Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17B Solutions
EXERCISE – 17B
(Q1) The figure shows two concentric circles and AD is a chord of larger circle.
Prove that : AB = CD
Solution:
To Prove:
AB = CD
Proof:
In smallest circle,
BM = MC —– (i)
In greater circle,
AM = DM —– (ii)
Subtracting equation (i) —- (ii) we get,
AM – BM = DM – MC
AB = CD (Hence Proved)
(Q2) A straight line is drawn cutting two equal circles and passing through the midpoint M of the line joining their centres O and O’.
Prove that chords AB and CD, which are intercepted by the two circle are equal.
Solution:
To Prove:
AB = CD
Proof:
In △OPM and △O’ QM
∠OPB = ∠O’QC (Each 90°)
∠OMP = ∠O’MQ (vertically opposite angle)
OM = MO’ (∵ Given)
∴ △OPM ≅ △O’QM (By angle – angled – side) (AAS)
OP = O’Q
Then, AB = CD
{Chords are equidistant from centre are equal in length} (Hence Proved)
(Q3) M and N are the mid-points of two equal chords AB and CD respectively of a circle wit centre O. Prove that:
(i) ∠BMN = ∠DNM
(ii) ∠AMN = ∠CNM
Solution:
Given:
AB = CD
To Prove:
(i) ∠BMN = ∠DNM
(ii) ∠AMN = ∠CNM
Proof: CM = ON
∠OMN = ∠ONM —– (i) (Angle opposite to equal sides are also equal)
Also, ∠BMO = ∠DNO —- (ii) (Each 90°)
Subtracting equation (ii) —– (i),
We get,
∠BMO – ∠OMN = ∠DNO – ∠ONM
∠BMN = ∠DNM (Hence Proved)
(ii) We know, ∠ AMO = ∠CNO —– (iii)
Adding equation (iii) + (i),
∠AMO + ∠OMN = ∠CNO + ∠ONM
∠AMN = ∠CNM (Hence Proved)
(Q4) In the following figure ; P and Q are the points of intersection of two circles with centres O and O’. If straight lines APB and CQD are parallel to OO’. Prove that
(i) OO’ = 1/2 AB
(ii) AB = CD
Solution:
Given:
AB Parallel OO’ parallel CD (AB//OO’//CD)
Photo
To Prove:
(i) OO’ = 1/2 AB
(ii) AB = CD
Proof: OO’ = MN
(i) MP + PN
= 1/2 AP + 1/2 PB
OO’ = 1/2 (AP + PB)
OO’ = 1/2 AB —— (i)
Also,
OO’ = xy
OO’ = XQ + QY
OO’ = 1/2 CQ + 1/2 QD
OO’ = 1/2 (CQ + QD)
OO’ = 1/2 CD — (ii)
From equation (i) and (ii),
1/2 AB = 1/2 CD
∴ AB = CD (Hence Proved)
(Q5) Two equal chords AB and CD of a circle with centre O, intersect each other at a point P inside the circle prove that:
(i) AP = CP
(ii) BP = DP
Solution:
Given: AB = CD
To Prove:
In △ONP and △OMP,
∠N = ∠M (∵ each 90°)
ON = OM
OP = OP (Common side)
∴ △ONP ≅ △OMP (By right hand side) (RHS)
=> PN = PM (by CPCTC)
Also, PM = PN —- (i)
AB = CD
2AM = 2CN
AM = CN —– (ii)
Subtracting equation (ii) — (i),
AM – PM = CN – PN
AP = CP (Hence Proved)
Also, AB = CD
2BM = 2DN
BM = DN —– (iii)
Adding equation (iii) + (i),
BM + PM = DN + PN
BP = DP (Hence Proved)
Here is your solution of Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17B
Dear Student, I appreciate your efforts and hard work that you all had put in. Thank you for being concerned with us and I wish you for your continued success.
For more solutions, See below ⇓