NCERT Solutions for Class 9 Maths Exercise 10.5 Chapter 10 Circles

*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 9.

NCERT Solutions for Class 9 Maths Chapter 10 – Circles Exercise 10.5 is provided here, in downloadable PDF format. Click on the specified link given below to download it. These solutions have been designed by our Maths experts, to make each student understand the concepts in a better way. The answers given for each question in NCERT solutions for 9th Maths subject are as per the NCERT syllabus and guidelines. No other out-of-syllabus question has been included in this. Solving the problems, and taking these solutions as reference material, will help students to score well in the board exam.

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

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Access Other Exercise Solutions of Class 9 Maths Chapter 10 – Circles

Exercise 10.1 Solutions 2 Question (2 Short)
Exercise 10.2 Solutions 2 Question (2 long)
Exercise 10.3 Solutions 3 Question (3 long)
Exercise 10.4 Solutions 6 Question (6 long)
Exercise 10.6 Solutions 10 Questions (10 long)

Access Answers to NCERT Class 9 Maths Chapter 10 – Circles Exercise 10.5

1. In Fig. 10.36, A,B and C are three points on a circle with centre O such that ∠BOC = 30° and ∠AOB = 60°. If D is a point on the circle other than the arc ABC, find ∠ADC.

Solution:

It is given that,

∠AOC = ∠AOB+∠BOC

So, ∠AOC = 60°+30°

∴ ∠AOC = 90°

It is known that an angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.

So,

∠ADC = (½)∠AOC

= (½)× 90° = 45°

2. A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Solution:

Here, the chord AB is equal to the radius of the circle. In the above diagram, OA and OB are the two radii of the circle.

Now, consider ΔOAB. Here,

AB = OA = OB = radius of the circle.

So, it can be said that ΔOAB has all equal sides and thus, it is an equilateral triangle.

∴ ∠AOC = 60°

And, ∠ACB = ½ ∠AOB

So, ∠ACB = ½ × 60° = 30°

Now, since ACBD is a cyclic quadrilateral,

∠ADB +∠ACB = 180° (Since they are the opposite angles of a cyclic quadrilateral)

So, ∠ADB = 180°-30° = 150°

So, the angles subtended by the chord at a point on the minor arc and also at a point on the major arc are 150° and 30°, respectively.

3. In Fig. 10.37, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

Solution:

The angle which is subtended by an arc at the centre of the circle is double the angle subtended by that arc at any point on the remaining part of the circle.

So, the reflex ∠POR = 2×∠PQR

We know the values of angle PQR as 100°

So, ∠POR = 2×100° = 200°

∴ ∠POR = 360°-200° = 160°

Now, in ΔOPR,

OP and OR are the radii of the circle

So, OP = OR

Also, ∠OPR = ∠ORP

Now, we know the sum of the angles in a triangle is equal to 180 degrees

So,

∠POR+∠OPR+∠ORP = 180°

∠OPR+∠OPR = 180°-160°

As ∠OPR = ∠ORP

2∠OPR = 20°

Thus, ∠OPR = 10°

4. In Fig. 10.38, ∠ABC = 69°, ∠ACB = 31°, find ∠BDC.

Solution:

We know that angles in the segment of the circle are equal, so,

∠BAC = ∠BDC

Now in the ΔABC, the sum of all the interior angles will be 180°

So, ∠ABC+∠BAC+∠ACB = 180°

Now, by putting the values,

∠BAC = 180°-69°-31°

So, ∠BAC = 80°

∴ ∠BDC = 80°

5. In Fig. 10.39, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠ BEC = 130° and ∠ ECD = 20°. Find BAC.

Solution:

We know that the angles in the segment of the circle are equal.

So,

∠ BAC = ∠ CDE

Now, by using the exterior angles property of the triangle

In ΔCDE we get,

∠ CEB = ∠ CDE+∠ DCE

We know that ∠ DCE is equal to 20°

So, ∠ CDE = 110°

∠ BAC and ∠ CDE are equal

∴ ∠ BAC = 110°

6. ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠ DBC = 70°, ∠ BAC is 30°, find ∠ BCD. Further, if AB = BC, find ∠ ECD.

