EXERCISE 10.1 Q.1) 𝐴𝐷 is a diameter of a circle and 𝐴𝐵 is a chord. If 𝐴𝐷 = 34 𝑐𝑚, 𝐴𝐵 = 30 𝑐𝑚, the
distance of 𝐴𝐵 from the centre of the circle is : (A) 17 cm (B) 15 cm (C) 4 cm (D) 8 cm
Sol.1) (d) Given, 𝐴𝐷 = 34 𝑐𝑚 and 𝐴𝐵 = 30 𝑐𝑚 In figure, draw 𝑂𝐿 ⊥ 𝐴𝐵. Since, the perpendicular from the centre of a circle to a chord bisects the chord.
∴ 𝐴𝐿 = 𝐿𝐵 =1
2𝐴𝐵 = 15 𝑐𝑚
In right angles, ∆𝑂𝐿𝐴, 𝑂𝐴2 = 𝑂𝐿2 + 𝐴𝐿2 (17)2 = 𝑂𝐿2 + (15)2 289 = 𝑂𝐿2 + 225 𝑂𝐿2 = 289 − 225 = 64 𝑂𝐿 = 8𝑐𝑚 Hence, the distance of the chord from the centre is 8𝑐𝑚.
Q.2) In Fig. 10.3, if 𝑂𝐴 = 5 𝑐𝑚, 𝐴𝐵 = 8 𝑐𝑚 and 𝑂𝐷 is perpendicular to 𝐴𝐵, then 𝐶𝐷 is equal to: (A) 2 cm (B) 3 cm (C) 4 cm (D) 5 cm
Sol.2) (a) We know that, the perpendicular from the centre of a circle to a chord bisects the chord. 𝐴𝐶 = 𝐶𝐵 = ½ 𝐴𝐵 = ½ × 8 = 4 𝑐𝑚 Given 𝑂𝐴 = 5 𝑐𝑚 𝐴𝑂 = 𝐴𝐶 + 𝑂𝐶 (5)2 = (4)2 + 𝑂𝐶2 25 = 16 + 𝑂𝐶 𝑂𝐶2 = 25 − 16 = 9 𝑂𝐶 = 3 𝑐𝑚 [taking positive square root, because length is always positive] 𝑂𝐴 = 𝑂𝐷 [same radius of a circle] 𝑂𝐷 = 5 𝑐𝑚 𝐶𝐷 = 𝑂𝐷 – 𝑂𝐶 = 5 – 3 = 2 𝑐𝑚
Q.3) If 𝐴𝐵 = 12 𝑐𝑚, 𝐵𝐶 = 16 𝑐𝑚 and 𝐴𝐵 is perpendicular to BC, then the radius of the circle passing through the points A, B and C is : (A) 6 cm (B) 8 cm (C) 10 cm (D) 12 cm
Sol.3) (c) Given, 𝐴𝐵 = 12 𝑐𝑚, 𝐵𝐶 = 16 𝑐𝑚 In a circle, 𝐵𝐶 ⊥ 𝐴𝐵, it means that 𝐴𝐶 will be a diameter of circle. Use Pythagoras theorem, 𝐴𝐶2 = 𝐴𝐵2 + 𝐵𝐶2 𝐴𝐶2 = (12)2 + (16)2 𝐴𝐶2 = 144 + 256 = 400 𝐴𝐶 = 200 𝑐𝑚
Q.4) In figure, if ∠𝐴𝐵𝐶 = 20°, then ∠𝐴𝑂𝐶 is equal to
(a) 20° (b) 40° (c) 60° (d) 10°
Sol.4) (b) Given, ∠𝐴𝐵𝐶 = 20° We know that, angle subtended at the centre by an arc is twice the angle subtended by it at the remaining part of circle. ∠𝐴𝑂𝐶 = 2∠𝐴𝐵𝐶 = 2 × 20° = 40°
Q.5) In figure, if 𝐴𝑂𝐵 is a diameter of the circle and 𝐴𝐶 = 𝐵𝐶, then ∠𝐶𝐴𝐵 is equal to
(a) 30° (b) 60° (c) 90° (d) 45°
Sol.5) d) We know that, diameter subtends a right angle to the circle.
∴ ∠𝐵𝐶𝐴 = 90° … (i) Given, 𝐴𝐶 = 𝐵𝐶 ∠𝐴𝐵𝐶 = ∠𝐶𝐴𝐵 … (ii) In ∆𝐴𝐵𝐶,
∠𝐶𝐴𝐵 + ∠𝐴𝐵𝐶 + ∠𝐵𝐶𝐴 = 180° [from eqs. (i) & (ii)]
∠𝐶𝐴𝐵 + ∠𝐶𝐴𝐵 + ∠90° = 180 2∠𝐶𝐴𝐵 = 180 − 90°
∠𝐶𝐴𝐵 =90°
2= 45°
∠𝐶𝐴𝐵 = 45 Q.6) In figure, if ∠𝑂𝐴𝐵 = 40°, then ∠𝐴𝐶𝐵 is equal to from eq. (i) ∠𝐴𝐶𝐵 = ∠𝐴𝐷𝐵 = 70° .
