Circle — Exercise 16(B)

In the given figure, arc APB = arc CQD, then :

1. AB = CD

2. AB > CD

3. AB < CD

4. none of the above

Answer

We know that,

Equal arcs subtends equal chords.

Given,

⇒ arc APB = arc CQD

⇒ AB = CD.

Hence, Option 1 is the correct option.

In the given figure, O is center of the circle and ∠COD is greater than ∠AOB, then :

1. AB > CD

2. AB < CD

3. AB = CD

4. AB + CD = AD

Answer

We know that,

Greater the chord, greater is the angle subtended by it at the center.

Given,

⇒ ∠AOB < ∠COD

⇒ AB < CD

Hence, Option 2 is the correct option.

In a circle, O is its center and AB, CD are its two chords. If AB : CD = 3 : 2, then ratio between ∠AOB and ∠COD is :

1. 1 : 1

2. 3 : 2

3. 2 : 5

4. 3 : 5

Answer

Ratio of the angles subtended by the chords on the center is equal to the ratio of the chords.

⇒ $\dfrac{∠AOB}{∠COD} = \dfrac{AB}{CD} = \dfrac{3}{2}$

⇒ ∠AOB : ∠COD = 3 : 2.

Hence, Option 2 is the correct option.

In the given figure, O is the center of the circle and ABC is an equilateral triangle, then ∠AOB is equal to :

1. 105°

2. 90°

3. 60°

4. 120°

Answer

Since, ABC is an equilateral triangle.

∴ AB = BC = AC.

We know that,

Equal chords subtend equal angles at the center.

∴ ∠AOB = ∠BOC = ∠COA = x (let)

We know that,

Angles around a point add to 360 °.

∴ ∠AOB + ∠BOC + ∠COA = 360°

⇒ x + x + x = 360°

⇒ 3x = 360°

⇒ x = $\dfrac{360°}{3}$ = 120°

⇒ ∠AOB = 120°.

Hence, Option 4 is the correct option.

In the given figure, O is the center of the circle and chord AB : chord CD = 5 : 3. If angle DOC = 60°; then ∠AOB is :

1. 120°

2. 75°

3. 100°

4. 80°

Answer

Ratio of the angles subtended by the chords on the center is equal to the ratio of the chords.

$\Rightarrow \dfrac{∠AOB}{∠COD} = \dfrac{AB}{CD} \\[1em] \Rightarrow \dfrac{∠AOB}{60°} = \dfrac{5}{3} \\[1em] \Rightarrow ∠AOB = \dfrac{5}{3} \times 60° = \dfrac{300°}{3} = 100°.$

Hence, Option 3 is the correct option.

In the given figure, a square is inscribed in a circle with center O. Find :

(i) ∠BOC

(ii) ∠OCB

(iii) ∠COD

(iv) ∠BOD

Is BD a diameter of the circle?

Answer

Join OA and OD.

(i) We know that,

Diagonals of a square bisect each other at 90°.

∴ ∠BOC = 90°.

Hence, ∠BOC = 90°.

(ii) From figure,

OB and OC are the radius of the circle.

In △ OBC,

⇒ OB = OC

⇒ ∠OCB = ∠OBC = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ x + x + 90° = 180°

⇒ 2x + 90° = 180°

⇒ 2x = 180° - 90°

⇒ 2x = 90°

⇒ x = $\dfrac{90°}{2}$

⇒ x = 45°

⇒ ∠OCB = 45°.

Hence, ∠OCB = 45°.

(iii) We know that,

Diagonals of a square bisect each other at 90°.

∴ ∠COD = 90°.

Hence, ∠COD = 90°.

(iv) From figure,

⇒ ∠BOD = ∠BOC + ∠COD = 90° + 90° = 180°.

Hence, ∠BOD = 180° and BD is the diameter of the circle.

In the given figure, AB is a side of a regular pentagon and BC is a side of a regular hexagon.

(i) ∠AOB

(ii) ∠BOC

(iii) ∠AOC

(iv) ∠OBA

(v) ∠OBC

(vi) ∠ABC

Answer

We know that,

The angle subtended by each side of an n-sided regular polygon at the center of circle = $\dfrac{360°}{n}$.

(i) Given,

AB is the side of the pentagon.

