**Q) A circle with centre O and radius 8 cm is inscribed in a quadrilateral ABCD in which P, Q, R, S are the points of contact as shown. If AD is perpendicular to DC, BC = 30 cm and BS = 24 cm, then find the length DC.**

**Ans: **To find length of DC, we need to find the values of DQ and QC

**Step 1:** Let’s start for DQ.

Let’s connect point O with points P and Q, analyse quadrilateral OPDQ

DP = DQ (being tangents drawn to same circle from an external point)

OP = OQ (being radii of same circle)

∠ ADC = 90^{0 }(being AD perpendicular to DC)

Now, by tangents property, we know that the tangent make right angle to the radius of the circle at the point of contact.

Hence, ∠ OQD = 90^{0 }and ∠ OPD = 90^{0}

We know that sum of all angles of a quadrilateral will be equal to 360^{0}.

Therefore, ∠ OPD + ∠ PDQ + ∠ OQD + ∠ POQ = 360^{0}

90^{0} + 90^{0} + 90^{0} + ∠ POQ = 360^{0}

∠ POQ = 90^{0}

Since all the angles of the quadrilateral are equal to 90^{0} and the adjacent sides are also equal,

Therefore the quadrilateral OPDQ is a square.

∴ DQ = OP

Since we are given radius r of the circle = 8 cm

∴ DQ = 8 cm …. (i)

**Step 2: **Now let’s take for QC:

From point B, BS & BR are tangents to the circle,

∴ BS = BR = 24 cm (given)

Since BC = BR + CR = 30 cm (given)

∴ 24 + CR = 30

∴ CR = 30 – 24 = 6 cm

From point C, CR & CQ are tangents to the circle,

∴ CQ = CR = 6 cm (calculated above) ………… (ii)

**Step 3:** Now DC = DQ + CQ

from equations (i) and (ii),

DC = 8 + 6 = 14 cm

**Therefore, length of DC is 14 cm.**

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