(a) 50° (b) 40° (c) 60° (d) 70°
Sol.6) (a) In 𝛥𝑄𝐴𝐵, 𝑂𝐴 = 𝑂𝐵 [both are the radius of a circle] ∠𝑂𝐴𝐵 = ∠𝑂𝐵𝐴 ⇒ ∠𝑂𝐵𝐴 = 40° [angles opposite to equal sides are equal]
Also, ∠𝐴𝑂𝐵 + ∠𝑂𝐵𝐴 + ∠𝐵𝐴𝑂 = 180° [by angle sum property of a triangle]
∠𝐴𝑂𝐵 + 40° + 40° = 180° ⇒ ∠𝐴𝑂𝐵 = 180° – 80° = 100° We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. ∠𝐴𝑂𝐵 = 2 ∠𝐴𝐶𝐵 ⇒ 100° = 2 ∠𝐴𝐶𝐵
∠𝐴𝐶𝐵 =100°
2 = 50°
Q.7) In figure, if ∠𝐷𝐴𝐵 = 60° , ∠𝐴𝐵𝐷 = 50°, then ∠𝐴𝐶𝐵 is equal to
Sol.7) (c) Given, ∠𝐷𝐴𝐵 = 60°, ∠𝐴𝐵𝐷 = 50° Since, ∠𝐴𝐷𝐵 = ∠𝐴𝐶𝐵 …(i) [angles in same segment of a circle are equal] In 𝛥𝐴𝐵𝐷, ∠𝐴𝐵𝐷 + ∠𝐴𝐷𝐵 + ∠𝐷𝐴𝐵 = 180° [by angle sum property of a triangle]
(C) Short Answer Questions with Reasoning Exercise 10.2
Write True or False and justify your answer in each of the following:
Q.1) Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then 𝐴𝐵 = 𝐶𝐷.
Sol.1) True Because, the chords equidistant from the centre of circle are equal in length.
Q.2) Two chords 𝐴𝐵 and 𝐴𝐶 of a circle with centre O are on the opposite sides of 𝑂𝐴.
Then ∠𝑂𝐴𝐵 = ∠𝑂𝐴𝐶 .
Sol.2) False In figure, 𝐴𝐵 and 𝐴𝐶 are two chords of a circle. Join 𝑂𝐵 and 𝑂𝐶.
In 𝛥𝑂𝐴𝐵 and 𝛥𝑂𝐴𝐶, 𝑂𝐴 = 𝑂𝐴 𝑂𝐵 = 𝑂𝐶 Here, we are not able to show that either the any angle or third side is equal & ∆𝑂𝐴𝐵 is not congruent to ∆𝑂𝐴𝐶. ∴ ∠𝑂𝐴𝐵 ≠ ∠𝑂𝐴𝐶
Q.3) The congruent circles with centres 𝑂 and 𝑂′ intersect at two points A and B. Then, ∠𝐴𝑂𝐵 = ∠𝐴𝑂’𝐵.
Sol.3) True Join 𝐴𝐵, 𝑂𝐴 and 𝑂𝐵, 𝑂’𝐴 and 𝐵𝑂’. In 𝛥𝐴𝑂𝐵 and 𝛥𝐴𝑂’𝐵, 𝑂𝐴 = 𝐴𝑂’ [both circles have same radius] 𝑂𝐵 = 𝐵𝑂’ [both circles have same radius] and 𝐴𝐵 = 𝐴𝐵 [common chord] ∆𝐴𝑂𝐵 = ∆𝐴𝑂′𝐵 [by SSS congruence rule] ∠𝐴𝑂𝐵 = ∠𝐴𝑂’𝐵. [by CPCT]
Q.4) Through three collinear points a circle can be drawn.
Sol.4) False Because, circle can pass through only two collinear points but not through three collinear points.
Q.5) A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.
Sol.5) True Suppose, we consider diameter of a circle is 𝐴𝐵 = 66𝑚. AB
Then, radius of a circle =𝐴𝐵
2=
6
2 = 3 𝑐𝑚, which is true.
Q.6) If 𝐴𝑂𝐵 is a diameter of a circle and 𝐶 is a point on the circle, then 𝐴𝐶2 + 𝐵𝐶2 = 𝐴𝐵2 .
Sol.6) True
Since, any diameter of the circle subtends a right angle to any point on the circle. If 𝐴𝑂𝐵 is a diameter of a circle and 𝐶 is a point on the circle, then 𝛥𝐴𝐶𝐵 is right angled at 𝐶. In right angled 𝛥𝐴𝐶𝐵, [use Pythagoras theorem] 𝐴𝐶2 + 𝐵𝐶2 = 𝐴𝐵2
Q.7) 𝐴𝐵𝐶𝐷 is a cyclic quadrilateral such that ∠𝐴 = 90°, ∠𝐵 = 70°, ∠𝐶 = 95° and ∠𝐷 = 105°.
Sol.7) False In a cyclic quadrilateral, the sum of opposite angles is 180°. Now, ∠𝐴 + ∠𝐶 = 90° + 95° = 185° ≠ 180° and ∠𝐵 + ∠𝐷 = 70° + 105° = 175° ≠ 180° Here, we see that, the sum of opposite angles is not equal to 180°. So, it is not a cyclic quadrilateral.
Q.8) If 𝐴, 𝐵, 𝐶 and D are four points such that ∠𝐵𝐴𝐶 = 30° and ∠𝐵𝐷𝐶 = 60°, then 𝐷 is the centre of the circle through 𝐴, 𝐵 and 𝐶.
Sol.8) False
Because, there can be many points D, such that ∠𝐵𝐷𝐶 = 60° and each such point cannot be the centre of the circle through 𝐴, 𝐵 and 𝐶.
Q.9) If 𝐴, 𝐵, 𝐶 and 𝐷 are four points such that ∠𝐵𝐴𝐶 = 45°and ∠𝐵𝐷𝐶 = 45°, then 𝐴, 𝐵, 𝐶 and 𝐷 are concyclic.
Sol.9) True
Since, ∠𝐵𝐴𝐶 = 45° and ∠𝐵𝐷𝐶 = 45°
As we know, angles in the same segment of a circle are equal. Hence, A, B, C and D are concyclic.
Q.10) In figure, if 𝐴𝑂𝐵 is a diameter and ∠𝐴𝐷𝐶 = 120°, then ∠𝐶𝐴𝐵 = 30°.
Sol.10) True
Join 𝐶𝐴 and 𝐶𝐵 Since, 𝐴𝐷𝐶𝐵 is a cyclic quadrilateral.
