The height of the isosceles triangle, drawn to the base, is 32 cm, the radius of the circle

| September 4, 2021 | education

The height of the isosceles triangle, drawn to the base, is 32 cm, the radius of the circle inscribed in the triangle is 12 cm. Find the radius of the circle circumscribed about this triangle.

Let the lateral side of the triangle ABC be equal to X cm, and the base of the AC – Y cm.

Then the area of ​​the triangle ABC will be equal to Savs = AC * BN / 2 = Y * 32/2 = 16 * Y.

The semi-perimeter of the triangle will be equal to: p = (AB + BC + AC) / 2 = (2 * X + Y) / 2.

Then the radius of the inscribed circle is: r = S / p = 16 * Y / (2 * X + Y) / 2.

12 = 32 * Y / (2 * X + Y).

3 * (2 * X + Y) = 8 * Y.

6 * X = 5 * Y.

X = 5 * Y / 6. (1).

In a right-angled triangle ABN, according to the Pythagorean theorem, we express the leg AN.

AH ^ 2 = AB ^ 2 – AH ^ 2.

(Y / 2) ^ 2 = X ^ 2 – 32 ^ 2.

Y ^ 2/4 = (25 * Y ^ 2/36) – 1024.

(25 * Y ^ 2 – 9 * Y ^ 2) / 36 = 1024.

16 * Y ^ 2 = 36864.

Y ^ 2 = 2304.

Y = AC = 48 cm.

X = AB = BC = 5 * 48/6 = 40 cm.

Determine the area of ​​the triangle ABC. Savs = 16 * Y = 16 * 48 = 768 cm2.

Determine the radius of the circumscribed circle.

R = AB * BC * AC / 4 * Savs = 40 * 40 * 48/4 * 768 = 25 cm.

Answer: The radius of the circumscribed circle is 25 cm.



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