Angle subtended by each arm of the pentagon at the center of the circle is $\dfrac{360°}{5}$ = 72°.

Hence, ∠AOB = 72°.

(ii) Given,

BC is the side of the hexagon.

Angle subtended by each arm of the hexagon at the center of the circle is $\dfrac{360°}{6}$ = 60°.

Hence, ∠BOC = 60°.

(iii) From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 72° + 60° = 132°.

Hence, ∠AOC = 132°.

(iv) In △ OAB,

⇒ OA = OB

⇒ ∠OBA = ∠OAB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 72° = 180°

⇒ 2x + 72° = 180°

⇒ 2x = 180° - 72°

⇒ 2x = 108°

⇒ x = $\dfrac{108°}{2}$

⇒ x = 54°

⇒ ∠OBA = 54°.

Hence, ∠OBA = 54°.

(v) In △ OBC,

⇒ OC = OB

⇒ ∠OBC = ∠OCB = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + 60° = 180°

⇒ 2y + 60° = 180°

⇒ 2y = 180° - 60°

⇒ 2y = 120°

⇒ y = $\dfrac{120°}{2}$

⇒ y = 60°

⇒ ∠OBC = 60°.

Hence, ∠OBC = 60°.

(vi) From figure,

⇒ ∠ABC = ∠OBA + ∠OBC = 54° + 60° = 114°.

Hence, ∠ABC = 114°.

In the given figure, arc AB and arc BC are equal in length.

If ∠AOB = 48°, find :

(i) ∠BOC

(ii) ∠OBC

(iii) ∠AOC

(iv) ∠OAC

Answer

(i) We know that,

If two arcs are of a circle are equal, they subtend equal angles at the center.

∴ ∠BOC = ∠AOB = 48°.

Hence, ∠BOC = 48°.

(ii) Join BC.

In △ OBC,

⇒ OC = OB (Radius of same circle)

⇒ ∠OBC = ∠OCB = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ y + y + 48° = 180°

⇒ 2y + 48° = 180°

⇒ 2y = 180° - 48°

⇒ 2y = 132°

⇒ y = $\dfrac{132°}{2}$

⇒ y = 66°

⇒ ∠OBC = 66°.

Hence, ∠OBC = 66°.

(iii) From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 48° + 48° = 96°.

Hence, ∠AOC = 96°.

(iv) Join AC.

In △ AOC,

⇒ OC = OA (Radius of same circle)

⇒ ∠OAC = ∠OCA = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 96° = 180°

⇒ 2x + 96° = 180°

⇒ 2x = 180° - 96°

⇒ 2x = 84°

⇒ x = $\dfrac{84°}{2}$

⇒ x = 42°

⇒ ∠OAC = 42°.

Hence, ∠OAC = 42°.

In the given figure, the lengths of arcs AB and BC are in the ratio 3 : 2.

If ∠AOB = 96°, find :

(i) ∠BOC

(ii) ∠ABC

Answer

(i) We know that,

Ratio of the angles subtended by the chords on the center is equal to the ratio of the chords.

$\Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{96°}{∠BOC} = \dfrac{3}{2} \\[1em] \Rightarrow ∠BOC = \dfrac{2}{3} \times 96° = 2 \times 32° = 64°.$

Hence, ∠BOC = 64°.

(ii) In △ AOB,

⇒ OA = OB (Radius of same circle)

⇒ ∠OBA = ∠OAB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 96° = 180°

⇒ 2x + 96° = 180°

⇒ 2x = 180° - 96°

⇒ 2x = 84°

⇒ x = $\dfrac{84°}{2}$

⇒ x = 42°

⇒ ∠OBA = 42°.

In △ BOC,

⇒ OB = OC (Radius of same circle)

⇒ ∠OCB = ∠OBC = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OCB + ∠OBC + ∠BOC = 180°

⇒ y + y + 64° = 180°

⇒ 2y + 64° = 180°

⇒ 2y = 180° - 64°

⇒ 2y = 116°

⇒ y = $\dfrac{116°}{2}$

⇒ y = 58°

⇒ ∠OBC = 58°.

From figure,

⇒ ∠ABC = ∠OBA + ∠OBC = 42° + 58° = 100°.

Hence, ∠ABC = 100°.