Solution:

Consider the following diagram.

Consider the chord CD,

We know that angles in the same segment are equal.

So, ∠ CBD = ∠ CAD

∴ ∠ CAD = 70°

Now, ∠ BAD will be equal to the sum of angles BAC and CAD.

So, ∠ BAD = ∠ BAC+∠ CAD

= 30°+70°

∴ ∠ BAD = 100°

We know that the opposite angles of a cyclic quadrilateral sum up to 180 degrees.

So,

∠ BCD+∠ BAD = 180°

It is known that ∠ BAD = 100°

So, ∠ BCD = 80°

Now consider the ΔABC.

Here, it is given that AB = BC

Also, ∠ BCA = ∠ CAB (They are the angles opposite to equal sides of a triangle)

∠ BCA = 30°

also, ∠ BCD = 80°

∠ BCA +∠ ACD = 80°

Thus, ∠ ACD = 50° and ∠ ECD = 50°

7. If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:

Draw a cyclic quadrilateral ABCD inside a circle with centre O such that its diagonal AC and BD are two diameters of the circle.

We know that the angles in the semi-circle are equal.

So, ∠ ABC = ∠ BCD = ∠ CDA = ∠ DAB = 90°

So, as each internal angle is 90°, it can be said that the quadrilateral ABCD is a rectangle.

8. If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:

9. Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40). Prove that ∠ ACP = ∠ QCD.

Solution:

Construction:

Join the chords AP and DQ.

For chord AP, we know that angles in the same segment are equal.

So, ∠ PBA = ∠ ACP — (i)

Similarly, for chord DQ,

∠ DBQ = ∠ QCD — (ii)

It is known that ABD and PBQ are two line segments which intersect at B.

At B, the vertically opposite angles will be equal.

∴ ∠ PBA = ∠ DBQ — (iii)

From equation (i), equation (ii) and equation (iii) we get,

∠ ACP = ∠ QCD

10. If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:

First, draw a triangle ABC and then two circles having diameters AB and AC, respectively.

We will have to prove now that D lies on BC and BDC is a straight line.

Proof:

We know that angles in the semi-circle are equal

So, ∠ ADB = ∠ ADC = 90°

Hence, ∠ ADB+∠ ADC = 180°

∴ ∠ BDC is straight line.

So, it can be said that D lies on the line BC.

11. ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠ CAD = ∠CBD.

Solution:

We know that AC is the common hypotenuse and ∠ B = ∠ D = 90°.

Now, it has to be proven that ∠ CAD = ∠ CBD

Since ∠ ABC and ∠ ADC are 90°, it can be said that they lie in the semi-circle.

So, triangles ABC and ADC are in the semi-circle, and the points A, B, C and D are concyclic.

Hence, CD is the chord of the circle with centre O.

We know that the angles which are in the same segment of the circle are equal.

∴ ∠ CAD = ∠ CBD

12. Prove that a cyclic parallelogram is a rectangle.

Solution:

It is given that ABCD is a cyclic parallelogram, and we will have to prove that ABCD is a rectangle.

Proof:

Thus, ABCD is a rectangle.

Exercise 10.5 has twelve questions based on the topics and theorems given below.

Angle Subtended by an Arc of a Circle
Theorem 1: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the left part of the circle.
Theorem 2: Chords which are equal in distance from the centre of a circle are also equal in length.
Theorem 3: Angles in the same segment of a circle are equal.
Theorem 4: If a line segment joining two points subtends equal angles at two other points lying on the same side of the line containing the line segment, the four points lie on a circle.

Cyclic Quadrilaterals
Theorem 1: The sum of both pairs of opposite angles of a cyclic quadrilateral is 180º.
Theorem 2: If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.

The questions in this exercise have long answers based on constructions and proofs. Solve each exercise problems for chapter 10 of Maths Class 9 here with detailed answers. Also, get our advanced learning materials, such as notes and tips and tricks, to prepare for CBSE exams. For all the classes from 6 to 12, NCERT solutions are provided here, chapter-wise and exercise-wise, in PDFs. Download and learn offline as well.