The height CD drawn to the base AB of an isosceles triangle ABC is 3 cm, AB = 8 cm

The height CD drawn to the base AB of an isosceles triangle ABC is 3 cm, AB = 8 cm. Find the radii of the inscribed and circumscribed circles around the triangle.

1. The radius of the circumscribed circle:
R = a ^ 2 / 2h,
where a is the lateral side of an isosceles triangle, h is the height drawn to the base of the isosceles triangle.
Since the triangle ABC is isosceles, AB is the base, then CD is not only the height, but also the median, therefore the segments AD and DB are equal to half of AB:
AD = DB = AB / 2 = 8/2 = 4 (cm).
Let’s find AC (according to the Pythagorean theorem):
AC = √ (CD ^ 2 + AD ^ 2) = √ (3 ^ 2 + 4 ^ 2) = √ (9 + 16) = √25 = 5 (cm).
AC = BC = 5 cm.
R = 5 ^ 2/2 * 3 = 25/6 = 4 whole 1/3 (cm).
2. The radius of the inscribed circle:
r = S / p,
where S is the area of ​​the triangle in which the circle is inscribed, p is the semiperimeter of the triangle.
p = (2a + b) / 2 = (2 * 5 + 8) / 2 = 18/2 = 9 (cm).
Area of ​​a triangle according to Heron’s formula:
S = √ (p (p – a) ^ 2 * (p – b)) = √ (9 (9 – 5) ^ 2 * (9 – 8)) = √ (9 * 4 ^ 2 * 1) = √ 9 * 16 * 1 = √144 = 12 (cm square).
r = 12/9 = 4/3 = 1 whole 1/3 (cm).
Answer: r = 1 whole 1/3 cm, R = 4 whole 1/3 cm.