In the given figure, AB = BC = DC and ∠AOB = 50°. Find :

(i) ∠AOC

(ii) ∠AOD

(iii) ∠BOD

(iv) ∠OAC

(v) ∠ODA

Answer

(i) We know that,

Equal chords subtend equal angles at the center.

Since, AB = BC = DC

∴ ∠BOC = ∠AOB = 50°.

From figure,

⇒ ∠AOC = ∠BOC + ∠AOB = 50° + 50° = 100°.

Hence, ∠AOC = 100°.

(ii) We know that,

Equal chords subtend equal angles at the center.

Since, AB = BC = DC

∴ ∠DOC = ∠AOB = 50°.

From figure,

⇒ ∠AOD = ∠AOC + ∠DOC = 100° + 50° = 150°.

Hence, ∠AOD = 150°.

(iii) From figure,

⇒ ∠BOD = ∠BOC + ∠DOC = 50° + 50° = 100°.

Hence, ∠BOD = 100°.

(iv) Join AC.

In △ AOC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OAC = ∠OCA = y (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ y + y + 100° = 180°

⇒ 2y + 100° = 180°

⇒ 2y = 180° - 100°

⇒ 2y = 80°

⇒ y = $\dfrac{80°}{2}$

⇒ y = 40°

⇒ ∠OAC = 40°.

Hence, ∠OAC = 40°.

(v) Join AD.

In △ OAD,

⇒ OA = OD (Radius of same circle)

⇒ ∠ODA = ∠OAD = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAD + ∠ODA + ∠AOD = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°

⇒ ∠ODA = 15°.

Hence, ∠OAD = 15°.

In the given figure, AB is a side of a regular hexagon and AC is a side of regular eight sided polygon. Find :

(i) ∠AOB

(ii) ∠AOC

(iii) ∠BOC

(iv) ∠OBC

Answer

We know that,

The angle subtended by each side of an n-sided regular polygon at the center of circle = $\dfrac{360°}{n}$.

(i) Given,

AB is the side of the hexagon.

Angle subtended by each arm of the hexagon at the center of the circle is $\dfrac{360°}{6}$ = 60°.

Hence, ∠AOB = 60°.

(ii) Given,

AC is the side of a regular eight sided polygon.

Angle subtended by each arm of the regular eight sided polygon at the center of the circle is $\dfrac{360°}{8}$ = 45°.

Hence, ∠AOC = 45°.

(iii) From figure,

⇒ ∠BOC = ∠AOB + ∠AOC = 60° + 45° = 105°.

Hence, ∠BOC = 105°.

(iv) In △ OBC,

⇒ OB = OC (Radius of same circle)

⇒ ∠OBC = ∠OCB = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OBC + ∠OCB + ∠BOC = 180°

⇒ x + x + 105° = 180°

⇒ 2x + 105° = 180°

⇒ 2x = 180° - 105°

⇒ 2x = 75°

⇒ x = $\dfrac{75°}{2}$

⇒ x = 37.5°

⇒ ∠OBC = 37.5°.

Hence, ∠OBC = 37.5°=37°30'.

In the given figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC.

If ∠AOB = 100°, find :

(i) ∠BOC

(ii) ∠OAC

Answer

(i) We know that,

Ratio of the angles subtended by the chords on the center is equal to the ratio of the chords.

$\Rightarrow \dfrac{∠AOB}{∠BOC} = \dfrac{AB}{BC} \\[1em] \Rightarrow \dfrac{100°}{∠BOC} = \dfrac{2}{1} \\[1em] \Rightarrow ∠BOC = \dfrac{1}{2} \times 100° = 50°.$

Hence, ∠BOC = 50°.

(ii) Join AC.

From figure,

⇒ ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.

In △ OAC,

⇒ OA = OC (Radius of same circle)

⇒ ∠OCA = ∠OAC = x (let) [Angle opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠OAC + ∠OCA + ∠AOC = 180°

⇒ x + x + 150° = 180°

⇒ 2x + 150° = 180°

⇒ 2x = 180° - 150°

⇒ 2x = 30°

⇒ x = $\dfrac{30°}{2}$

⇒ x = 15°

⇒ ∠OAC = 15°.

Hence, ∠OAC = 15°.