10.009 In an isosceles triangle, the base is 16, and a side is 10

10.009. In an isosceles triangle, the base is 16 cm, and a side is 10 cm. Find the radii of the inscribed and circumscribed circles and the distance between their centers.

Let’s consider an isosceles triangle ABC. Let AC be the base, which is 16 cm. The equal sides are AB and BC, both 10 cm. We first need to find the area of this triangle.

In an isosceles triangle, the altitude from the vertex between the equal sides to the base also bisects the base. Let BD be the altitude from vertex B to the base AC. Thus, D is the midpoint of AC, and AD equals half of AC, which is 16 / 2 = 8 cm.

Now, we can find the length of the altitude BD using the Pythagorean theorem in the right-angled triangle ADB:

BD = \sqrt{AB^2 - AD^2}

Substitute the values: AB = 10 and AD = 8:

$BD = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6$ cm.

Now we can calculate the area of triangle ABC, denoted as $S_{\triangle ABC}$ , using the formula: Area = $\frac{1}{2}$ * base * height.

S_{\triangle ABC} = \frac{1}{2} \cdot AC \cdot BD

Substitute AC = 16 and BD = 6:

$S_{\triangle ABC} = \frac{1}{2} \cdot 16 \cdot 6 = 8 \cdot 6 = 48$ cm².

Next, we need to find the radii of the circumscribed circle (R) and the inscribed circle (r).

For the circumscribed circle, the radius R can be found using the formula involving the product of the side lengths and the area:

R = \frac{AB \cdot BC \cdot AC}{4S}

Substitute the side lengths (10, 10, 16) and the area (48):

R = \frac{10 \cdot 10 \cdot 16}{4 \cdot 48} = \frac{1600}{192}

After simplifying the fraction, we get:

$R = \frac{25}{3}$ cm.

For the inscribed circle, the radius r can be found using the formula: Area = semi-perimeter * radius, or $r = \frac{S}{p}$ , where ‘p’ is the semi-perimeter of the triangle.

First, let’s calculate the semi-perimeter ‘p’. The perimeter is 10 + 10 + 16 = 36 cm. The semi-perimeter ‘p’ is half of the perimeter:

$p = \frac{36}{2} = 18$ cm.

Now, substitute the area (48) and the semi-perimeter (18) into the formula for r:

r = \frac{48}{18}

After simplifying the fraction, we get:

$r = \frac{8}{3}$ cm.

Finally, we need to find the distance between the centers of the inscribed and circumscribed circles.
Since both centers lie on the same vertical line (the altitude BD), and the height of the triangle is 6 cm, we can determine their relative positions. The incenter is located ‘r’ (which is $\frac{8}{3}$ cm) above the base AC. The circumcenter is located ‘R’ (which is $\frac{25}{3}$ cm) from the top vertex B. Therefore, the distance between them is given by the formula:

\text{Distance} = R + r - \text{height}

Plugging in the values we found: R = $\frac{25}{3}$ , r = $\frac{8}{3}$ , and the height BD = 6:

\text{Distance} = \frac{25}{3} + \frac{8}{3} - 6

First, sum the fractions:

\frac{25}{3} + \frac{8}{3} = \frac{33}{3} = 11

Now, subtract 6:

$\text{Distance} = 11 - 6 = 5$ cm.

Answer: The radius of the inscribed circle is $\frac{8}{3}$ cm, the radius of the circumscribed circle is $\frac{25}{3}$ cm, and the distance between their centers is 5 cm.

Related Posts