To find the radius $r$ of the inscribed circle (or incircle) of an isosceles triangle with side lengths $a = 10.0$ inches, $b = 10.0$ inches, and $c = 2w$ inches, we can use the following formula for the radius of the inscribed circle in any triangle:

$r = \frac{A}{s}$

where:

• $A$ is the area of the triangle,
• $s$ is the semi-perimeter of the triangle, calculated as $s = \frac{a + b + c}{2}$.

We’ll solve for each case: $w = 5.00$, $w = 6.00$, and $w = 8.00$.

Step 1: Calculate the Semi-Perimeter $s$

1. For $w = 5.00$: $c = 2w = 2 \times 5 = 10 \text{ inches}$ $s = \frac{10 + 10 + 10}{2} = 15 \text{ inches}$

2. For $w = 6.00$: $c = 2w = 2 \times 6 = 12 \text{ inches}$ $s = \frac{10 + 10 + 12}{2} = 16 \text{ inches}$

3. For $w = 8.00$: $c = 2w = 2 \times 8 = 16 \text{ inches}$ $s = \frac{10 + 10 + 16}{2} = 18 \text{ inches}$

Step 2: Calculate the Area $A$ Using Heron's Formula

The area $A$ of a triangle with sides $a$, $b$, and $c$ is given by Heron's formula: $A = \sqrt{s(s - a)(s - b)(s - c)}$

Let’s calculate $A$ for each value of $w$:

1. For $w = 5.00$: $A = \sqrt{15 \times (15 - 10) \times (15 - 10) \times (15 - 10)} = \sqrt{15 \times 5 \times 5 \times 5} = \sqrt{1875} \approx 43.30 \text{ square inches}$

2. For $w = 6.00$: $A = \sqrt{16 \times (16 - 10) \times (16 - 10) \times (16 - 12)} = \sqrt{16 \times 6 \times 6 \times 4} = \sqrt{2304} = 48.0 \text{ square inches}$

3. For $w = 8.00$: $A = \sqrt{18 \times (18 - 10) \times (18 - 10) \times (18 - 16)} = \sqrt{18 \times 8 \times 8 \times 2} = \sqrt{2304} = 48.0 \text{ square inches}$

Step 3: Calculate the Radius $r$ of the Inscribed Circle

Now we can find $r = \frac{A}{s}$ for each case.

1. For $w = 5.00$: $r = \frac{43.30}{15} \approx 2.89 \text{ inches}$

2. For $w = 6.00$: $r = \frac{48.0}{16} = 3.0 \text{ inches}$

3. For $w = 8.00$: $r = \frac{48.0}{18} \approx 2.67 \text{ inches}$

Final Answers

• For $w = 5.00$: $r \approx 2.89$ inches
• For $w = 6.00$: $r = 3.0$ inches
• For $w = 8.00$: $r \approx 2.67$ inches

Would you like more details on any step?

Here are related questions:

1. How does changing $w$ affect the area of the triangle?
2. What happens to the radius $r$ if the triangle is equilateral?
3. How would the solution change for a scalene triangle with different side lengths?
4. Can we generalize the result for any isosceles triangle?
5. How is the radius of the circumcircle calculated for this triangle?

Tip: The radius of the incircle depends directly on both the area and semi-perimeter, so maximizing $A$ relative to $s$ increases $r$.