∠𝐴𝐷𝐶 + ∠𝐶𝐵𝐴 = 180°. [sum of opposite angles of cyclic quadrilateral is 180°]
In 𝛥𝐴𝐶𝐵, ∠𝐶𝐴𝐵 + ∠𝐶𝐵𝐴 + ∠𝐴𝐶𝐵 = 180° [by angle sum property of a triangle]
∠𝐶𝐴𝐵 + 60° + 90° = 180° [triangle formed from diameter to the circle is 90° i.e.,
∠𝐴𝐶𝐵 = 90°) ⇒ ∠𝐶𝐴𝐵 = 180° – 150° = 30°.
Short Answer Type Questions Exercise 10.3
Q.1) If arcs 𝐴𝑋𝐵 and 𝐶𝑌𝐷 of a circle are congruent, find the ratio of 𝐴𝐵 and 𝐶𝐷
Sol.1) Let 𝐴𝑋𝐵 and 𝐶𝑌𝐷 are arcs of circle whose centre and radius are 𝑂 and 𝑟 𝑢𝑛𝑖𝑡𝑠, respectively. Hence, the ratio of 𝐴𝐵 and 𝐶𝐷 is 1: 1.
Q.2) If the perpendicular bisector of a chord 𝐴𝐵 of a circle 𝑃𝑋𝐴𝑄𝐵𝑌 intersects the circle at 𝑃 and 𝑄, prove that 𝑎𝑟𝑐 𝑃𝑋𝐴 = 𝑎𝑟𝑐 𝑃𝑌𝐵.
Sol.2) Let 𝐴𝐵 be a chord a circle having at 𝑂𝑃𝑄 be the perpendicular bisector of the chord 𝐴𝐵, which intersects at 𝑀 & it always passes through 𝑂. To prove 𝑎𝑟𝑐 𝑃𝑋𝐴 ≅ 𝑎𝑟𝑐 𝑃𝑌𝐵 Construction Join 𝐴𝑃 and 𝐵𝑃. Proof In 𝛥𝐴𝑃𝑀 and 𝛥𝐵𝑃𝑀, 𝐴𝑀 = 𝑀𝐵 ∠𝑃𝑀𝐴 = ∠𝑃𝑀𝐵 𝑃𝑀 = 𝑃𝑀 ∴ 𝛥𝐴𝑃𝑀 = 𝛥𝐵𝑃𝑀 ∴ 𝑃𝐴 = 𝑃𝐵 ⇒ 𝑎𝑟𝑐 𝑃𝑋𝐴 ≅ 𝑎𝑟𝑐 𝑃𝑌𝐵
Q.3) 𝐴, 𝐵 and 𝐶 are three points on a circle. Prove that the perpendicular bisectors of 𝐴𝐵, 𝐵𝐶 and 𝐶𝐴 are concurrent.
Sol.3) Given A circle passing through three points 𝐴𝐵, 𝐵 & 𝐶 Construction Draw a perpendicular bisectors of 𝐴𝐵 & 𝐴𝐶 & they meet at a point 𝑂. Join 𝑂𝐴, 𝑂𝐵 & 𝑂𝐶. To prove Perpendicular bisector of 𝐵𝐶, also passes through 𝑂, i.e., 𝐿𝑂, 𝑂𝑁 & 𝑂𝑀 are concurrent. Proof In ∆𝑂𝐸𝐴 and ∆𝑂𝐸𝐵, 𝐴𝐸 = 𝐵𝐸 [OL is perpendicular bisector of 𝐴𝐵] ∠𝐴𝐸𝑂 = ∠𝐵𝐸𝑂 & 𝑂𝐸 = 𝑂𝐸 [common side] ∴ ∆𝑂𝐸𝐴 ≅ ∆𝑂𝐸𝐵 [by SAS congruence rule] 𝑂𝐴 = 𝑂𝐵 Similarly, ∆𝑂𝐹𝐴 ≅ ∆𝑂𝐹𝐶 [by SAS congruence rule] 𝑂𝐴 = 𝑂𝐶 [by CPCT] 𝑂𝐴 = 𝑂𝐵 = 𝑂𝐶 = 𝑟 [say] Now, we draw perpendicular from 𝑂 to the 𝐵𝐶 & join them. In ∆𝑂𝑀𝐵 and ∆𝑂𝑀𝐶, 𝑂𝐵 = 𝑂𝐶 [proved above] 𝑂𝑀 = 𝑂𝑀 [common side] & ∠𝑂𝑀𝐵 = ∠𝑂𝑀𝐶
∆𝑂𝑀𝐵 ≅ ∆𝑂𝑀𝐶 [by RHS congruence rule] 𝐵𝑀 = 𝑀𝐶 [by CPCT] Hence, 𝑂𝑀 is the perpendicular bisector of 𝐵𝐶 Hence, 𝑂𝐿, 𝑂𝑁 & 𝑂𝑀 are concurrent.
Q.4) AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.
Sol.4) Given 𝐴𝑆 and 𝐴𝐶 are two equal chords whose centre is 𝑂. To prove Centre 𝑂 lies on the bisector of ∠𝐵𝐴𝐶. Construction Join SC, draw bisector 𝐴𝐷 of ∠𝐵𝐴𝐶. Proof In ∆𝑆𝐴𝑀 & ∆𝐶𝐴𝑀 𝐴𝑆 = 𝐴𝐶 [given] ∠𝐵𝐴𝑀 = ∠𝐶𝐴𝑀 [given] & 𝐴𝑀 = 𝐴𝑀 [common side] ∆𝐵𝐴𝑀 ≅ ∆𝐶𝐴𝑀 [by SAS congruence rule] 𝐵𝑀 = 𝐶𝑀 [by CPCT] & ∠𝐵𝑀𝐴 = ∠𝐶𝑀𝐴 [by CPCT]
So, 𝐵𝑀 = 𝐶𝑀 & ∠𝐵𝑀𝐴 = ∠𝐶𝑀𝐴 = 90° So, 𝐴𝑀 is the perpendicular bisector of the chord 𝐵𝐶. Hence, bisector of ∠𝐵𝐴𝐶 i.e, 𝐴𝑀 passes through the centre 𝑂. Hence proved.
Q.5) If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.
Sol.5) Given 𝐴𝐵 and 𝐶𝐷 are two chords of a circle whose centre is 𝑂 & 𝑃𝑄 is a diameter bisecting the chord 𝐴𝐵 & 𝐶𝐷 at 𝐿 & 𝑀, respectively & the diameter 𝑃𝑄 passes through the centre 𝑂 of the circle. To prove 𝐴𝐵|| CD Proof Since, 𝐿 is the mid point of 𝐴𝐵. ∴ 𝑂𝐿 ⊥ 𝐴𝐵 ⇒ ∠𝐴𝐿𝑂 = 90 … (i) Similarly, 𝑂𝑀 ⊥ 𝐶𝐷 ∴ ∠𝑂𝑀𝐷 = 90 … (ii) From eq. (i) & (ii), we get ∠𝐴𝐿𝑂 = ∠𝑂𝑀𝐷 = 90 But these are alternating angles. So, 𝐴𝐵|| CD Hence proved.
Q.6) 𝐴𝐵𝐶𝐷 is such a quadrilateral that A is the centre of the circle passing through 𝐵, 𝐶 and 𝐷. Prove that ∠𝐶𝐵𝐷 + ∠𝐶𝐷𝐵 = ½ ∠𝐵𝐴𝐷.
Sol.6) Given In a circle, 𝐴𝐵𝐶𝐷 is a quadrilateral having centre 𝐴. To prove ∠𝐶𝐵𝐷 + ∠𝐶𝐷𝐵 = ½ ∠𝐵𝐴𝐷 Construction Join 𝐴𝐶 and 𝐵𝐷. Proof Since, arc 𝐷𝐶 subtends ∠𝐷𝐴𝐶 at the centre & ∠𝐶𝐵𝐷 at a point 𝐵 in the remaining part of the circle. ∴ ∠𝐷𝐴𝐶 = 2∠𝐶𝐵𝐷 … (i) In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. Similarly, arc 𝐵𝐶 subtends ∠𝐶𝐴𝐵 at the centre & ∠𝐶𝐷𝐵 at a point 𝐷 in the remaining part of the circle. ∴ ∠𝐶𝐴𝐵 = 2∠𝐶𝐷𝐵 … (ii)
In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. On adding eqs. (i) & (ii), we get ∠𝐷𝐴𝐶 + ∠𝐶𝐴𝐵 = 2∠𝐶𝐵𝐷 + 2∠𝐶𝐷𝐵 ⇒ ∠𝐵𝐴𝐷 = 2(∠𝐶𝐵𝐷 + ∠𝐶𝐷𝐵)
⇒ ∠𝐶𝐷𝐵 + ∠𝐶𝐵𝐷 =1
2∠𝐵𝐴𝐷 Hence proved.
Q.7) 𝑂 is the circumcentre of the 𝛥𝐴𝐵𝐶 and 𝐷 is the mid-point of the base 𝐵𝐶. Prove that ∠𝐵𝑂𝐷 = ∠𝐴.
Sol.7) Given In a ∆𝐴𝐵𝐶 is a circle circumscribed having centre 𝑂. Also, 𝐷 is the mid point of 𝐵𝐶 To prove ∠𝐵𝑂𝐷 = ∠𝐴 or ∠𝐵𝑂𝐷 = ∠𝐵𝐴𝐶 Construction Join 𝑂𝐵, 𝑂𝐷 and 𝑂𝐶. 𝑂𝐵 = 𝑂𝐶 [both are the radius of the circle] 𝐵𝐷 = 𝐷𝐶 [D is the mid point of BC] & 𝑂𝐷 = 𝑂𝐷 [common side] ∴ ∆𝐵𝑂𝐷 ≅ ∆𝐶𝑂𝐷 [by SSS congruence rule] ∴ ∠𝐵𝑂𝐷 = ∠𝐶𝑂𝐷 [by CPCT] … (i) In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. 2∠𝐵𝐴𝐶 = ∠𝐵𝑂𝐶
∠𝐵𝐴𝐶 =2
3∠𝐵𝑂𝐷 [∠𝐵𝑂𝐶 = 2∠𝐵𝑂𝐷] from eq. (i)
∠𝐵𝐴𝐶 = ∠𝐵𝑂𝐷 Hence proved.
Q.8) On a common hypotenuse 𝐴𝐵, two right angled triangles, 𝐴𝐶𝐵 and 𝐴𝐷𝐵 are situated on opposite sides. Prove that ∠𝐵𝐴𝐶 = ∠𝐵𝐷𝐶.
Sol.8) Given, ∆𝐴𝐶𝐵 & ∆𝐴𝐷𝐵 are two right angles triangles with common hypotenuse 𝐴𝐵. To prove ∠𝐵𝐴𝐶 = ∠𝐵𝐷𝐶 Construction Join 𝐶𝐷. Proof Let 𝑂 be the mid point of 𝐴𝐵. Then, 𝑂𝐴 = 𝑂𝐵 = 𝑂𝐶 = 𝑂𝐷 Since, mid point of the hypotenuse of a right triangle is equidistant from its vertices. Now, draw a circle to pass through the point 𝐴, 𝐵, 𝐶 and 𝐷 𝑤𝑖𝑡ℎ O as centre & radius equal to 𝑂𝐴. Angles in the same segment of a circle are equal. From the figure ∠𝐵𝐴𝐶 & ∠𝐵𝐷𝐶 are angles of the same segment 𝐵𝐶. ∴ ∠𝐵𝐴𝐶 = ∠𝐵𝐷𝐶 Hence proved.
Q.9) Two chords 𝐴𝐵 and 𝐴𝐶 of a circle subtends angles equal to 90° and 150°, respectively at the centre. Find ∠𝐵𝐴𝐶, if 𝐴𝐵 and 𝐴𝐶 lie on the opposite sides of the centre.
Sol.9) In ∆𝐵𝑂𝐴, 𝑂𝐵 = 𝑂𝐴 [both are the radius of circle] ∴ ∠𝑂𝐴𝐵 = ∠𝑂𝐵𝐴 … (i) In ∆𝑂𝐴𝐵,
Q.10) If 𝐵𝑀 and 𝐶𝑁 are the perpendiculars drawn on the sides AC and AB of the ΔABC, prove that the points B, C, M and N are concyclic.
Sol.10) Given In 𝛥𝐴𝐵𝐶, 𝐵𝑀 ⊥ 𝐴𝐶 and 𝐶𝑁 ⊥ 𝐴𝐵. To prove Points 𝐵, 𝐶, 𝑀 and 𝑁 are con-cyclic.
Construction Draw a circle passing through the points 𝐵, 𝐶, 𝑀 and 𝑁. Proof Suppose, we consider SC as a diameter of the circle. Also, we know that 𝑆𝐶 subtends a 90° to the circle. So, the points 𝑀 and 𝑁 should be on a circle. Hence, 𝐵𝐶𝑀𝑁 form a con-cyclic quadrilateral. Hence proved.
Q.11) If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic.
Sol.11) Given 𝛥𝐴𝐵𝐶 is an isosceles triangle such that 𝐴𝐵 = 𝐴𝐶 and also 𝐷𝐸 || 𝑆𝐶. To prove Quadrilateral 𝐵𝐶𝐷𝐸 is a cyclic quadrilateral. Proof In ∆𝐴𝐵𝐶, 𝐴𝐵 = 𝐴𝐶 ∠𝐴𝐶𝐵 = ∠𝐴𝐵𝐶 … (i) Since, 𝐷𝐸 || 𝐵𝐶 ∠𝐴𝐷𝐸 = ∠𝐴𝐶𝐵 [corresponding angles] … (ii) On adding both sides by ∠𝐸𝐷𝐶 in eq. (ii), we get ∠𝐴𝐷𝐸 + ∠𝐸𝐷𝐶 = ∠𝐴𝐶𝐵 + ∠𝐸𝐷𝐶 180° = ∠𝐴𝐶𝐵 + ∠𝐸𝐷𝐶 ∠𝐸𝐷𝐶 + ∠𝐴𝐵𝐶 = 180° [from eq. (i)] Hence, 𝐵𝐷𝐶𝐸 is a cyclic quadrilateral, because sum of the opposite angles is 180.
Q.12) If a pair of opposite sides of a cyclic quadrilateral are equal, then prove that its diagonals are also equal.
Sol.12) Given Let ABCD be a cyclic quadrilateral and 𝐴𝐷 = 𝐵𝐶. Join 𝐴𝐶 and 𝐵𝐷. To prove 𝐴𝐶 = 𝐵𝐷 Proof In 𝛥𝐴𝑂𝐷 and 𝛥𝐵𝑂𝐶, ∠𝑂𝐴𝐷 = ∠𝑂𝐵𝐶 and ∠𝑂𝐷𝐴 = ∠𝑂𝐶𝐵 [since, same segments subtends equal angle to the circle] 𝐴𝐵 = 𝐵𝐶 [given] 𝛥𝐴𝑂𝐷 = 𝛥𝐵𝑂𝐶 [by ASA congruence rule] Adding is 𝐷𝑂𝐶 on both sides, we get 𝛥𝐴𝑂𝐷 + 𝛥𝐷𝑂𝐶 ≅ 𝛥𝐵𝑂𝐶 + 𝛥𝐷𝑂𝐶 ⇒ 𝛥𝐴𝐷𝐶 ≅ 𝛥𝐵𝐶𝐷 𝐴𝐶 = 𝐵𝐷 [by CPCT]
Q.13) The circumcentre of the 𝛥𝐴𝐵𝐶 is 𝑂. Prove that ∠𝑂𝐵𝐶 + ∠𝐵𝐴𝐶 = 90°.
Sol.13) Given A circle is circumcised on a ∆𝐴𝐵𝐶 having centre 𝑂.
To prove ∠𝑂𝐵𝐶 + ∠𝐵𝐴𝐶 = 90° Construction Join 𝐵𝑂 & 𝐶𝑂. Proof Let ∠𝑂𝐵𝐶 = ∠𝑂𝐶𝐵 = 𝜃 … (i)
In ∆𝑂𝐵𝐶, ∠𝐵𝑂𝐶 + ∠𝑂𝐶𝐵 + ∠𝐶𝐵𝑂 = 180°
∠𝐵𝑂𝐶 + 𝜃 + 𝜃 = 180° ∠𝐵𝑂𝐶 = 180° − 2𝜃 In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. ∠𝐵𝑂𝐶 = 2∠𝐵𝐴𝐶
∠𝐵𝐴𝐶 =∠𝐵𝑂𝐶
2=
180°−2𝜃
2= 90° − 𝜃
∠𝐵𝐴𝐶 + 𝜃 = 90° [from eq. (i)]
∴ ∠𝐵𝐴𝐶 + ∠𝑂𝐵𝐶 = 90° Hence proved.
Q.14) A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.
Sol.14) Given, 𝐴𝐵 is a chord of a circle, which is equal to the radius of the circle, i.e., 𝐴𝐵 = 𝐵𝑂 …(i) Join 𝑂𝐴, 𝐴𝐶 and 𝐵𝐶. Since, 𝑂𝐴 = 𝑂𝐵 = Radius of circle 𝑂𝐴 = 𝐴𝑆 = 𝐵𝑂 Thus, ∆𝑂𝐴𝐵 is an equilateral triangle.
∠𝐴𝑂𝐵 = 60° By using the theorem, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. i.e., ∠𝐴𝑂𝐵 = 2∠𝐴𝐶𝐵
∠𝐴𝐶𝐵 =60°
2= 30°
Q.15) In figure, ∠𝐴𝐷𝐶 = 130° and chord 𝐵𝐶 = chord BE. Find ∠𝐶𝐵𝐸.
Sol.15) We have, ∠𝐴𝐷𝐶 = 130° and chord BC =chord 𝐵𝐸. Suppose, we consider the points A, B, C and D form a cyclic quadrilateral.
