Std 8 Mathematics · Chapter 17

17. Circle: Chord and Arc

Solutions to Practice Sets 17.1 and 17.2 — perpendicular from the centre to a chord, and congruent arcs and chords, with figures. Maharashtra State Board (Balbharati), Standard Eight Mathematics.

Practice Set 17.1

Properties used: The perpendicular from the centre to a chord bisects the chord. The distance of a chord from the centre is the length of this perpendicular, and radius, half-chord and distance form a right angled triangle.

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

PQ ⊥ AB, so Q is the midpoint of AB.

seg PQ ⊥ chord AB ∴ Q is the midpoint of chord AB.
l(QB) = 12 × l(AB) = 12 × 13 = 6.5

Answer: l(QB) = 6.5 cm

2The 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.

OP ⊥ CD; △OPD is right angled at P.

seg OP ⊥ chord CD ∴ l(PD) = 12 × 48 = 24 cm
In right angled △OPD, by Pythagoras theorem,
l(OP)² = l(OD)² − l(PD)² = 25² − 24² = (25 + 24)(25 − 24) = 49
∴ l(OP) = 7 cm

Answer: The chord is 7 cm from the centre.

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

OM ⊥ AB and OM = 9 cm.

Let seg OM ⊥ chord AB. Then l(MB) = 12 × 24 = 12 cm and l(OM) = 9 cm.
In right angled △OMB,
l(OB)² = l(OM)² + l(MB)² = 9² + 12² = 81 + 144 = 225
∴ l(OB) = 15 cm

Answer: The radius is 15 cm.

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

CM ⊥ XY, where XY is the chord of length 12 cm.

Let XY be the chord and seg CM ⊥ chord XY. Then l(MY) = 12 × 12 = 6 cm.
In right angled △CMY,
l(CM)² = l(CY)² − l(MY)² = 10² − 6² = 100 − 36 = 64
∴ l(CM) = 8 cm

Answer: The chord is 8 cm from the centre.

Practice Set 17.2

Properties used: the measure of a minor arc equals the measure of its central angle; arcs of equal measure are congruent; chords corresponding to congruent arcs are congruent.

1The 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.

The perpendicular diameters make four central angles of 90° each.

The diameters are perpendicular, so ∠PCS = ∠SCQ = ∠QCR = ∠RCP = 90°.
m(arc PS) = m∠PCS = 90° and m(arc SQ) = m∠SCQ = 90°.
The arcs have equal measures, so arc PS ≅ arc SQ.
Similarly m(arc PR) = m(arc RQ) = 90°.

Answer: arc PS ≅ arc SQ because each measures 90°. Also arc PS ≅ arc PR ≅ arc RQ.

2In 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: (1) m∠AOB and m∠COD (2) Show that arc AB ≅ arc CD. (3) Show that chord AB ≅ chord CD.

MN is a diameter, so ∠MON is a straight angle (180°).

(1) m∠AOB and m∠COD

MN is a diameter, so the angles on each side of it add up to 180°.
m∠AOB = 180° − (100° + 35°) = 180° − 135° = 45°
m∠COD = 180° − (100° + 35°) = 45°

Answer: m∠AOB = m∠COD = 45°

(2) Show that arc AB ≅ arc CD.

m(arc AB) = m∠AOB = 45° and m(arc CD) = m∠COD = 45°.
The measures of the arcs are equal ∴ arc AB ≅ arc CD.

(3) Show that chord AB ≅ chord CD.

arc AB ≅ arc CD (proved above).
Chords corresponding to congruent arcs are congruent ∴ chord AB ≅ chord CD.

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