It is proved that the sum of the two right sides of a right triangle is equal to the sum of the hypotenuse and the diameter of the inscribed circle

It is proved that the sum of the two right sides of a right triangle is equal to the sum of the hypotenuse and the diameter of the inscribed circle

Let △ ABC, C = 90 °, ab = C, BC = a, AC = B. his inscribed circle O is tangent to AB, BC, AC at points F, D, e respectively, and its radius is r
Connecting od and OE, we prove that odcf is a square (omitted), so EC = DC = R,
So BD = A-R, AE = b-r
BD = BF, AE = AF, so BF = A-R, AF = b-r
So C = BF + AF = A-R + b-r
So r = (a + B-C) / 2

Given that the two right sides of a right triangle are 3 and 4 respectively, then the radius of its circumscribed circle is, and the diameter of its inscribed circle is

In a right triangle, the radius of the inscribed circle = (a + B-C) / 2, where C is the circumscribed circle of the hypotenuse. The center of the circle is the midpoint of the hypotenuse

If the lengths of three sides of an obtuse triangle are a, B, C (a > b > C), and the radii of circumscribed circle and inscribed circle are R, R, respectively, then the minimum radius of the circular paper that can cover the triangle is ()
A. RB. rC. a2D. c2

∵ the circle which can cover the triangle and has the smallest radius should be circumscribed circle, ∵ the minimum radius of the circular paper which can cover the triangle is R. so select a