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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.
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Solution
Given: AD ⊥ DC
BC = 38, QB = 27, DC = 25
To find: Radius of the circle, i.e., OP.
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° ...[Remaining 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.
