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Question
In the figure, a circle touches all the sides of quadrilateral ABCD from the inside. The center of the circle is O. If AD⊥ DC and BC = 38, QB = 27, DC = 25, then find the radius of the circle.
Solution
Given: AD ⊥ DC
BC = 38, QB = 27, DC = 25
To find: Radius of the circle, i.e., OP.
Solution:
BC = 38 ......[Given]
∴ BQ + QC = 38 ......[B–Q–C]
∴ 27 + QC = 38 .......[Given]
∴ QC = 38 – 27
∴ QC = 11 units ......(i)
Now, QC = SC ......[Tangent segment theorem]
∴ SC = 11 units .....(ii) [From (i)]
DC = 25 .......[Given]
∴ DS + SC = 25 ......[D–S–C]
∴ DS + 11 = 25 ......[From (ii)]
∴ DS = 25 – 11
∴ DS = 14 units ......(iii)
In ▢DSOP,
∠P = ∠S = 90° ......[Tangent theorem]
∠D = 90° ......[Given]
∴ ∠O = 90° ......[Remaning angle of ▢DSOP]
∴ ▢DSOP is a rectangle.
Also, OP = OS ......[Radii of the same circle]
∴ ▢DSOP is a square .......`[("A rectangle is square if its"),("adjacent sides are congruent")]`
∴ OS = DS = DP = PO .....(iv) [Sides of the square]
∴ OP = 14 units ......[From (iii) and (iv)]
∴ The radius of the circle is 14 units.
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