NCERT Solutions for Exercise 10.5 Class 9 Maths Chapter 10 - Circles

# NCERT Solutions for Exercise 10.5 Class 9 Maths Chapter 10 - Circles

Edited By Vishal kumar | Updated on Oct 10, 2023 09:41 AM IST

## NCERT Solutions for Class 9 Maths Chapter 10 Circles Ex 10.5- Download Free PDF

NCERT Solutions Class 9 Maths Exercise 10.5 – Chapter 10 in Class 9 Maths is all about circles. 9th class maths exercise 10.5 answers is a part of it and it has questions about things like tangents and chords in circles. These class 9 maths chapter 10 exercise 10.5 solutions are like your helpful guides. They explain the answers to these questions in clear and simple language. You can even download them as a free PDF and use them offline.

Along with NCERT book Class 9 Maths chapter, 10 exercise 10.5 the following exercises are also present.

**Please be aware that this chapter has been renumbered as Chapter 9 in the CBSE Syllabus for the academic year 2023-24.

## Circles Class 9 Chapter 10 Exercise: 10.5

$\angle$ AOC = $\angle$ AOB + $\angle$ BOC= $60 \degree+30 \degree=90 \degree$

$\angle$ AOC = 2 $\angle$ ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)

$\angle ADC=\frac{1}{2}\angle AOC$

$\Rightarrow \angle ADC=\frac{1}{2}90 \degree= 45 \degree$

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and $\angle$ ACB.

Solution :

In $\triangle$ OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

$\Rightarrow$ ABC is a equilateral triangle.

So, $\angle$ AOB = $60 \degree$

$\angle$ AOB = 2 $\angle$ ADB

$\Rightarrow \angle ADB=\frac{1}{2}\angle AOB$

$\Rightarrow \angle ADB=\frac{1}{2}60 \degree=30$

ACBD is a cyclic quadrilateral .

So, $\angle$ ACB+ $\angle$ ADB = $180 \degree$

$\Rightarrow \angle ACB+30 \degree= 180 \degree$

$\Rightarrow \angle ACB= 180 \degree-30 \degree=150 \degree$

Construction: Join PS and RS.

So, $\angle$ PSR + $\angle$ PQR = $180 \degree$

$\Rightarrow \angle PSR+100 \degree=180 \degree$

$\Rightarrow \angle PSR=180 \degree-100 \degree=80 \degree$

Here, $\angle$ POR = 2 $\angle$ PSR

$\Rightarrow \angle POR=2\times 80 \degree=160 \degree$

In $\triangle$ OPR ,

$\angle$ ORP = $\angle$ OPR (the angles opposite to equal sides)

In $\triangle$ OPR ,

$\angle$ OPR+ $\angle$ ORP+ $\angle$ POR= $180 \degree$

$\Rightarrow 2\angle OPR+160 \degree= 180 \degree$

$\Rightarrow 2\angle OPR=180 \degree- 160 \degree$

$\Rightarrow 2\angle OPR=20 \degree$

$\Rightarrow \angle OPR=10 \degree$

In $\triangle$ ABC,

$\angle$ A+ $\angle$ ABC+ $\angle$ ACB= $180\degree$

$\Rightarrow \angle A+69 \degree+31 \degree=180\degree$

$\Rightarrow \angle A+100 \degree=180\degree$

$\Rightarrow \angle A=180 \degree-100\degree$

$\Rightarrow \angle A=80 \degree$

$\angle$ A = $\angle$ BDC = $80 \degree$ (Angles in same segment)

.

$\angle$ DEC+ $\angle$ BEC = $180 \degree$ (linear pairs)

$\Rightarrow$ $\angle$ DEC+ $130 \degree$ = $180 \degree$ ( $\angle$ BEC = $130 \degree$ )

$\Rightarrow$ $\angle$ DEC = $180 \degree$ - $130 \degree$

$\Rightarrow$ $\angle$ DEC = $50 \degree$

In $\triangle$ DEC,

$\angle$ D+ $\angle$ DEC+ $\angle$ DCE = $180 \degree$

$\Rightarrow \angle D+50 \degree+20 \degree= 180 \degree$

$\Rightarrow \angle D+70 \degree= 180 \degree$

$\Rightarrow \angle D= 180 \degree-70 \degree=110 \degree$

$\angle$ D = $\angle$ BAC (angles in same segment are equal )

$\angle$ BAC = $110 \degree$

$\angle BDC=\angle BAC$ (angles in the same segment are equal )

$\angle BDC= 30 \degree$

In $\triangle BDC,$

$\angle BCD+\angle BDC+\angle DBC= 180 \degree$

$\Rightarrow \angle BCD+30 \degree+70 \degree= 180 \degree$

$\Rightarrow \angle BCD+100 \degree= 180 \degree$

$\Rightarrow \angle BCD=180 \degree- 100 \degree=80 \degree$

If AB = BC ,then

$\angle BCA=\angle BAC$

$\Rightarrow \angle BCA=30 \degree$

Here, $\angle ECD+\angle BCE=\angle BCD$

$\Rightarrow \angle ECD+30 \degree=80 \degree$

$\Rightarrow \angle ECD=80 \degree-30 \degree=50 \degree$

AC is the diameter of the circle.

