Please solve the problem as soon as possible.
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This can be solved with help of trigonometric identities.
first of all triangle is an equilateral triangle .
so, all angles are 60° .
I attached rough daigram for finding relation between a and r .
Here it is clear from ∆OAB
tan30° = r/AB
so, AB = r/tan30° = √3r
[ Here you have to know that line passing through centre of circle bisects the angle made by tangents at exterior points. That's why ∠OAB = half of angle 60° = 30°]
Similarly we should apply for other side,
CD = √3r
∵ a = CD + r + r + AB
a = √3r + 2r + √3r = (2 + 2√3)r
Hence, a = 2(1 + √3)r
first of all triangle is an equilateral triangle .
so, all angles are 60° .
I attached rough daigram for finding relation between a and r .
Here it is clear from ∆OAB
tan30° = r/AB
so, AB = r/tan30° = √3r
[ Here you have to know that line passing through centre of circle bisects the angle made by tangents at exterior points. That's why ∠OAB = half of angle 60° = 30°]
Similarly we should apply for other side,
CD = √3r
∵ a = CD + r + r + AB
a = √3r + 2r + √3r = (2 + 2√3)r
Hence, a = 2(1 + √3)r
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