RS AGGARWAL CLASS 9 CHAPTER 12 CIRCLES MCQ

 MULTIPLE CHOICE QUESTIONS 

PAGE NO-489

Question 1:

The radius of a circle is 13 cm and the length of one of its chords is 10 cm. The distance of the chord from the centre is
(a) 11.5 cm
(b) 12 cm
(c) $\sqrt{69} \mathrm{cm}$
(d) 23 cm

Answer 1:

(b) 12 cm
Let AB be the chord of the given circle with centre O and a radius of 13 cm.
Then, AB = 10 cm and OB = 13 cm

From O, draw OM perpendicular to AB.
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ BM = $\left(\frac{10}{2}\right)\mathrm{cm}=5\mathrm{cm}$
From the right ΔOMB, we have:
OB² = OM² + MB²
⇒ 13² = OM² + 5²
⇒ 169 = OM² + 25
⇒ OM² = (169 - 25) = 144
⇒ $\mathrm{OM}=\sqrt{144}\mathrm{cm}=12\mathrm{cm}$
Hence, the distance of the chord from the centre is 12 cm.

Question 2:

A chord is at a distance of 8 cm from the centre of a circle of radius 17 cm. The length of the chord is
(a) 25 cm
(b) 12.5 cm
(c) 30 cm
(d) 9 cm

Answer 2:

(c) 30 cm

Let AB be the chord of the given circle with centre O and a radius of 17 cm.
From O, draw OM perpendicular to AB.
Then OM = 8 cm and OB = 17 cm

From the right ΔOMB, we have:
OB² = OM² + MB²
⇒ 17² = 8² + MB²
⇒ 289 = 64 + MB²
⇒ MB² = (289 - 64) = 225
⇒ $\mathrm{MB}=\sqrt{225}\mathrm{cm}=15\mathrm{cm}$
The perpendicular from the centre of a circle to a chord bisects the chord.
∴ AB = 2 × MB = (2 x 15) cm = 30 cm
Hence, the required length of the chord is 30 cm.

Question 3:

In the given figure, BOC is a diameter of a circle and AB = AC. Then, ∠ABC = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 3:

(b) 45°

Since an angle in a semicircle is a right angle, ∠BAC = 90°
∴ ∠ABC + ∠ACB = 90°

Now, AB = AC       (Given)
⇒ ∠ABC = ∠ACB = 45°

Question 4:

In the given figure, O is the centre of a circle and ∠ACB = 30°. Then, ∠AOB = ?
(a) 30°
(b) 15°
(c) 60°
(d) 90°
Figure

Answer 4:

(c) 60°
We know that the angle at the centre of a circle is twice the angle at any point on the remaining part of the circumference.
Thus, ∠AOB = (2 × ∠ACB) = (2 × 30°) = 60°

Question 5:

In the given figure, O is the centre of a circle. If ∠OAB = 40° and C is a point on the circle, then ∠ACB = ?
(a) 40°
(b) 50°
(c) 80°
(d) 100°

Answer 5:

(b) 50°
OA = OB
⇒ ∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
∴ $\angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 100\right)°=50°$

Question 6:

In the given figure, AOB is a diameter of a circle with centre O such that AB = 34 cm and CD is a chord of length 30 cm. Then the distance of CD from AB is
(a) 8 cm
(b) 15 cm
(c) 18 cm
(d) 6 cm

Answer 6:

(a) 8 cm
Join OC. Then OC = radius = 17 cm

$\mathrm{CL}=\frac{1}{2}\mathrm{CD}=\left(\frac{1}{2}\times 30\right)\mathrm{cm}=15\mathrm{cm}$
In right ΔOLC, we have:
OL² = OC² - CL² = (17)² - (15)² = (289 - 225) = 64
$\Rightarrow \mathrm{OL}=\sqrt{64}=8\mathrm{cm}$
∴ Distance of CD from AB = 8 cm

Question 7:

