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 = 900 (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 = 900 and ∠ OPD = 900
We know that sum of all angles of a quadrilateral will be equal to 3600.
Therefore, ∠ OPD + ∠ PDQ + ∠ OQD + ∠ POQ = 3600
900 + 900 + 900 + ∠ POQ = 3600
∠ POQ = 900
Since all the angles of the quadrilateral are equal to 900 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.
Please do press “Heart” button if you liked the solution.