Tangent Perpendicular to Radius at Point of Contact
A circle of r...
Question
A circle of radius r is inscribed in a triangle of area ′Δ′. If the semi-perimeter of the triangle is s, then the correct relation is
A
2r=Δs
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B
r=Δs
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C
r=sΔ
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D
2s=Δr
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Solution
The correct option is Br=Δs
We have, a circle of radius r is inscribed in a triangle of area ′Δ′
Here, AB, AC and BC are tangent to circle to circle and makes 90∘ with radius We know that, Area of ΔABC =Area of ΔOBA +Area of ΔOBC +Area of ΔOAC =12×r×AB+12×r×BC+12×r×AC=12×r(AB+BC+CA)=12×r×2s⇒Δ=rs∴r=Δs