AB and CD are two equal chords of a circle with centre O such that ∠AOB = 80°, then ∠COD = ?
(a) 100°
(b) 80°
(c) 120°
(d) 40°

Answer 7:

(b) 80°
Given: AB = CD
We know that equal chords of a circle subtend equal angles at the centre.
∴ ∠COD = ∠AOB = 80°

PAGE NO-490

Question 8:

In the given figure, CD is the diameter of a circle with centre O and CD is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, then radius of the circles is
(a) 6 cm
(b) 9 cm
(c) 7.5 cm
(d) 8 cm

Answer 8:

(c) 7.5 cm
Let OA = OC = r cm.
Then OE = (r - 3) cm and $\mathrm{AE}=\frac{1}{2}\mathrm{AB}=6\mathrm{cm}$
Now, in right ΔOAE, we have:
OA² = OE² +AE² 
⇒ (r)² = (r - 3)² + 6²
⇒ r² = r² + 9 - 6r + 36
⇒ 6r = 45
⇒ $r=\frac{45}{6}=7.5$ cm
Hence, the required radius of the circle is 7.5 cm.

Question 9:

In the given figure, O is the centre of a circle and diameter AB bisects the chord CD at a point E such that CE = ED = 8 cm and EB = 4 cm. The radius of the circle is
(a) 10 cm
(b) 12 cm
(c) 6 cm
(d) 8 cm

Answer 9:

(a) 10 cm
Let the radius of the circle be r cm.
Let OD = OB = r cm.
Then OE = (r - 4) cm and ED = 8 cm
Now, in right ΔOED, we have:
OD² = OE² +ED² 
⇒ (r)² = (r - 4)² + 8²
⇒ r² = r² + 16 - 8r + 64
⇒ 8r = 80
⇒ r = 10 cm
Hence, the required radius of the circle is 10 cm.

Question 10:

In the given figure, BOC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. If AB = 10 cm, then CD = ?
(a) 5 cm
(b) 12.5 cm
(c) 15 cm
(d) 10 cm

Answer 10:

(d) 10 cm



Draw OE ⊥ AB and OF ⊥ CD.
In Δ OEB and ΔOFC, we have:
OB =  OC              (Radius of a circle)
∠BOE = ∠COF     (Vertically opposite angles)
∠OEB = ∠OFC     (90° each)
∴ ΔOEB ≅ ΔOFC (By AAS congruency rule)
∴ OE = OF
Chords equidistant from the centre are equal.
∴ CD = AB = 10 cm

Question 11:

In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠ACD = 25°, then ∠AOD = ?
(a) 50°
(b) 75°
(c) 90°
(d) 100°

Answer 11:

(b) 75°
OB = BC (Given)
⇒ ∠BOC = ∠BCO = 25°
Exterior ∠OBA = ∠BOC + ∠BCO = (25° + 25°) = 50°
OA = OB (Radius of a circle)
⇒ ∠OAB = ∠OBA  = 50°
In Δ AOC, side CO has been produced to D.
∴ Exterior ∠AOD = ∠OAC  + ∠ACO
                           = ∠OAB + ∠BCO
                           = (50° + 25°) = 75°

Question 12:

In the given figure, AB is a chord of a circle with centre O and BOC is a diameter. If OD ⊥ AB such that OD = 6 cm, then AC = ?
(a) 9 cm
(b) 12 cm
(c) 15 cm
(d) 7.5 cm

Answer 12:

(b) 12 cm
OD ⊥ AB
i.e., D is the mid point of AB.
Also, O is the mid point of  BC.
Now, in Δ BAC, D is the mid point of AB and O is the mid point of BC.
∴ $\mathrm{OD}=\frac{1}{2}\mathrm{AC}$  (By mid point theorem)
⇒ AC = 2OD = (2 × 6) cm = 12 cm

Question 13:

An equilateral triangle of side 9 cm is inscribed in a circle. The radius of the circle is
(a) 3 cm
(b) $3\sqrt{2} \mathrm{cm}$
(c) $3\sqrt{3} \mathrm{cm}$
(d) 6 cm
Figure

Answer 13:

(c) $3\sqrt{3} \mathrm{cm}$

Let ΔABC be an equilateral triangle of side 9 cm.
Let AD be one of its medians.

