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Given a right triangle, its circumscribed circle radius = 6, inscribed circle radius = 2, find the perimeter of the triangle
Because the radius of circumscribed circle is 6, the length of hypotenuse is 12. Make a vertical line of three sides on the center of inscribed circle, and divide the three sides into two parts. One part of each side will be equal to one part of the adjacent side, so one part of the two right angle sides will be equal to the hypotenuse, so 12 + 12 = 24, and the half of the two right angle sides will be equal
The length of the hypotenuse of a right triangle is 8, and the radius of the inscribed circle is 1
emergency
18
After three vertical lines are drawn from the center of the circle, the two right angles are divided into two sections. The right angle and two radii drawn from the center of the circle form a square. The sum of the remaining lines of the two right angles is just the length of the hypotenuse (two pairs of similar triangles)
So perimeter = 1 + 1 + 8 + 8 = 18
The length of the hypotenuse of a right triangle is 12, and the radius of the inscribed circle is 1
The triangle perimeter: 26 in the triangle ABC, hypotenuse AB, tangent D, right angle side AC, tangent e, right angle side BC, tangent F, then: ad = AE, BD = BF, CE = CF = r = 1, so AE + BF = AD + BD = AB = 12, perimeter: C = AB + AC + BC = AD + DB + AE + CE + CF + BF = 12 + 12 + 1 + 1 = 26
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