Solutions
Practice Set 17.1
Question 1.1. In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find l(QB).
Geometric Property Used : The perpendicular drawn from the centre of a circle to its chord bisects the chord.
Since seg PQ ⊥ chord AB from the centre P, point Q is the midpoint of chord AB.
Because Q bisects the chord, the length of QB is exactly half the length of the entire chord AB.
l(AB) = 13 …. (Given)
∴ l(QB) = \(\frac{1}{2}\) × l(AB)
∴ l(QB) = \(\frac{1}{2}\) × 13 cm
∴ l(QB) = 6.5 cm
The length of segment QB is 6.5 cm.
Question 1.2. Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.
Given :
- l(OD) = 25 cm
- l(CD) = 48 cm
- seg OP ⊥ chord CD
The perpendicular drawn from the centre of the circle to the chord bisects the chord.
l(PD) = \(\frac{1}{2}\) l(CD)
∴ l(PD) = \(\frac{1}{2}\) × 48
∴ l(PD) = 24 cm
In right angled Δ OPD,
by Pythagoras theorem,
l(OD)2 = l(OP)2 + l(PD)2
∴ 252 = l(OP)2 + 242
∴ 625 = l(OP)2 + 576
∴ l(OP)2 = 625 - 576
∴ l(OP)2 = 49
∴ l(OP) = \(\sqrt{49}\)
∴ l(OP) = 7 cm
Answer : Distance of the chord from the centre is 7 cm.
Question 1.3. O is 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.
l(AB) = 24 cm
Let seg OP ⊥ chord AB
l(OP) = 9 cm
The perpendicular drawn from the centre of the circle to the chord bisects the chord.
∴ l(AP) = \(\frac{1}{2}\) l(AB)
∴ l(AP) = \(\frac{1}{2}\) × 24
∴ l(AP) = 12 cm
In right angled Δ OPA,
by Pythagoras theorem,
l(OA)2 = l(OP)2 + l(AP)2
∴ l(OA)2 = 92 + 122
∴ l(OA)2 = 81 + 144
∴ l(OA)2 = 225
∴ l(OA) = \(\sqrt{225}\)
∴ l(OA) = 15 cm.
Answer: The radius of the circle is 15 cm.
Question 1.4. 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.
Let seg AB be the chord and seg CM ⊥ chord AB as shown.
l(AM) = \(\frac{1}{2}\) l(AB) …[The perpendicular drawn from the centre of the circle to the chord bisects the chord]
∴ l (AM) = \(\frac{1}{2}\) × 12
∴ l (AM) = 6 cm
In right angled A CMA,
By Pythagoras theorem,
l(CA)2 = l(CM)2 + l(AM)2
∴ 102 = l(CM)2 + 62
∴ 100 = l(CM)2 + 36
∴ l(CM)2 = 100 - 36
∴ l(CM)2 = 64
∴ l(CM) = \(\sqrt{64}\)
∴ l(CM) = 8 cm
Answer: Distance of the chord from the centre is 8 cm.
Practice Set 17.2
Question 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
The diameters PQ and RS of a circle are perpendicular to each other at the centre C.
∴ m∠ PCS = m∠ SCQ = m∠ RCQ = m∠ PCR = 90°
m (arc PS) = m∠ PCS = 90° ... (i)
m (arc QS) = m∠ SCQ = 90° ... (ii)
m (arc RQ) = m∠ RCQ = 90° ... (iii)
m (arc PR) = m∠ PCR = 90° ... (iv)
∴ m (arc PS) = m (arc SQ) ... [From (i) and (ii)]
∴ arc PS ≅ arc SQ
From (i), (iii) and (iv) arcs RQ and PR are also congruent to arc PS
∴ arc PS ≅ arc PQ ≅ arc PR
Question 2.2. 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
(1) Seg MN is the diameter
∴ m∠ MOA + m∠ AON = 180° ... (Angles in a linear pair)
∴ m∠ MOA + (m∠ AOB + m∠ BON) = 180°
∴ 100° + m∠ AOB + 35° = 180°
∴ m∠ AOB + 135° = 180°
∴ m∠ AOB = 180 - 135
∴ m∠ AOB = 45°
Similarly, m∠ COD = 45°
(2) m (arc AB) = m/AOB=45°
m (arc CD) = m∠ COD = 45°
∴ m (arc AB) ≅ m (arc CD)
∴ arc AB ≅ arc CD ……[. Arcs are of same measures]
(3) Arc AB ≅ arc CD
∴ chord AB ≅ chord CD ... [Corresponding chords of congruent arcs are congruent]