Extra Practice Questions
Circles (R.D. Sharma Level – Class 9)
1. In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively. Calculate the distance between the chords if they are on opposite sides of the centre.
Let O be the centre. Draw perpendiculars OM to AB and ON to CD. M and N are midpoints of the chords.
AM = AB/2 = 16/2 = 8 cm. CN = CD/2 = 12/2 = 6 cm.
In right ΔOMA, OM² = OA² – AM² = 10² – 8² = 100 – 64 = 36 => OM = 6 cm.
In right ΔONC, ON² = OC² – CN² = 10² – 6² = 100 – 36 = 64 => ON = 8 cm.
Since the chords are on opposite sides, the distance between them is OM + ON = 6 + 8 = 14 cm.
Answer: The distance between the chords is 14 cm.
2. Prove that the quadrilateral formed by the angle bisectors of a cyclic quadrilateral is also cyclic.
Given: A cyclic quadrilateral ABCD. The angle bisectors of ∠A, ∠B, ∠C, ∠D form a quadrilateral PQRS.
To Prove: PQRS is a cyclic quadrilateral.
In ΔAPB, ∠P = 180° – (∠PAB + ∠PBA) = 180° – (1/2 ∠A + 1/2 ∠B).
Similarly, in ΔCRD, ∠R = 180° – (1/2 ∠C + 1/2 ∠D).
Sum of opposite angles of PQRS: ∠P + ∠R = 360° – 1/2(∠A + ∠B + ∠C + ∠D).
Since the sum of angles in quadrilateral ABCD is 360°, (∠A + ∠B + ∠C + ∠D) = 360°.
∠P + ∠R = 360° – 1/2(360°) = 360° – 180° = 180°.
Since the sum of opposite angles of quadrilateral PQRS is 180°, it is a cyclic quadrilateral. (Proved)
… (8 more questions follow)
3. The radius of a circle is 13 cm and the length of one of its chords is 10 cm. Find the distance of the chord from the centre.
Let the chord be AB and the centre be O. Draw perpendicular OM from O to AB. M is the midpoint of AB.
AM = AB/2 = 10/2 = 5 cm. OA (radius) = 13 cm.
In right-angled ΔOMA, by Pythagoras theorem:
OM² + AM² = OA² => OM² + 5² = 13² => OM² + 25 = 169.
OM² = 144 => OM = 12 cm.
Answer: The distance of the chord from the centre is 12 cm.
4. AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on the opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.
Let the radius be r. Let the perpendicular distance from the centre to AB be x, and to CD be y. So, x+y=17.
In the triangle formed by radius and half of chord AB: r² = x² + (10/2)² = x² + 25. —(1)
In the triangle formed by radius and half of chord CD: r² = y² + (24/2)² = y² + 144. —(2)
From (1) and (2), x² + 25 = y² + 144. From x+y=17, y=17-x.
x² + 25 = (17-x)² + 144 => x² + 25 = 289 – 34x + x² + 144.
25 = 433 – 34x => 34x = 408 => x = 12 cm.
Now find the radius using eq(1): r² = 12² + 25 = 144 + 25 = 169 => r = 13 cm.
Answer: The radius of the circle is 13 cm.
5. In a cyclic quadrilateral ABCD, if ∠D = 100°, find ∠B. If ∠A = (3x+10)° and ∠C = (4x+5)°, find the value of x.
In a cyclic quadrilateral, the sum of opposite angles is 180°.
Part 1: Find ∠B.
∠B + ∠D = 180° => ∠B + 100° = 180° => ∠B = 80°.
Part 2: Find x.
∠A + ∠C = 180° => (3x + 10) + (4x + 5) = 180°.
7x + 15 = 180 => 7x = 165 => x = 165/7.
Answer: ∠B = 80° and x = 165/7.
6. Prove that the angle in a semicircle is a right angle.
Given: A semicircle with diameter AB and centre O. C is any point on the semicircle.
To Prove: ∠ACB = 90°.
Proof:
The arc ADB subtends the angle ∠AOB at the centre O.
Since AOB is a straight line, ∠AOB = 180°.
The angle subtended by an arc at any point on the remaining part of the circle is half the angle subtended by it at the centre.
∠ACB = (1/2) × ∠AOB = (1/2) × 180° = 90°.
Hence, the angle in a semicircle is a right angle. (Proved)
7. O is the centre of the circle. If ∠AOC = 130°, find ∠ABC.
The angle subtended by arc AC at the centre is ∠AOC = 130°.
The angle subtended by the same arc AC at any point on the remaining part of the circle (like B) is half the angle at the centre.
This applies to the reflex angle at the centre.
Reflex ∠AOC = 360° – 130° = 230°.
∠ABC = (1/2) × Reflex ∠AOC = (1/2) × 230° = 115°.
Answer: ∠ABC = 115°.
8. ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
Given ABCD is a cyclic trapezium, so sum of opposite angles is 180°.
∠B + ∠D = 180° => 70° + ∠D = 180° => ∠D = 110°.
Since AD || BC, the consecutive interior angles sum to 180°.
∠A + ∠B = 180° => ∠A + 70° = 180° => ∠A = 110°.
Also, ∠A + ∠C = 180° (cyclic property).
110° + ∠C = 180° => ∠C = 70°.
Answer: The angles are ∠A=110°, ∠C=70°, ∠D=110°.
9. Two chords AB and CD of a circle are parallel and a line ‘l’ is the perpendicular bisector of AB. Show that ‘l’ bisects CD also.
Given: Chords AB || CD. Line ‘l’ is the perpendicular bisector of AB.
To Prove: ‘l’ bisects CD.
Proof:
Since ‘l’ is the perpendicular bisector of chord AB, it must pass through the centre O of the circle.
Now, we have a line ‘l’ passing through the centre and perpendicular to chord AB. Since AB || CD, the line ‘l’ must also be perpendicular to chord CD.
We know that a perpendicular drawn from the centre of a circle to a chord bisects the chord.
Therefore, the line ‘l’ bisects the chord CD. (Proved)
10. In a circle with centre O, an equilateral triangle ABC is inscribed. Find the measure of ∠BOC.
Since ΔABC is equilateral, all its sides are equal chords of the circle (AB = BC = CA).
We know that equal chords of a circle subtend equal angles at the centre.
Therefore, ∠AOB = ∠BOC = ∠COA.
The sum of angles at the centre is 360°.
∠AOB + ∠BOC + ∠COA = 360°
3 × ∠BOC = 360°
∠BOC = 360° / 3 = 120°.
Answer: ∠BOC = 120°.
