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

NCERT Solutions for Class 9 Maths Chapter 10 â€“ 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 pattern has been included in this. Solving the problems, taking these solutions as reference material, will help students to score well in the first and second term exam.

### 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 â€“ Circles Exercise 10.5

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

Solution:

It is given that,

âˆ AOC = âˆ AOB+âˆ BOC

So, âˆ AOC = 60Â°+30Â°

âˆ´ âˆ AOC = 90Â°

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,

âˆ ADC = (Â½)âˆ AOC

= (Â½)Ã— 90Â° = 45Â°

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 the Î”OAB. Here,

AB = OA = OB = radius of the circle.

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

âˆ´ âˆ AOC = 60Â°

And, âˆ ACB = Â½ âˆ AOB

So, âˆ ACB = Â½ Ã— 60Â° = 30Â°

Now, since ACBD is a cyclic quadrilateral,

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

So, âˆ ADB = 180Â°-30Â° = 150Â°

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

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

Solution:

Since 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 âˆ POR = 2Ã—âˆ PQR

We know the values of angle PQR as 100Â°

So, âˆ POR = 2Ã—100Â° = 200Â°

âˆ´ âˆ POR = 360Â°-200Â° = 160Â°

Now, in Î”OPR,

OP and OR are the radii of the circle

So, OP = OR

Also, âˆ OPR = âˆ ORP

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

So,

âˆ POR+âˆ OPR+âˆ ORP = 180Â°

âˆ OPR+âˆ OPR = 180Â°-160Â°

As âˆ OPR = âˆ ORP

2âˆ OPR = 20Â°

Thus, âˆ OPR = 10Â°

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

Solution:

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

âˆ BAC = âˆ BDC

Now in the in Î”ABC, the sum of all the interior angles will be 180Â°

So, âˆ ABC+âˆ BAC+âˆ ACB = 180Â°

Now, by putting the values,

âˆ BAC = 180Â°-69Â°-31Â°

So, âˆ BAC = 80Â°

âˆ´ âˆ BDC =Â 80Â°

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 âˆ  BEC = 130Â° and âˆ  ECD = 20Â°. Find BAC.

Solution:

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

So,

âˆ  BAC = âˆ  CDE

Now, by using the exterior angles property of the triangle

In Î”CDE we get,

âˆ  CEB = âˆ  CDE+âˆ  DCE

We know that âˆ  DCE is equal to 20Â°

So, âˆ  CDE = 110Â°

âˆ  BAC and âˆ  CDE are equal

âˆ´ âˆ  BAC = 110Â°

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

Solution:

Consider the following diagram.

Consider the chord CD,

We know that angles in the same segment are equal.

So, âˆ  CBD = âˆ  CAD

âˆ´ âˆ  CAD = 70Â°

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

So, âˆ  BAD = âˆ  BAC+âˆ  CAD

= 30Â°+70Â°

âˆ´ âˆ  BAD = 100Â°

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

So,

âˆ  BCD+âˆ  BAD = 180Â°

It is known that âˆ  BAD = 100Â°

So, âˆ  BCD = 80Â°

Now consider the Î”ABC.

Here, it is given that AB = BC

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

âˆ  BCA = 30Â°

also, âˆ  BCD = 80Â°

âˆ  BCA +âˆ  ACD = 80Â°

Thus, âˆ  ACD = 50Â° and âˆ  ECD = 50Â°

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 center 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, âˆ  ABC = âˆ  BCD = âˆ  CDA = âˆ  DAB = 90Â°

So, as each internal angle is 90Â°, 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 âˆ  ACP = âˆ  QCD.

Solution:

Construction:

Join the chords AP and DQ.

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

So, âˆ  PBA = âˆ  ACP â€” (i)

Similarly for chord DQ,

âˆ  DBQ = âˆ  QCD â€” (ii)

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

At B, the vertically opposite angles will be equal.

âˆ´ âˆ  PBA = âˆ  DBQ â€” (iii)

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

âˆ  ACP = âˆ  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 diameter as AB and AC respectively.

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

Proof:

We know that angle in the semi-circle are equal

So, âˆ  ADB = âˆ  ADC = 90Â°

Hence, âˆ  ADB+âˆ  ADC = 180Â°

âˆ´ âˆ  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 âˆ  CAD = âˆ CBD.

Solution:

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

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

Since, âˆ  ABC and âˆ  ADC are 90Â°, 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 center O.

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

âˆ´ âˆ  CAD = âˆ  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 as below; Angle Subtended by an Arc of a Circle
Theorem 1: The angle subtended by an arc at the center 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 center 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 pair of opposite angles of a cyclic quadrilateral is 180Âº.
Theorem 2: If the sum of a pair of opposite angles of a quadrilateral is 180Âº, 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 first and second term 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.

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1. Prakash kosre

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