Thus, $\angle ADC=90 \degree$ and $\angle ABC=90 \degree$ ............................1(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, $\angle BAD=90 \degree$ and $\angle BCD=90 \degree$ ............................2(Angle in a semi-circle is a right angle)

From 1 and 2, we get

$\angle BCD=\angle ADC=\angle ABC=\angle BAD =90 \degree$

Hence, ABCD is a rectangle.

Given: ABCD is a trapezium.

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of parallelogram )

so, BE = BC

In $\triangle$ EBC,

BE = BC (Proved above )

Thus, $\angle C = \angle 2$ ...........1(angles opposite to equal sides )

$\angle A= \angle 1$ ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

$\angle 1+\angle 2=180 \degree$ (linear pair)

$\Rightarrow \angle A+\angle C=180 \degree$

Thus, ABED is a cyclic quadrilateral.

$\angle ABP=\angle QBD$ ................1(vertically opposite angles)

$\angle ACP=\angle ABP$ ..................2(Angles in the same segment are equal)

$\angle QBD=\angle QCD$ .................3(angles in the same segment are equal)

From 1,2,3 ,we get

$\angle ACP=\angle QCD$

Given: circles are drawn taking two sides of a triangle as diameters.

Proof: AB is the diameter of the circle and $\angle$ ADB is formed in a semi-circle.

$\angle$ ADB = $90 \degree$ ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and $\angle$ ADC is formed in a semi-circle.

$\angle$ ADC = $90 \degree$ ........................2(angle in a semi-circle)

From 1 and 2, we have

$\angle$ ADB+ $\angle$ ADC= $90 \degree$ + $90 \degree$ = $180 \degree$

$\angle$ ADB and $\angle$ ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : $\small \angle CAD =\angle CBD$

Proof :

Triangle ABC and ADC are on common base BC and $\angle$ BAC = $\angle$ BDC.

Thus, point A, B, C, D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

$\angle$ CAD = $\angle$ CBD (Angles in the same segment are equal)

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

$\angle A + \angle C = 180 \degree$ .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

$\angle A = \angle C$ ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

$\angle A + \angle A = 180 \degree$

$\Rightarrow 2\angle A = 180 \degree$

$\Rightarrow \angle A = 90 \degree$

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

This exercise 10.5 class 9 maths includes some important points like When two circle chords are equal, their corresponding arcs are congruent, and when two arcs are congruent, their corresponding chords are equal.

A circle's congruent arcs (or equal arcs) subtend equal angles at the centre.

The angle of an arc at its centre is double that of any point on the circle's remainder.

The above theorem establishes the link between the angles subtended by an object at its centre and at a point on the circle.

NCERT Solutions for Class 9 Maths Exercise 10.5 uses the concepts-

If all four vertices of a quadrilateral ABCD lie on a circle, it is called cyclic.

The sum of any pair of opposite angles of a cyclic quadrilateral is 180°.

Theorem 12: If a quadrilateral's sum of opposite angles is 180°, the quadrilateral is cyclic.

Also, Read| Circles Class 9 Notes

Benefits of NCERT Solutions for Class 9 Maths Exercise 10

• Equal Chords and Their Distances from the Center is the subject of exercise 10.5 Class 9 Maths.

• Theorems relating to equal chords in circles are introduced in chapter 10 exercise 10.5 of Class 9 Maths.

• Understanding the concepts from chapter 10 exercise 10.5 in Class 9 Math will help us understand the theorems of equal chords and their distance.

## Key Features of Class 9 Maths Chapter 10 Exercise 10.5

1. Comprehensive Coverage: 9th class maths exercise 10.5 answers is a comprehensive solution that covers various concepts related to circles, including tangents, secants, and chords.

2. Diverse Problem Set: class 9 maths ex 10.5 offers a diverse set of problems with varying levels of complexity, allowing students to practice and strengthen their understanding of circle properties.

3. Step-by-Step Solutions: Solutions to the problems in ex 10.5 class 9 are presented in a step-by-step format. This helps students follow the solution process and understand the concepts involved.

4. Conceptual Clarity: The primary objective of this exercise is to help students develop a clear understanding of circle theorems related to tangents, secants, and chords.

5. PDF Availability: Solutions for class 9 ex 10.5 are often available in PDF format. This enables students to download and access them offline for convenient studying.

6. Aligned with Syllabus: Exercise 10.5 aligns with the prescribed syllabus, covering all relevant topics and concepts related to circles in Class 9 Maths.

Also, See

## Subject Wise NCERT Exemplar Solutions

The quadrilateral with all of its vertices on the same circle is known as a cyclic quadrilateral.

2. What are the theorems for cyclic quadrilaterals?

If the sum of opposite angles in a quadrilateral is 180 degrees, the quadrilateral is said to be cyclic.

3. What is the angle subtended by diameter in the semicircle?

Diameter subtends right angle.

4. When a circle is drawn, it divides the plane into how many parts?

A circle divides the plane into 3 parts.

5. What is the condition for two circles to be congruent?

Circles should have the same radius.

6. What is the number of circles that can pass through 3 non-collinear points?

Only one circle can pass through 3 noncollinear points.

7. What is the relation between the diameter and radius?

The diameter is double the radius.

8. Define segment in circle?

A segment is defined as the region between an arc and the chord of the circle.

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