Circle (वर्तुळ) MCQ Test — Class 9 Maharashtra SSC

Solution:

A segment of a circle is defined as the region bounded by a chord and its corresponding arc.

Option A describes a sector, option C describes the circumference, and option D describes a chord.

Solution:

Theorem 1: Equal chords of a circle subtend equal angles at the center. So, statement B is true.

Theorem 2: Equal chords of a circle are equidistant from the center. So, statement C is true.

Therefore, both B and C are correct statements, making option D the best choice.

Solution:

Let the chord be AB = 16 cm. The radius OA = 10 cm.

The perpendicular from the center O to the chord AB (let's call the intersection point M) bisects the chord. So, AM = AB / 2 = 16 / 2 = 8 cm.

In the right-angled triangle OMA, by Pythagoras theorem: OA² = OM² + AM².

Substitute the values: 10² = OM² + 8² => 100 = OM² + 64 => OM² = 36 => OM = 6 cm.

Solution:

Ravi's task was to prove the statement: 'If a line bisects a chord, then it is perpendicular to the chord.' Here, 'a line bisects a chord' is the given condition, and 'it is perpendicular to the chord' is the conclusion.

By starting with 'assuming the line is perpendicular', Ravi began his proof by assuming the conclusion he was supposed to reach.

This is a fundamental logical error known as circular reasoning, where the conclusion is used as a premise.

Solution:

First, find the radius using Chord P: Half chord length = 8/2 = 4 cm. Using Pythagoras theorem (radius² = distance² + (half chord)²): radius² = 3² + 4² = 9 + 16 = 25. So, radius = 5 cm.

Next, find the distance (d) of Chord Q (length 6 cm) from the center. Half chord length = 6/2 = 3 cm.

Using Pythagoras theorem: 5² = d² + 3² => 25 = d² + 9 => d² = 16 => d = 4 cm.

Since 4 cm (distance of Chord Q) is greater than 3 cm (distance of Chord P), Chord Q is farther from the center.

Solution:

Statement A is false: No circle can pass through three collinear points. The perpendicular bisectors of the segments formed by these points would be parallel and never intersect.

Statement B is true: For three non-collinear points, the perpendicular bisectors of the segments joining them will intersect at a unique point (the circumcenter), which is equidistant from all three points, thus defining a unique circle.

Statement C is false: An infinite number of circles can pass through two distinct points; their centers lie on the perpendicular bisector of the segment joining them.

Statement D is false: A unique circle cannot generally be drawn through any four points; for this to be possible, the four points must be concyclic (lie on the same circle).

Solution:

Arc AB subtends ∠AOB at the center O.

Arc AB also subtends ∠ACB at point C on the remaining part of the circle.

According to the theorem, the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. So, ∠AOB = 2 × ∠ACB.

Given ∠AOB = 80°. Therefore, 80° = 2 × ∠ACB => ∠ACB = 80° / 2 = 40°.

Solution:

Observe that both ∠CAD and ∠CBD are angles subtended by the same arc CD on the circumference of the circle.

According to the theorem, angles in the same segment of a circle are equal.

Since ∠CAD and ∠CBD are angles in the same segment (formed by arc CD), they must be equal. Given ∠CAD = 35°, therefore ∠CBD = 35°.

Solution:

AB is the diameter of the circular park.

Point C is on the boundary (circumference) of the park.

The angle ∠ACB is subtended by the diameter AB at point C on the circumference. According to a key theorem, the angle subtended by a diameter (or a semicircle) at any point on the circumference is always a right angle (90°).

Solution:

Let the radius (r) = 13 cm. Let the distance of the chord from the center (d) = 5 cm.

The perpendicular from the center to the chord bisects the chord. This forms a right-angled triangle with the radius as the hypotenuse, the distance from the center as one leg, and half the chord length as the other leg.

Let half the length of the chord be x. Using the Pythagorean theorem: r² = d² + x².

Substitute the values: 13² = 5² + x² => 169 = 25 + x² => x² = 144.

Solving for x: x = √144 = 12 cm. The full length of the chord is 2x = 2 × 12 = 24 cm.