Maharashtra Board Textbook Solutions for Standard Eight

Chapter 17 – Circle : Chord and Arc

Practice set 17.1

1. In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⟂ chord AB, then find l(QB).

IMG 20230307 080608 Chapter 17 – Circle - Chord and Arc

Given:

In a circle with centre P, 

l(AB) = 13 cm

seg PQ ⟂ chord AB

 

To find:

l(QB)

 

Solution:

We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.

 

∴ l(QB) = \(\large \frac {1}{2}\) × l(AB)

∴ l(QB) = \(\large \frac {1}{2}\) × 13

∴ l(QB) = 6.5 cm

 

Ans: l(QB) = 6.5 cm

2. Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if the length of the chord is 48 cm.

IMG 20230307 080632 Chapter 17 – Circle - Chord and Arc

Given:

In a circle with centre O, 

l(OD) = 25 cm

l(CD) = 48 cm

 

To find:

l(OP)

 

Solution:

We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.

 

∴ l(PD) = \(\large \frac {1}{2}\) × l(CD)

∴ l(PD) = \(\large \frac {1}{2}\) × 48

∴ l(PD) = 24 cm

 

In △OPD, 

∠OPD = 90° …[Given]

∴ OD² = OP² + PD² …[By Pythagoras theorem]

∴ 25² = OP² + 24²

∴ 625 = OP² + 576

∴ 625 – 576 = OP²

∴ OP² = 49

∴ \( \sqrt {OP²}\) = \( \sqrt {49}\) …[By Taking square root of both sides] 

∴ OP = 7 cm

 

Ans: Distance of the chord from the centre of the circle is 7 cm.

3. O is the centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the circle. 

IMG 20230307 080705 Chapter 17 – Circle - Chord and Arc

Given:

In a circle with centre O, 

l(AB) = 10 cm

Distance of chord from the centre of the circle = 9 cm

 

To find:

Radius

 

Solution:

Join seg OA which is the radius of the circle.

 

We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.

 

∴ l(AM) = \(\large \frac {1}{2}\) × l(AB)

∴ l(AM) = \(\large \frac {1}{2}\) × 24

∴ l(AM) = 12 cm

 

In △OMA, 

∠OMA = 90° …[Given]

∴ OA² = OM² + AM² …[By Pythagoras theorem]

∴ OA² = 9² + 12²

∴ OA² = 81 + 144

∴ OA² = 225

∴ \( \sqrt {OA²}\) = \( \sqrt {225}\) …[By Taking square root of both sides] 

∴ OP = 15 cm

 

Ans: Radius of the circle is 15 cm.

4. C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.

IMG 20230307 080549 Chapter 17 – Circle - Chord and Arc

Given:

In a circle with centre C, 

Radius = 10 cm

Chord = 12 cm

 

To find:

Distance of chord from the centre of the circle

 

Solution:

Let seg AB be the chord of the circle with centre C. 

 

Draw seg CD ⊥ chord AB.

 

We know that the perpendicular drawn from the centre of the circle to the chord bisects the chord.

 

∴ l(AD) = \(\large \frac {1}{2}\) × l(AB)

∴ l(AD) = \(\large \frac {1}{2}\) × 12

∴ l(AD) = 6 cm

 

In △ADC, 

∠ADC = 90° …[Given]

∴ AC² = AD² + DC² …[By Pythagoras theorem]

∴ 10² = 6² + DC²

∴ 100 = 36 + DC²

∴ 100 – 36 = DC²

∴ DC² = 64

∴ \( \sqrt {DC²}\) = \( \sqrt {64}\) …[By Taking square root of both sides] 

∴ DC = 8 cm

 

Ans: Distance of the chord from the centre of the circle is 8 cm.

Practice set 17.2

1. The diameters PQ and RS of the circle with centre C are perpendicular to each other at C. State, why arc PS and arc SQ are congruent. Write the other arcs which are congruent to arc PS.

IMG 20230307 080858 Chapter 17 – Circle - Chord and Arc

Given:

In a circle with centre C, 

Diameter PQ ⊥ Diameter RS

 

To find:

Why arc PS ≅ arc SQ

Other arcs which are congruent to arc PS

 

Solution:

Diameter PQ ⊥ Diameter RS …[Given]  

∴ m∠PCS = m∠SCQ = m∠PCR = m∠RCQ = 90°  

 

We know that the measure of the angle subtended at the centre by an arc is the measure of the arc.

 

∴ m∠PCS = m(arc PS)

∵ m∠PCS = 90° 

∴ m(arc PS) = 90° …(i)  

 

∴ m∠SCQ = m(arc SQ)  

∵ m∠SCQ = 90° 

∴ m(arc SQ) = 90°

 

From (i) and (ii)

∴ m(arc PS) ≅ m(arc SQ) …[If the measures of two arcs of a circle are same, then the two arcs are congruent]  

 

Now, 

∴ m∠PCR = m(arc PR) 

∵ m∠PCR = 90° 

∴ m(arc PR) = 90° …(iii)  

 

∴ m∠RCQ = m(arc RQ) 

∵ m∠RCQ = 90° 

∴ m(arc RQ) = 90° …(iv)  

 

From (i), (ii), (iii) and (iv)

arc PS ≅ arc PR ≅ arc RQ ≅ arc SQ …[If the measures of two arcs of a circle are same, then the two arcs are congruent] 

 

Ans: arc PS ≅ arc PR ≅ arc RQ ≅ arc SQ

2. In the adjoining figure O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure. Hence find the following

(1) m ∠AOB and m ∠COD

(2) Show that arc AB ≅ arc CD.

(3) Show that chord AB ≅ chord CD

IMG 20230307 081000 Chapter 17 – Circle - Chord and Arc

Given:

In a circle with centre O, 

MN is the diameter

m∠AOM = 100⁰

m∠MOD = 100⁰

m∠BON = 35⁰

m∠NOC = 35⁰

 

To find:

m∠AOB and m∠COD

 

To prove:

arc AB ≅ arc CD

chord AB ≅ chord CD

 

Solution:  

m∠AOM + m∠AON = 180° …[Angles in a linear pair]  

∴ m∠AOM + (m∠AOB + m∠BON) = 180° …[Angle addition property]  

∴ 100° + m∠AOB + 35° = 180° …[Given]  

∴ m∠AOB + 135° = 180°  

∴ m∠AOB = 180° – 135°  

∴ m∠AOB = 45° …(i) 

 

Also, 

m∠DOM + m∠DON = 180° …[Angles in a linear pair]  

∴ m∠DOM + (m∠COD + m∠CON) = 180° …[Angle addition property]  

∴ 100° + m∠COD + 35° = 180° …[Given]  

∴ m∠COD + 135° = 180°  

∴ m∠COD = 180° – 135°  

∴ m∠COD = 45° …(ii)

 

Now,

∴ m∠AOB = m(arc AB) 

∵ m∠AOB = 90° 

∴ m(arc AB) = 90° …(iii)

 

∴ m∠DOC = m(arc DC) 

∵ m∠DOC = 90° 

∴ m(arc DC) = 90° …(iv)

 

∴ m(arc AB) = m(arc DC) …[From (iii) and (iv)]  

∴ arc AB ≅ arc CD …[If the measures of two arcs of a circle are same, then the two arcs are congruent] 

 

And,

chord AB ≅ chord CD ….[The chords corresponding to congruent arcs are congruent]

 

Hence Proved

 

Ans: m∠AOB = 45° and m∠COD = 45°.