Then AD ⊥ BC and BD = 4.5 cm
∴ $\mathrm{AD}=\sqrt{{\mathrm{AB}}^2-{\mathrm{BD}}^2}=\sqrt{{\left(9\right)}^2-{\left(\frac{9}{2}\right)}^2}=\sqrt{81-\frac{81}{4}}=\sqrt{\frac{324-81}{4}}=\sqrt{\frac{243}{4}}=\frac{9\sqrt{3}}{2}\mathrm{cm}$
Let G be the centroid of ΔABC.
Then AG : GD = 2 : 1
∴ Radius =  AG = $\frac{2}{3}\mathrm{AD}=\left(\frac{2}{3}\times \frac{9\sqrt{3}}{2}\right)\mathrm{cm}=3\sqrt{3}\mathrm{cm}$

Question 14:

The angle in a semicircle measures
(a) 45°
(b) 60°
(c) 90°
(d) 36°
Figure

Answer 14:

(c) 90°
The angle in a semicircle measures 90°.

Question 15:

Angles in the same segment of a circle area are
(a) equal
(b) complementary
(c) supplementary
(d) none of these
Figure

Answer 15:

(a) equal
The angles in the same segment of a circle are equal.

PAGE NO-491

Question 16:

In the given figure, ∆ABC and ∆DBC are inscribed in a circle such that ∠BAC = 60° and ∠DBC = 50°.
(a) 50°
(b) 60°
(c) 70°
(d) 80°

Answer 16:

(c) 70°
∠BDC = ∠BAC = 60°   (Angles in the same segment of a circle)
In Δ BDC, we have:
∠DBC + ∠BDC + ∠BCD = 180°    (Angle sum property of a triangle)
∴ 50° + 60° + ∠BCD  = 180°
⇒ ∠BCD = 180° - (50° + 60°) = (180° - 110°) = 70°

Question 17:

In the given figure, BOC is a diameter of a circle with centre O. If ∠BCA = 30°, then ∠CDA = ?
(a) 30°
(b) 45°
(c) 60°
(d) 50°

Answer 17:

(c) 60°
Angles in a semi circle measure 90°.
∴ ∠BAC = 90°
In Δ ABC, we have:
∠BAC + ∠ABC + ∠BCA = 180° (Angle sum property of a triangle)
∴ 90° + ∠ABC + 30° = 180°
⇒ ∠ABC = (180° - 120°) = 60°
∴ ∠CDA = ∠ABC = 60° (Angles in the same segment of a circle)

Question 18:

In the given figure, O is the centre of a circle. If ∠OAC = 50°, then ∠ODB = ?
(a) 40°
(b) 50°
(c) 60°
(d) 75°

Answer 18:

(b) 50°
∠ODB =∠OAC = 50° (Angles in the same segment of a circle)

Question 19:

In the given figure, O is the centre of a circle in which ∠OBA = 20° and ∠OCA = 30°. Then, ∠BOC = ?
(a) 50°
(b) 90°
(c) 100°
(d) 130°

Answer 19:

(c) 100°
In Δ OAB, we have:
OA = OB          (Radii of a circle)
⇒ ∠OAB = ∠OBA = 20°
In ΔOAC, we have:
OA = OC         (Radii of a circle)
⇒ ∠OAC = ∠OCA = 30°
Now, ∠BAC = (20° + 30°) = 50°
∴ ∠BOC = (2 × ∠BAC) = (2 × 50°) = 100°

Question 20:

In the given figure, O is the centre of a circle. If ∠AOB = 100° and ∠AOC = 90°, then ∠BAC = ?
(a) 85°
(b) 80°
(c) 95°
(d) 75°

