prove that ratio of indiameter and circumdiameter of right angled triangle is 1:2 using tangents
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The question is wrongly formed. It is an equilateral triangle in which the ratio is 1:2.
see diagram of the equilateral triangle.
The line segments BD, BF are tangents from B on to the in circle. Hence they are same. ID is the perpendicular on to the BC as it is radius at the point D. IA, IB and IC are radii of the circumcircle. From symmetry BD = DC = a/2.
R² = r² + (a/2)²
We know that in any triangle : Δ = s r = (a+b+c) r /2 = 3 a r /2
we know area of an equilateral triangle = √3/4 a²
r = a/(2√3)
Hence, R² = a²/12 + a²/4
= a²/3
=> R = a/√3
Ratio r: R = 1 : 2 (ratio of diameters is same)
======================================
For information the ratio r : R in a right angled triangle is : a b / [c(a+b+c) ]
where c² = a² + b² , r = ab/(a+b+c) and R = c/2
In an isosceles right angle triangle (a = b, c = a √2 ), the ratio will be :
r : R = 1/(√2+1)
see diagram of the equilateral triangle.
The line segments BD, BF are tangents from B on to the in circle. Hence they are same. ID is the perpendicular on to the BC as it is radius at the point D. IA, IB and IC are radii of the circumcircle. From symmetry BD = DC = a/2.
R² = r² + (a/2)²
We know that in any triangle : Δ = s r = (a+b+c) r /2 = 3 a r /2
we know area of an equilateral triangle = √3/4 a²
r = a/(2√3)
Hence, R² = a²/12 + a²/4
= a²/3
=> R = a/√3
Ratio r: R = 1 : 2 (ratio of diameters is same)
======================================
For information the ratio r : R in a right angled triangle is : a b / [c(a+b+c) ]
where c² = a² + b² , r = ab/(a+b+c) and R = c/2
In an isosceles right angle triangle (a = b, c = a √2 ), the ratio will be :
r : R = 1/(√2+1)
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