Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17D Solutions
EXERCISE – 17D
(Q1) The radius of a circle is 13cm and the length of one of its chords is 24cm. Find the distance of the chord from the centres.
Solution:
In △ACO, by Pythagoras theorem,
(hypotenuse)2 = (Base)2 + (height)2
(OA)2 = (OC)2 + (AC)2
(13)2 = OC2 + (12)2
169 = OC2 + 144
169 – 144 = OC2
25 = OC2
Taking square root on both sides,
√25 = √OC2
5cm = OC
(Q2) Prove that equal chords of congruent circles subtend equal angles at their centre.
Solution:
In △AOB and △CO’ D
AO = CO’ (radii of circle)
AB = CD (∵ Given)
OB = O’D (radii of circle)
Then, △AOB ≅ △CO’D (by side-side-side) (SSS)
= ∠AOB = ∠CO’D (Hence Proved)
(Q5) Given two equal chords AB and CD of a circle, with centre O intersecting each other at point P. Prove that:
(i) AP = CP
(ii) BP = DP
Solution:
To Prove:
(i) AP = CP
(ii) BP = DP
Proof:
AB = CD —- (iii)
AE + BE = CF + DF
AE + AE = CF + CF
2AE = 2CF
AE = CF — (i)
In △OEP and △OFP
∠OEP = ∠OFP (each 90°)
OE = OF (Distance of the chord from the centre given)
OP = OP (common)
∴ △OEP ≅ △OFP (by Right hand side R.H.S)
Then, EP = PF —- (ii) (by CPCTC)
Adding equation (i) and (ii),
AE + EP = CF + PF
AP = CP —– (iv) (Hence Proved)
Subtracting equation (iii) — (iv),
AB – AP = CD – CP
BP = DP (Hence Proved)
Here is your solution of Selina Concise Class 9 Maths Chapter 17 Circles Exercise 17D
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