Answer 20:

(a) 85°
We have:
∠BOC + ∠BOA + ∠AOC = 360°
⇒ ∠BOC + 100° + 90° = 360°
⇒ ∠BOC = (360° - 190°) = 170°
∴ $\angle \mathrm{BAC}=\left(\frac{1}{2}\times \angle \mathrm{BOC}\right)=\left(\frac{1}{2}\times 170°\right)=85°$

Question 21:

In the given figure, O is the centre of a circle. Then, ∠OAB = ?
(a) 50°
(b) 60°
(c) 55°
(d) 65°

Answer 21:

(d) 65°
We have:
OA = OB (Radii of a circle)
Let ∠ OAB = ∠ OBA = x°
In Δ OAB, we have:
x° + x° + 50° = 180°   (Angle sum property of a triangle)
⇒ 2x° = (180° - 50°) = 130°
⇒ $x=\left(\frac{130}{2}\right)°=65°$
Hence, ∠OAB = 65°

Question 22:

In the given figure, O is the centre of a circle and ∠AOC = 120°. Then, ∠BDC = ?
(a) 60°
(b) 45°
(c) 30°
(d) 15°

Answer 22:

(c) 30°

∠COB = 180° - 120° = 60°  (Linear pair)
Now, arc BC subtends ∠COB at the centre and ∠BDC at the point D of the remaining part of the circle.
∴ ∠COB = 2∠BDC
⇒ $\angle \mathrm{BDC}=\frac{1}{2}\angle \mathrm{COB}=\left(\frac{1}{2}\times 60°\right)=30°$

PAGE NO-492

Question 23:

In the given figure, O is the centre of a circle and ∠OAB = 50°. Then , ∠CDA = ?
(a) 40°
(b) 50°
(c) 75°
(d) 25°

Answer 23:

(b) 50°
We have:
OA = OB (Radii of a circle)
⇒ ∠OBA = ∠OAB = 50°  
∴ ∠CDA = ∠OBA = 50°   (Angles in the same segment of  a circle)

Question 24:

In the give figure, AB and CD are two intersecting chords of a circle. If ∠CAB = 40° and ∠BCD = 80°, then ∠CBD = ?
(a) 80°
(b) 60°
(c) 50°
(d) 70°
Figure

Answer 24:

(b) 60°
We have:
∠CDB = ∠CAB = 40°  (Angles in the same segment of a circle)
In Δ CBD, we have:
∠CDB + ∠BCD +∠CBD = 180°   (Angle sum property of a triangle)
⇒ 40° + 80° + ∠CBD = 180°
⇒ ∠CBD = (180° - 120°) = 60°

Question 25:

In the given figure, O is the centre of a circle and chords AC and BD intersect at E. If ∠AEB = 110° and ∠CBE = 30°, then ∠ADB = ?
(a) 70°
(b) 60°
(c) 80°
(d) 90°

Answer 25:

(c) 80°
We have:
∠AEB + ∠CEB = 180°     (Linear pair angles)
⇒ 110° + ∠CEB = 180° 
⇒ ∠CEB = (180° - 110°) = 70°
In ΔCEB, we have:
∠CEB + ∠EBC + ∠ECB = 180°   (Angle sum property of a triangle)
⇒ 70° +  30° + ∠ECB = 180° 
⇒ ∠ECB = (180° - 100°) = 80°

The angles in the same segment are equal.
Thus, ∠ADB  = ∠ECB = 80°

Question 26:

In the given figure, O is the centre of a circle in which ∠OAB = 20° and ∠OCB = 50°. Then, ∠AOC = ?
(a) 50°
(b) 70°
(c) 20°
(d) 60°

Answer 26:

(d) 60°
We have:
OA = OB (Radii of a circle)
⇒ ∠OBA= ∠OAB = 20°
In ΔOAB, we have:
∠OAB + ∠OBA + ∠AOB = 180°  (Angle sum property of a triangle)
⇒ 20° + 20° + ∠AOB = 180° 
⇒ ∠AOB = (180° - 40°) = 140°

