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NCERT exemplar Class 9 Maths solutions chapter 10 provides the circle’s properties, which are crucial and applicable in higher Mathematics. The solutions delivered for NCERT exemplar Class 9 Maths chapter 10 solutions are prepared by our experts to provide precise answers while attempting NCERT Class 9 Maths questions. These NCERT exemplar Class 9 Maths chapter 10 solutions helps the students to understand the concepts in a gradual manner leading to high proficiency in solving the problems based on circles. The CBSE Syllabus for Class 9 is followed for these NCERT exemplar Class 9 Maths solutions chapter 10.
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Question:1
AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the center of the circle is:
(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm
Answer:
(D) 8 cmQuestion:2
In Fig. 10.3, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:
Fig. 10.3
(A) 2 cm
(B) 3 cm
(C) 4 cm
(D) 5 cm
Answer:
(A) 2 cmQuestion:3
If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm
Answer:
(C) 10 cmQuestion:4
In Fig.10.4, if , then is equal to:
Answer:
(B)
Solution:
Given,
We know that,
Angle subtended at the center by an arc is twice the angle subtended by it at any part of the circle.
Hence,
Therefore option (B) is correct.
Question:5
In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then is equal to:
(A) 30º
(B) 60º
(C) 90º
(D) 45º
Answer:
(D)Question:6
In Fig. 10.6, if , then is equal to:
Fig. 10.6
(A) 50º
(B) 40º
(C) 60º
(D) 70°
Answer:
(A) 50ºQuestion:7
In Fig. 10.7, if , then is equal to:
Fig. 10.7
(A) 60º
(B) 50º
(C) 70º
(D) 80º
Answer:
(C) 70°Question:8
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ADC = 140º, then BAC is equal to:
(A) 80º
(B) 50º
(C) 40º
(D) 30º
Answer:
(B) 50°Question:9
In Fig. 10.8, BC is a diameter of the circle and BAO = 60º. Then ADC is equal to:
Fig. 10.8
(A) 30º
(B) 45º
(C) 60º
(D) 120º
Answer:
(C)Question:10
In Fig. 10.9, AOB = 90º and ABC = 30º, then CAO is equal to:
(A) 30º
(B) 45º
(C) 90º
(D) 60º
Answer:
(D)Question:1
Answer:
TrueQuestion:2
Answer:
FalseQuestion:3
Answer:
TrueQuestion:4
Answer: FalseQuestion:5
Answer:
TrueQuestion:6
Answer:
TrueQuestion:7
Answer:
False.Question:8
Answer:
TrueQuestion:9
Answer:
TrueQuestion:10
Answer:
TrueQuestion:1
If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.
Answer:
Given that arcs AXB and CYD are congruent arcs of a circleQuestion:2
Answer:
Given that the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and QQuestion:3
Answer:
Given: A, B, C are 3 points on a circleQuestion:4
Answer:
Given: AB and AC are two equal chords of the circle.Question:5
Answer:
Given: M and N are midpoints of chords AB and CD respectively.Question:6
Answer:
Given: A circle passing through points B, C, D and ABCD is a quadrilateral having centre at A.Question:7
O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that BOD = A.
Answer:
Given: O is the circumcentre of the triangle ABC and D is the mid-point of the base BC.Question:8
Answer:
Given that on a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides.Question:9
Answer:
60°Question:10
Answer:
Given: In DABC, MB AC and CN AB.Question:11
Answer:
Given: A line is drawn parallel to the base of an isosceles triangle to intersect its equal sidesQuestion:12
Answer:
Given: A pair of opposite sides of a cyclic quadrilateral are equal.Question:13
The circumcenter of the triangle ABC is O. Prove that OBC + BAC = 90º.
Answer:
Given that the circumcenter of the triangle ABC is O. So O will be the radius of the circle passing through the points A, B, CQuestion:14
Answer:
30^{o}Question:16
In Fig.10.14, ACB = 40º. Find OAB.
Answer:
50°Question:17
A quadrilateral ABCD is inscribed in a circle such that AB is diameter and ADC = 130º. Find BAC.
Answer:
40°Question:18
Answer:
Two circles with centres O and O′ intersect at two points A and B.Question:19
Answer:
270°Question:20
In Fig. 10.16, OAB = 30º and OCB = 57º. Find BOC and AOC.
Answer:
AOC = 54° and BOC = 66°Question:1
Answer:
Given: Two equal chords of a circle intersectQuestion:2
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
Answer:
Given: non-parallel sides of a trapezium are equalQuestion:3
Answer:
In ABC, P, Q and R are the midpoints of the sides BC, CA and AB respectivelyQuestion:4
Answer:
Given: ABCD is a parallelogram. A circle whose centre O passes through A, B has intersected AD at P and BC at Q.Question:5
Answer:
Given: Angle bisector of any angle of a triangle and perpendicular bisector of the opposite side intersect on the circumcircle of the triangle.Question:6
Answer:
Given: In circle AYDZBWCX two chords AB and CD intersect at right anglesQuestion:7
Answer:
Let DABC is an equilateral triangle inscribed in a circle with centre O.Question:8
Answer:
Given: AB and CD are two chords of a circle intersecting each other at point EQuestion:9
Answer:
Given: Bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and QQuestion:10
Answer:
A circle with radius cm is divided into two segments by a chord of length 2 cm.Question:11
Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD
Answer:
Given: Two equals chords AB and CD of a circle intersect at a point P.Question:12
Answer:
AB and AC are two chords of a circle of radius r such that AB = 2AC.Question:13
In Fig. 10.20, O is the centre of the circle, BCO = 30°. Find x and y.
Answer:
x = 30° and y = 15°Question:14
In Fig. 10.21, O is the centre of the circle, BD = OD and CD AB. Find CAB.
Answer:
30°Topics covered in NCERT exemplar Class 9 Maths solutions chapter 10 deals with the understanding of:
These Class 9 Maths NCERT exemplar chapter 10 solutions provide a basic knowledge of circles and are useful for Class 10 and 11 with Mathematics as a subject. The Class 9 Maths NCERT exemplar solutions chapter 10 Circles are highly detailed and provide a step-by-step solution to the problems, which helps clarify the queries of students in an extensive manner. These problems and solutions are enough to solve other books such as Mathematics Pearson Class 9, NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths, et cetera.
Students can download/view these solutions by clicking on NCERT exemplar Class 9 Maths solutions chapter 10 pdf download to access the pdf version of the chapter’s solutions while attempting NCERT exemplar Class 9 Maths chapter 10 in an offline environment.
Chapter No. | Chapter Name |
Chapter 1 | |
Chapter 2 | |
Chapter 3 | |
Chapter 4 | |
Chapter 5 | |
Chapter 6 | |
Chapter 7 | |
Chapter 8 | |
Chapter 9 | |
Chapter 10 | |
Chapter 11 | |
Chapter 12 | |
Chapter 13 | |
Chapter 14 | |
Chapter 15 |
No, it is not always possible to form a circle passing through four points given in any plane.
For three non-Collinear points, a unique circle can be formed but it is not necessary that it will pass through the fourth point.
We know that diameter subtends 90° at any point of the circle.
Therefore, the hypotenuse of the given right angle triangle will be diameter. The length of diameter will be the length of the hypotenuse.
The longest chord of any circle will be its diameter. Hence, its length will be twice the radius that is 2R.
Generally, you can expect 2-3 questions under the segment of concise answer type or long answer type questions from Circles. The NCERT exemplar Class 9 Maths solutions chapter 10 can help you score well in the exam.
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