Chapter 11: Circles (Exercise 11.1)

Chapter 11: Circles (Exercise 11.1)

Question 1:
In a circle of radius 5 cm, a chord AB is drawn at a distance of 3 cm from the center O. Find the length of the chord AB.

Solution:
* Let OM ⊥ AB be the perpendicular drawn from center O to chord AB.
* Given: Radius OA = 5 cm, Distance OM = 3 cm.

* In right-angled triangle OMA, by Pythagoras Theorem:
OA² = OM² + AM²
5² = 3² + AM²
25 = 9 + AM²
AM² = 16 ⇒ AM = 4 cm

* The perpendicular drawn from the center of a circle to a chord bisects the chord:
AB = 2 × AM = 2 × 4 = 8 cm

Question 2:
Two parallel chords AB and CD of lengths 10 cm and 24 cm respectively are drawn on opposite sides of the center of a circle of radius 13 cm. Find the distance between the two parallel chords.

Solution:
* Let O be the center and radius R = 13 cm.
* Draw OM ⊥ AB and ON ⊥ CD.

1. For chord AB = 10 cm:
AM = 10 / 2 = 5 cm.
In ΔOMA: OM = √(OA² - AM²) = √(13² - 5²) = √(169 - 25) = √144 = 12 cm.

2. For chord CD = 24 cm:
CN = 24 / 2 = 12 cm.
In ΔONC: ON = √(OC² - CN²) = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.

* Distance between chords MN = OM + ON:
MN = 12 + 5 = 17 cm.

Question 3:
Prove that equal chords of a circle subtend equal angles at the center.

Solution:
* Given: A circle with center O where chords AB = CD.
* To prove: ∠AOB = ∠COD.

* Proof:
In ΔAOB and ΔCOD:
1. OA = OC (Radii of the same circle)
2. OB = OD (Radii of the same circle)
3. AB = CD (Given equal chords)

* By SSS Congruence Rule:
ΔAOB ≅ ΔCOD

* By CPCT:
∠AOB = ∠COD
(Hence proved.)

Question 4:
If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Solution:
* Let chords AB and CD intersect at point E on diameter PQ such that ∠AEO = ∠CEO.
* Draw OL ⊥ AB and OM ⊥ CD.

* In ΔOLE and ΔOME:
1. ∠OLE = ∠OME = 90° (By construction)
2. ∠LEO = ∠MEO (Given equal angles)
3. OE = OE (Common hypotenuse)

* By AAS Congruence Rule:
ΔOLE ≅ ΔOME

* By CPCT:
OL = OM

* Since chords equidistant from the center are equal in length:
AB = CD
(Hence proved.)