Again, we have:
OB = OC   (Radii of a circle)
⇒ ∠OBC = ∠OCB = 50°
In ΔOCB, we have:
∠OCB + ∠OBC + ∠COB = 180°  (Angle sum property of a triangle)
⇒ 50° + 50° + ∠COB = 180° 
⇒ ∠COB = (180° - 100°) = 80°
Since ∠AOB = 140°, we have:
∠AOC + ∠COB  = 140°
⇒∠AOC + 80°  = 140°
⇒ ∠AOC = (180° - 80°) = 60°

Question 27:

In the given figure, AOB is a diameter and ABCD is a cyclic quadrilateral. If ∠ADC = 120°, then ∠BAC = ?
(a) 60°
(b) 30°
(c) 20°
(d) 45°

Answer 27:

(b) 30°
We have:
∠ABC + ∠ADC = 180°     (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC + 120° = 180°
⇒ ∠ABC = (180° - 120°) = 60°
Also, ∠ACB = 90°     (Angle in a semicircle)
In ΔABC, we have:
∠BAC + ∠ACB  + ∠ABC = 180°    (Angle sum property of a triangle)
⇒ ∠BAC + 90° + 60° = 180°
⇒ ∠BAC = (180° - 150°) = 30°

Question 28:

In the given figure ABCD is a cyclic quadrilateral in which AB || DC and ∠BAD = 100°. Then, ∠ABC = ?
(a) 80°
(b) 100°
(c) 50°
(d) 40°

Answer 28:

(b) 100°
Since ABCD is a cyclic quadrilateral, we have:
∠BAD + ∠BCD = 180°     (Opposite angles of a cyclic quadrilateral)
⇒ 100° + ∠BCD = 180° 
⇒ ∠BCD = (180° - 100°) = 80°
Now, AB || DC and CB is the transversal.
∴ ∠ABC + ∠BCD = 180° 
⇒ ∠ABC + 80° = 180°
⇒ ∠ABC = (180° - 80°) = 100°

Question 29:

In the given figure, O is the centre of a circle and ∠AOC = 130°. Then, ∠ABC = ?
(a) 50°
(b) 65°
(c) 115°
(d) 130°

Answer 29:

(c) 115°
Take a point D on the remaining part of the circumference.
Join AD and CD.

Then $\angle \mathrm{ADC}=\frac{1}{2}\angle \mathrm{AOC}=\left(\frac{1}{2}\times 130°\right)=65°$
In cyclic quadrilateral ABCD, we have:
∠ABC + ∠ADC = 180°     (Opposite angles of a cyclic quadrilateral)
⇒ ∠ABC + 65° = 180°
⇒ ∠ABC  = (180° - 65°) = 115°

PAGE NO-493

Question 30:

In the given figure, AOB is  a diameter of a circle and CD || AB. If ∠BAD = 30°, then ∠CAD = ?
(a) 30°
(b) 60°
(c) 45°
(d) 50°

Answer 30:

(a) 30°
∠ADC = ∠BAD = 30°     (Alternate angles)
∠ADB = 90°                    (Angle in semicircle)
∴ ∠CDB = (90° + 30°) = 120°
But ABCD being a cyclic quadrilateral, we have:
∠BAC + ∠CDB = 180°
⇒ ∠BAD + ∠CAD + ∠CDB = 180°
⇒ 30° + ∠CAD  + 120° = 180°
⇒ ∠CAD  = (180° - 150°) = 30°

Question 31:

In the given figure, O is the centre of a circle in which ∠AOC = 100°. Side AB of quad. OABC has been produced to D. Then, ∠CBD = ?
(a) 50°
(b) 40°
(c) 25°
(d) 80°

Answer 31:

(a) 50°
Take a point E on the remaining part of the circumference.
Join AE and CE.

