NCERT Solutions Class 9 Maths Chapter 5 I'm Up and Down, and Round and Round on Padhaiwise

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.1

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.1#

Theorem 1: There is a unique circle passing through three non-collinear points.

Its centre, the circumcentre, is where the perpendicular bisectors of the sides meet, because every point on the perpendicular bisector of $AB$ is equally far from $A$ and $B$.

To draw the circumcircle of a triangle: draw the perpendicular bisectors of any two sides; they meet at the circumcentre $O$. With centre $O$ and radius $OA$, draw the circle. It passes through all three vertices.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.2

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.2#

Theorem 2: Equal chords of a circle subtend equal angles at the centre.

Theorem 3: Chords of a circle that subtend equal angles at the centre are equal.

All radii of a circle are equal — this fact is used in almost every proof in this chapter.

1. Show that the triangle formed by a chord and the centre of the circle is isosceles. #

Solution

Let $AB$ be a chord of a circle with centre $C$ and radius $r$. Joining $C$ to $A$ and $B$ gives $\triangle CAB$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.3

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.3#

Theorem 4: The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.

1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord? (Hint: Use Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.) #

Solution

Given: a circle with centre $C$, a chord $AB$, and $CM \perp AB$, so $\angle CMA = \angle CMB = 90°$.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.4

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.4#

Theorem 6: Chords of a circle having the same length are all at the same distance from the centre.

Theorem 7: Chords of a circle that are equidistant from the centre have equal length.

The distance of a chord from the centre is the length of the perpendicular from the centre to the chord. By Theorem 5, this perpendicular meets the chord at its midpoint.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.5

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.5#

Theorem 8: If $AB > DE$ are two chords of a circle, then $AB$ is closer to the centre than $DE$. The longer chord is nearer the centre.

For a chord at distance $d$ from the centre of a circle of radius $r$, the perpendicular from the centre, half the chord and a radius form a right triangle.

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round Exercise Set 5.6

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.6#

Theorem 9: The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.

So angles in the same segment are equal, and the angle in a semicircle is 90°.

1. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB? #

Solution

$OA$ and $OB$ are radii, so $OA = OB = 12$ cm and $\triangle OAB$ is isosceles. Its base angles are equal, so:

NCERT Solutions for Class 9 Maths Ganita Manjari Chapter 5 I'm Up and Down, and Round and Round End-of-Chapter Exercises

Mon, 01 Jan 0001 00:00:00 +0000

NCERT Solutions for Class 9 Maths Chapter 5 — End-of-Chapter Exercises#

Results used in this exercise

The perpendicular from the centre to a chord bisects it, and the line from the centre to the midpoint of a chord is perpendicular to it (Theorems 4 and 5).

A chord at distance $d$ from the centre of a circle of radius $r$ has length $2\sqrt{r^2 - d^2}$.

Equal chords are equidistant from the centre, and the longer chord is nearer the centre (Theorems 6–8).

The angle at the centre is double the angle at the circle (Theorem 9); the angle in a semicircle is 90°.

Opposite angles of a cyclic quadrilateral add up to 180°, and the converse holds (Theorems 11 and 12).

1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord? #

Solution

Here $r = 13$ cm and $d = 5$ cm. We find the length of the chord: