NCERT Solutions Class 9 Maths Chapter 10 — Circles

Solution:

Step 1: A segment of a circle is defined as the region bounded by a chord and its corresponding arc.

Step 2: Option A describes a sector, option C describes the circumference, and option D describes a chord.

Solution:

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

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

Step 3: Therefore, both B and C are correct statements, making option D the best choice.

Solution:

Step 1: Let the chord be AB = 16 cm. The radius OA = 10 cm.

Step 2: 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.

Step 3: In the right-angled triangle OMA, by Pythagoras theorem: OA² = OM² + AM².

Step 4: Substitute the values: 10² = OM² + 8² => 100 = OM² + 64 => OM² = 36 => OM = 6 cm.

Solution:

Step 1: 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.

Step 2: By starting with 'assuming the line is perpendicular', Ravi began his proof by assuming the conclusion he was supposed to reach.

Step 3: This is a fundamental logical error known as circular reasoning, where the conclusion is used as a premise.

Solution:

Step 1: 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.

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

Step 3: Using Pythagoras theorem: 5² = d² + 3² => 25 = d² + 9 => d² = 16 => d = 4 cm.

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