Then $\angle \mathrm{AEC}=\frac{1}{2}\angle \mathrm{AOC}=\left(\frac{1}{2}\times 100°\right)=50°$
Now, side AB of the cyclic quadrilateral ABCE has been produced to D.
∴ Exterior ∠CBD = ∠AEC = 50°

Question 32:

In the given figure, O is the centre of a circle and ∠OAB = 50°. Then, ∠BOD = ?
(a) 130°
(b) 50°
(c) 100°
(d) 80°

Answer 32:

(c) 100°
OA = OB  (Radii of a circle)
⇒ ∠OBA = ∠OAB = 50°
In Δ OAB, we have:
∠ OAB + ∠OBA + ∠AOB = 180°    (Angle sum property of a triangle)
⇒ 50° + 50° + ∠AOB = 180°
⇒ ∠AOB = (180° - 100°) = 80°
Since ∠AOB + ∠BOD = 180°  (Linear pair)
∴ ∠BOD = (180° - 80°) = 100°

Question 33:

In the give figure, ABCD is a cyclic quadrilateral in which BC = CD and ∠CBD = 35°. Then, ∠BAD = ?
(a) 65°
(b) 70°
(c) 110°
(d) 90°

Answer 33:

(b) 70°
BC = CD (given)
⇒ ∠BDC = ∠CBD = 35°
In Δ BCD, we have:
∠BCD +  BDC + ∠CBD = 180°     (Angle sum property of a triangle)
⇒ ∠BCD + 35° + 35° = 180°
⇒ ∠BCD = (180° - 70°) = 110°
In cyclic quadrilateral ABCD, we have:
∠BAD + ∠BCD = 180°
⇒ ∠BAD + 110° = 180°
∴ ∠BAD = (180° - 110°) = 70°

Question 34:

In the given figure, equilateral ∆ABC is inscribed in a circle and ABCD is a quadrilateral, as shown. Then, ∠BDC = ?
(a) 90°
(b) 60°
(c) 120°
(d) 150°

Answer 34:

(c) 120°
Since ΔABC is an equilateral triangle, each of its angle is 60°.
∴ ∠BAC = 60°
In a cyclic quadrilateral ABCD, we have:
∠BAC + ∠BDC = 180°
⇒ 60° + ∠BDC = 180°
⇒ ∠BDC = (180° - 60°) = 120°

Question 35:

In the given figure, sides AB and AD of quad. ABCD are produced to E and F respectively. If ∠CBE = 100°, then ∠CDE = ?
(a) 100°
(b) 80°
(c) 130°
(d) 90°

Answer 35:

(b) 80°
In a cyclic quadrilateral ABCD, we have:
Interior opposite angle, ∠ADC = exterior ∠CBE = 100°
∴ ∠CDF = (180° - ∠ADC) = (180° - 100°) = 80°   (Linear pair)

Question 36:

In the given figure, O is the centre of a circle and ∠AOB = 140°. Then, ∠ACB = ?
(a) 70°
(b) 80°
(c) 110°
(d) 40°

Answer 36:

(c) 110°
Join AB.
Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle.

∴∠AOB = 2∠ADB
$\Rightarrow \angle \mathrm{ADB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 140°\right)=70°$
In the cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°
⇒ 70° + ∠ACB = 180°
∴∠ACB = (180° - 70°) = 110°

PAGE NO-494

Question 37:

In the given figure, O is the centre of a circle and ∠AOB = 130°. Then, ∠ACB = ?
(a) 50°
(b) 65°
(c) 115°
(d) 155°

Answer 37:

(c) 115°
Join AB.
Then chord AB subtends ∠AOB at the centre and ∠ADB at a point D of the remaining parts of a circle.

