is approximately $3.33 \, \text{cm}$.

Final Answer:

• Circumradius $R \approx 7.04 \, \text{cm}$,
• Inradius $r \approx 3.33 \, \text{cm}$.

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Would you like more details on how to derive these formulas or any other clarifications?

Follow-up Questions:

1. How does the Pythagorean theorem apply in this case of an isosceles triangle?
2. What is the role of the semi-perimeter in calculating the inradius?
3. How would the circumradius change if the triangle were a right triangle instead of an isosceles triangle?
4. Can the incircle and circumcircle radii be equal in any triangle?
5. How would the formulas change for an equilateral triangle?

Tip: The formula for the circumradius and inradius works for all triangles, but the symmetry in isosceles triangles (or equilateral triangles) simplifies the process of calculating their area and other properties.