Given the height of isosceles triangle is 6 and 8, how much is the circumference?

Let the waist length be x and the vertex angle be a (1). When the height on the bottom edge is 6, Sina = 8 / xsin (A / 2) = √ (X & # 178; - 6 & # 178;) / x, cos (A / 2) = 6 / X. from Sina = 2Sin (A / 2) cos (A / 2), we get 8x = 12 √ (X & # 178; - 6 & # 178;) and get x = 18 / √ 5, bottom length = 24 / √ 5, perimeter = 60 / √ 5 (2)

The middle line of a waist of an isosceles triangle divides the circumference into 6 and 15 parts to find the waist length of the triangle

Let AB = AC = 2A, BC = B,
D divides AC into ad = CD = a,
（1）AB+AD=2a+a=15
BC+DC=b+a=6,
∴a=35,b=1,
That is ab = AC = 10, BC = 1
(2)2a+a=6,
b+a=15
a=2,b=13,
If AB = AC = 4, B = 13 can not form a triangle, rounding off

The circumference of an isosceles triangle is 18 cm, the ratio of the waist to the bottom is 2:5, and the waist is () cm?

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