Since, the sum of opposite angles of a cyclic quadrilateral 𝐴𝐷𝐶𝐵 is 180°.
∠𝐴𝐷𝐶 + ∠𝑂𝐵𝐶 = 180° ⇒ 130° + ∠𝑂𝐵𝐶 = 180° ⇒ ∠𝑂𝐵𝐶 = 180° − 130° = 50° In 𝛥𝐵𝑂𝐶 and 𝛥𝐵𝑂𝐸, 𝐵𝐶 = 𝐵𝐸 [given equal chord] 𝑂𝐶 = 𝑂𝐸 [both are the radius of the circle] and 𝑂𝐵 = 𝑂𝐵 [common side] 𝛥𝐵𝑂𝐶 ≅ 𝛥𝐵𝑂𝐸 ∠𝑂𝐵𝐶 = ∠𝑂𝐵𝐸 = 50° [by CPCT] ∠𝐶𝐵𝐸 = ∠𝐶𝐵𝑂 + ∠𝐸𝐵𝑂 = 50° + 50° = 100°
Q.16) In figure, ∠𝐴𝐶𝐵 = 40°. Find ∠𝑂𝐴𝐵.
So.16) Given, ∠𝐴𝐶𝐵 = 40° A segment subtends an angle to the circle is half of the angle subtends to the centre. ∴ ∠𝐴𝑂𝐵 = 2∠𝐴𝐶𝐵
∠𝐴𝑂𝐶 + ∠𝐶𝑂𝐵 = 120° ∠𝐴𝑂𝐶 + 66° = 120° from eq. (ii)
∠𝐴𝑂𝐶 = 120° − 66° = 54° Long Answer Type Questions
Exercise 10.4 Q.1) If two equal chords of a circle intersect, prove that the parts of one chord are separately
equal to the parts of the other chord.
Sol.1) Given, consider 𝐴𝐵 & 𝐶𝐷 are two equal chords of a circle, which meet at point 𝐸. To prove 𝐴𝐸 = 𝐶𝐸 & 𝐵𝐸 = 𝐷𝐸 Construction Draw 𝑂𝑀 ⊥ 𝐴𝐵 & 𝑂𝑁 ⊥ 𝐶𝐷 & join 𝑂𝐸 where 𝑂 is the centre of circle. Proof In ∆𝑂𝑀𝐸 & ∆𝑂𝑁𝐸, 𝑂𝑀 = 𝑂𝑁 𝑂𝐸 = 𝑂𝐸 & ∠𝑂𝑀𝐸 = ∠𝑂𝑁𝐸 ∆𝑂𝑀𝐸 = ∆𝑂𝑁𝐸 𝐸𝑀 = 𝐸𝑁 [by CPCT]… (i) Now, 𝐴𝐵 = 𝐶𝐷
On adding equation (i) & (ii), we get 𝐸𝑀 + 𝐴𝑀 = 𝐸𝑁 + 𝐶𝑁 𝐴𝐸 = 𝐶𝐸 … (iii) Now, 𝐴𝐵 = 𝐶𝐷 On subtracting both sides by 𝐴𝐸, we get 𝐴𝐵 − 𝐴𝐸 = 𝐶𝐷 − 𝐴𝐸 𝐵𝐸 = 𝐶𝐷 − 𝐶𝐸 from eq. (iii) 𝐵𝐸 = 𝐷𝐸 Hence proved.
Q.2) If non-parallel sides of a trapezium are equal, prove that it is cyclic.
Sol.2) Given 𝐴𝐵𝐶𝐷 is a trapezium whose non-parallel sides 𝐴𝐷 and 𝐵𝐶 are equal. To prove trapezium 𝐴𝐵𝐶𝐷 is a cyclic. Join 𝐵𝐸, where 𝐵𝐸 || 𝐴𝐷. Proof Since, 𝐴𝐵 || 𝐷𝐸 and 𝐴𝐷 || 𝐵𝐸 Since, quadrilateral 𝐴𝐵𝐸𝐷 is a parallelogram. ∴ ∠𝐵𝐴𝐷 = ∠𝐵𝐸𝐷 … (i) & 𝐴𝐷 = 𝐵𝐸 … (ii) But 𝐴𝐷 = 𝐵𝐶 … (iii) [given] From eqs. (ii) & (iii), we get ∠𝐵𝐸𝐶 = ∠𝐵𝐶𝐸 … (iv)
Also, ∠𝐵𝐸𝐶 + ∠𝐵𝐸𝐷 = 180°
∠𝐵𝐶𝐸 + ∠𝐵𝐴𝐷 = 180° If sum of opposite angles of a quadrilateral is 180°, then quadrilateral is cyclic. Hence, trapezium 𝐴𝐵𝐶𝐷 is a cyclic. Hence proved.
Q.3) If P, Q and R are the mid-points of the sides, 𝐵𝐶, 𝐶𝐴 and 𝐴𝐵 of a triangle and AD is the perpendicular from 𝐴 on 𝐵𝐶, then prove that 𝑃, 𝑄, 𝑅 and 𝐷 are concyclic.
Sol.3) Given, In ∆𝐴𝐵𝐶, 𝑃, 𝑄 & 𝑅 are the mid points of the sides 𝐵𝐶, 𝐶𝐴 and 𝐴𝐵respectively. Also, 𝐴𝐷 ⊥ 𝐵𝐶 To prove 𝑃, 𝑄, 𝑅 and 𝐷 are concyclic. Construction Join 𝐷𝑅, 𝑅𝑄 and 𝑄𝑃 Proof In right angled ∆𝐴𝐷𝑃, 𝑅 is the mid point of 𝐴𝐵. ∴ 𝑅𝐵 = 𝑅𝐷 ∠2 = ∠1 … (i) Since, 𝑅 and 𝑄 are the mid-points of 𝐴𝐵 and 𝐴𝐶 , then 𝑅𝑄 || 𝐵𝐶 [by mid point theorem] or 𝑅𝑄|| 𝐵𝑃 since, 𝑄𝑃 || 𝑅𝐵, then quadrilateral 𝐵𝑃𝑄𝑅 is a parallelogram, ∠1 = ∠3 … (ii) From eqs. (i) & (ii), we get ∠2 = ∠3 But ∠2 + ∠4 = 180° [linear pair axiom]
∴ ∠3 + ∠4 = 180° Hence, quadrilateral 𝑃𝑄𝑅𝐷 is a cyclic quadrilateral. So, points 𝑃, 𝑄, 𝑅 and 𝐷 are concyclic. Hence proved.
