In the given figure, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively. Prove that – (1) seg AP || seg BQ, (2) ∆APR ~ ∆RQB, and (3) Find ∠ RQB if ∠ PAR = 35°
Chapter 3 – Circle – Text Book Solution
Practice Set 3.2 | Q 4 | Page 58
In the given figure, the circles with centres P and Q touch each other at R. A line passing through R meets the circles at A and B respectively.
Prove that –
(1) seg AP || seg BQ,
(2) ∆APR ~ ∆RQB, and
(3) Find ∠ RQB if ∠ PAR = 35°
(1) In ∆APR, AP = RP (Radii of the same circle)
∴ ∠ARP = ∠RAP .....(1) (In a triangle, equal sides have equal angles opposite to them)
In ∆BQR, QR = QB (Radii of the same circle)
∴ ∠RBQ = ∠BRQ .....(2) (In a triangle, equal sides have equal angles opposite to them)
Also, ∠ARP = ∠BRQ .....(3) (Vertically opposite angles)
From (1), (2) and (3), we have
∠RAP = ∠RBQ
∴ seg AP || seg BQ (If a transversal intersect two lines such that a pair of alternate interior angles is equal, then the two lines are parallel)
(2) In ∆APR and ∆RQB,
∠RAP = ∠RBQ (Proved)
∠ARP = ∠BRQ (Vertically opposite angles)
∴ ∆APR ~ ∆RQB (AA similarity criterion)
(3) ∠RBQ = ∠PAR = 35º
∴ ∠BRQ = ∠RBQ = 35º
In ∆RQB,
∠BRQ + ∠RBQ + ∠RQB = 180º (Angle sum property)
∴ 35º + 35º + ∠RQB = 180º
⇒ 70º + ∠RQB = 180º
⇒ ∠RQB = 180º − 70º = 110º
Thus, the measure of ∠RQB is 110º.
Explanation:-
(1) In triangle APR, AP = RP, as they are radii of the same circle. Hence, we have angle ARP = angle RAP by the property that in a triangle, equal sides have equal angles opposite to them. Similarly, in triangle BQR, QR = QB and we have angle RBQ = angle BRQ. Also, angle ARP = angle BRQ (vertical opposite angles). Therefore, we can conclude that angle RAP = angle RBQ and segment AP is parallel to segment BQ by the property that if a transversal intersects two lines such that a pair of alternate interior angles is equal, then the two lines are parallel.
(2) In triangles APR and RQB, we have angle RAP = angle RBQ (proved in step 1) and angle ARP = angle BRQ (vertical opposite angles). Therefore, we can apply the AA similarity criterion to conclude that triangles APR and RQB are similar.
(3) We are given that angle RBQ = 35 degrees. Therefore, using the angle sum property of triangles, we have:
angle BRQ + angle RBQ + angle RQB = 180 degrees
Substituting angle RBQ and simplifying, we get:
35 degrees + 35 degrees + angle RQB = 180 degrees
Solving for angle RQB, we get:
angle RQB = 110 degrees
Thus, the measure of angle RQB is 110 degrees.
Chapter 3 – Circle – Text Book Solution
Practice set 3.2 | Q 4 | Page 58
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