In Figure, ABC is a triangle in which ∠B = 90º, BC = 48 cm and AB = 14 cm. A circle is inscribed in the triangle, whose centre is O. Find radius r of in-circle.
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AC = √(AB2 + BC2)
=√(142 + 482)
= 50 cm
∠OQB = 90°
⇒ OPBQ is a square
⇒ BQ = r, QA = 14 – r = AR
Again PB = r,
PC = 48 – r
⇒ RC = 48 – r
AR + RC = AC
⇒ 14 – r + 48 – r = 50
⇒ r = 6 cm
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Answer:
AB=14cm
BC=48
Ac2=(14)(14)+(48)(48)
Ac2=196+2034
AC=50
Name the point of intersection of radius and side of triangle as d e f
Ae=ad(tangent property
CD=cf
Bf=be
Now do by ypurself
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