Q.4) 𝐴𝐵𝐶𝐷 is a parallelogram. A circle through 𝐴, 𝐵 is so drawn that it intersects 𝐴𝐷 at 𝑃 and 𝐵𝐶 at 𝑄. Prove that 𝑃, 𝑄, 𝐶 and 𝐷 are concyclic.
Sol.4) Given 𝐴𝐵𝐶𝐷 is a parallelogram. A circle whose centre 𝑂 passes through 𝐴, 𝐵 is so drawn that it intersect 𝐴𝐷 at 𝑃 and 𝐵𝐶 at 𝑄 To prove Points 𝑃, 𝑄, 𝐶 and D are con-cyclic. Construction Join 𝑃𝑄 Proof ∠1 = ∠𝐴 [exterior angle property of cyclic quadrilateral] But ∠𝐴 = ∠𝐶 [opposite angles of a parallelogram] ∴ ∠1 = ∠𝐶 ...(i)
But ∠𝐶 + ∠𝐷 = 180° [sum of cointerior angles on same side is 180°]
⇒ ∠1 + ∠𝐷 = 180° [from Eq. (i)]
Thus, the quadrilateral 𝑄𝐶𝐷𝑃 is cyclic. So, the points 𝑃, 𝑄, 𝐶 and 𝐷 are non-cyclic. Hence proved.
Q.5) Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side, if intersect they will intersect on the circumcircle of the triangle.
Sol.5) Given, ∆𝐴𝐵𝐶 is inscribed in a circle. Bisector ∠𝐴 & perpendicular bisector of 𝐵𝐶 intersect at point 𝑄. To prove 𝐴, 𝐵, 𝑄 & 𝐶 are con-cyclic. Construction Join 𝐵𝑄 & 𝑄𝐶. Proof We have assumed that 𝑄 lies outside the circle. In ∆𝐵𝑀𝑄 & ∆𝐶𝑀𝑄, 𝐵𝑀 = 𝐶𝑀 [𝑄𝑀 perpendicular bisector of 𝐵𝐶] ∠𝐵𝑀𝑄 = ∠𝐶𝑀𝑄 𝑀𝑄 = 𝑀𝑄 [common side] ∴ ∆𝐵𝑀𝑄 ≅ ∆𝐶𝑀𝑄 [by SAS congruence rule] ∴ 𝐵𝑄 = 𝐶𝑄 [by CPCT] … (i) Also, ∠𝐵𝐴𝑄 = ∠𝐶𝐴𝑄 [given] … (ii) From eqs. (i) & (ii), we can say that 𝑄 lies on the circle. Hence, 𝐴, 𝐵, 𝑄 and 𝐶 are con cyclic. Hence proved.
Q.6) If two chords 𝐴𝐵 and 𝐶𝐷 of a circle 𝐴𝑌𝐷𝑍𝐵𝑊𝐶𝑋 intersect at right angles, then prove that 𝑎𝑟𝑐 𝐶𝑋𝐴 + 𝑎𝑟𝑐 𝐷𝑍𝐵 = 𝑎𝑟𝑐 𝐴𝑌𝐷 + 𝑎𝑟𝑐 𝐵𝑊𝐶 = semi-circle.
Sol.6) Given In a circle 𝐴𝑌𝐷𝑍𝐵𝑊𝐶𝑋, two chords 𝐴𝐵 and 𝐶𝐷 intersect at right angles.
To prove 𝑎𝑟𝑐 𝐶𝑋𝐴 + 𝑎𝑟𝑐 𝐷𝑍𝐵 = 𝑎𝑟𝑐 𝐴𝑌𝐷 + 𝑎𝑟𝑐 𝐵𝑊𝐶 = Semi-circle. Construction Draw a diameter 𝐸𝐹 parallel to 𝐶𝐷 having centre 𝑀. Proof Since, 𝐶𝐷||𝐸𝐹 𝑎𝑟𝑐 𝐸𝐶 = 𝑎𝑟𝑐 𝑃𝐷 … (i) 𝑎𝑟𝑐 𝐸𝐶𝑋𝐴 = 𝑎𝑟𝑐 𝐸𝑊𝐵 [symmetrical about diameter of a circle] 𝑎𝑟𝑐 𝐴𝐹 = 𝑎𝑟𝑐 𝐵𝐹 …(ii) We know that, 𝑎𝑟𝑐 𝐸𝐶𝑋𝐴𝑌𝐷𝐹 =Semi-circle 𝑎𝑟𝑐 𝐸𝐴 + 𝑎𝑟𝑐 𝐴𝐹 = 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒 𝑎𝑟𝑐 𝐸𝐶 + 𝑎𝑟𝑐 𝐶𝑋𝐴 + 𝑎𝑟𝑐 𝐹𝐵 = 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒 𝑎𝑟𝑐 𝐷𝐹 + 𝑎𝑟𝑐 𝐶𝑋𝐴 + 𝑎𝑟𝑐 𝐹𝐵 = 𝑠𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒 𝑎𝑟𝑐 𝐷𝐹 + 𝑎𝑟𝑐 𝐹𝐵 + 𝑎𝑟𝑐 𝐶𝑋𝐴 = 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒
𝑎𝑟𝑐 𝐷𝑍𝐵 + 𝑎𝑟𝑐 𝐶 × 𝐴 = 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒 Circle divides itself in two semi circles therefore the remaining portion of the circle is also equal to the semicircle. ∴ 𝑎𝑟𝑐 𝐴𝑌𝐷 + 𝑎𝑟𝑐 𝐵𝑊𝐶 = 𝑠𝑒𝑚𝑖 𝑐𝑖𝑟𝑐𝑙𝑒 Hence proved.