∴∠AOB = 2∠ADB
$\Rightarrow \angle \mathrm{ADB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 130°\right)=65°$
In cyclic quadrilateral, we have:
∠ADB + ∠ACB = 180°
⇒ 65° + ∠ACB = 180°
∴∠ACB = (180° - 65°) = 115°

Question 38:

In the given figure, ABCD and ABEF are two cyclic   quadrilaterals. If ∠BCD = 110°, then ∠BEF = ?
(a) 55°
(b) 70°
(c) 90°
(d) 110°

Answer 38:

(d) 110°
Since ABCD is a cyclic quadrilateral, we have:
∠BAD + ∠BCD = 180°
⇒ ∠BAD + 110° = 180°
⇒∠BAD = (180° - 110°) = 70°
Similarly in ABEF, we have:
∠BAD + ∠BEF = 180°
⇒ 70° + ∠BEF = 180°
⇒ ∠BEF  = (180° - 70°) = 110°

Question 39:

In the given figure, ABCD is a cyclic quadrilateral in which DC is produced to E and CF is drawn parallel to AB such that ∠ADC = 95° and ∠ECF = 20°. Then, ∠BAD = ?
(a) 95°
(b) 85°
(c) 105°
(d) 75°

Answer 39:

(c) 105°
We have:
∠ABC + ∠ADC = 180°
⇒ ∠ABC + 95° = 180°
⇒∠ABC = (180° - 95°) = 85°
Now, CF || AB and CB is the transversal.
∴ ∠BCF = ∠ABC = 85°     (Alternate interior angles)
⇒ ∠BCE = (85° + 20°) = 105°
⇒ ∠DCB = (180° - 105°) = 75°
Now, ∠BAD + ∠BCD = 180°
⇒ ∠BAD + 75° = 180°
⇒ ∠BAD = (180° - 75°) = 105°

Question 40:

Two chords AB and CD of a circle intersect each other at a point E outside the circle. If AB = 11 cm, BE = 3 cm and DE = 3.5 cm, then CD = ?
(a) 10.5 cm
(b) 9.5 cm
(c) 8.5 cm
(d) 7.5 cm

Answer 40:

(c) 8.5 cm
Join AC.

Then AE : CE = DE : BE     (Intersecting secant theorem)
∴ AE × BE = DE × CE
Let CD = x cm
Then AE = (AB + BE) = (11 + 3) cm = 14 cm; BE = 3cm; CE = (x + 3.5) cm; DE = 3.5 cm
∴ 14 × 3 = (x + 3.5) × 3.5
$\Rightarrow x+3.5=\frac{14\times 3}{3.5}=\frac{42}{3.5}=12$
⇒ x = (12 - 3.5) cm = 8.5 cm
Hence, CD = 8.5 cm

Question 41:

In the given figure, A and B are the centres of two circles having radii 5 cm and 3 cm respectively and intersecting at points P and Q respectively. If AB = 4 cm, then the length of common chord PQ is
(a) 3 cm
(b) 6 cm
(c) 7.5 cm
(d) 9 cm

Answer 41:

(b) 6 cm
We know that the line joining their centres is the perpendicular bisector of the common chord.
Join AP.
Then AP = 5 cm; AB = 4 cm
Also, AP² = BP² + AB²   
Or BP²  = AP² - AB²  
Or BP²  = 5² - 4² 
Or BP = 3 cm
∴ ΔABP is a right angled and PQ = 2 × BP = (2 × 3) cm = 6 cm

Question 42:

In the given figure, ∠AOB = 90° and ∠ABC = 30°. Then, ∠CAO = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer 42:

(c) 60°
We have:
∠AOB = 2∠ACB
$\Rightarrow \angle \mathrm{ACB}=\frac{1}{2}\angle \mathrm{AOB}=\left(\frac{1}{2}\times 90°\right)=45°$
∠COA = 2∠CBA = (2 × 30°) = 60°
∴ ∠COD = 180° - ∠COA = (180° - 60°) = 120°
$\Rightarrow \angle \mathrm{CAO}=\frac{1}{2}\angle \mathrm{COD}=\left(\frac{1}{2}\times 120°\right)=60°$