Q.7) If ∆𝐴𝐵𝐶 is an equilateral triangle inscribed in a circle and 𝑃 be any point on the minor arc 𝐵𝐶 which does not coincide with 𝐵 or 𝐶, then prove that 𝑃𝐴 is angle bisector of ∠𝐵𝑃𝐶.
Sol.7) Given, ∆𝐴𝐵𝐶 is an equilateral triangle inscribed in a circle and 𝑃 be any point on the minor arc 𝐵𝐶 which does not coincide with 𝐵 or 𝐶. To prove 𝑃𝐴 is an angle bisector of ∠𝐵𝑃𝐶. Construction Join 𝑃𝐵 and 𝑃𝐶. Proof Since, ∆𝐴𝐵𝐶 is an equilateral triangle.
∠3 = ∠4 = 60° Now, ∠1 = ∠4 = 60° … (i)
∠2 = ∠3 = 60° … (ii)
∠1 = ∠2 = 60° Hence, 𝑃𝐴 is the bisector of ∠𝐵𝑃𝐶. Hence proved.
Q.8) In figure, AB and CD are two chords of a circle intersecting each other at point E. Prove that ∠𝐴𝐸𝐶 = ½ (angle subtended by 𝑎𝑟𝑐 𝐶 × 𝐴 at centre + angle subtended by 𝑎𝑟𝑐 𝐷𝑌𝐵 at the centre).
Sol.8) Given In a figure, two chords 𝐴𝐵 and 𝐶𝐷 intersecting each other at point 𝐸.
To prove ∠𝐴𝐸𝐶 =1
2
Construction extend the line 𝐷𝑂 & 𝐵𝑂 at the points 𝑙 & 𝐻 on the circle. Also, join 𝐴𝐶. Proof In a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle. ∴ ∠1 = 2∠6 … (i) & ∠3 = 2∠7 … (ii) In ∆𝐴𝑂𝐶, 𝑂𝐶 = 𝑂𝐴 ∠𝑂𝐶𝐴 = ∠4 Also, ∠𝐴𝑂𝐶 + ∠𝑂𝐶𝐴 + ∠4 = 180°
Q.9) If bisectors of opposite angles of a cyclic quadrilateral 𝐴𝐵𝐶𝐷 intersect the circle, circumscribing it at the points 𝑃 and 𝑄, prove that 𝑃𝑄 is a diameter of the circle.
Sol.9) Given, 𝐴𝐵𝐶𝐷 is a cyclic quadrilateral. 𝐷𝑃 and QB are the bisectors of ∠𝐷 and ∠𝐵 respectively. To prove 𝑃𝑄 is the diameter of a circle. Construction Join 𝑄𝐷 & 𝑄𝐶 Proof Since, 𝐴𝐵𝐶𝐷 is a cyclic quadrilateral
∴ ∠𝐶𝐷𝐴 + ∠𝐶𝐵𝐴 = 180° On dividing both sides by 2, we get
1
2∠𝐶𝐷𝐴 +
1
2∠𝐶𝐵𝐴 =
1
2× 180° = 90°
∠1 + ∠2 = 90° … (i) But ∠2 = ∠3 … (ii) ∠1 + ∠3 From eqs. (i) & (ii), we get
∠𝑃𝑄𝐷 = 90° Hence, 𝑃𝑄 is a diameter of a circle, because diameter of the circle. Subtends a right angle at the circumference.
Q.10) A circle has radius √2 𝑐𝑚. It is divided into two segments by a chord of length 2 cm.
Prove that the angle subtended by the chord at a point in major segment is 45°.
Sol.10) Draw a circle having centre 𝑂. Let 𝐴𝐵 = 2𝑐𝑚 be a chord of a circle. A chord 𝐴𝐵 is divided by the line 𝑂𝑀 in two equal segments. To prove ∠𝐴𝑃𝐵 = 45 Here, 𝐴𝑁 = 𝑁𝐵 = 1𝑐𝑚
𝐵𝐿 = 𝐷𝑀 … (ii) On subtracting eq. (ii) & (i), we get 𝐿𝑃 − 𝐵𝐿 = 𝑀𝑃 − 𝐷𝑀 𝑃𝐵 = 𝑃𝐷 Hence proved.
Q.12) 𝐴𝐵 and 𝐴𝐶 are two chords of a circle of radius r such that 𝐴𝐵 = 2𝐴𝐶. If 𝑝 and 𝑞 are the distances of 𝐴𝐵 and 𝐴𝐶 from the centre, prove that 4𝑞2 = 𝑝2 + 3𝑟2.
Sol.12) Given In a circle of radius 𝑟, there are two chords 𝐴𝐵 & 𝐴𝐶 such that 𝐴𝐵 = 2𝐴𝐶. Also, the distance of 𝐴𝐵 & 𝐴𝐶 from the centre are 𝑝 & 𝑞, respectively. To prove 4𝑞2 = 𝑝2 + 3𝑟2 Proof Let 𝐴𝐶 = 𝑎, then 𝐴𝐵 = 2𝑎 From centre 𝑂, perpendicular is drawn to the chords 𝐴𝐶 & 𝐴𝐵 at 𝑀 & 𝑁 respectively.
∴ 𝐴𝑀 = 𝑀𝐶 =𝑎
2
𝐴𝑁 = 𝑁𝐵 = 𝑎 In ∆𝑂𝐴𝑀, 𝐴𝑂2 = 𝐴𝑀2 + 𝑀𝑂2
⇒ 𝐴𝑂2 = (𝑎
2)
2+ 𝑞2 … (i)
In ∆𝑂𝐴𝑁, 𝐴𝑂2 = (𝐴𝑁)2 + (𝑁)2 𝐴𝑂2 = (𝑎)2 + 𝑝2 … (ii) From eqs. (i) & (ii), we get
