NCERT Solutions Class 9 Maths Chapter 9 Circles
Circles Class 9: NCERT Solutions for Chapter 9 Maths
The Circles Class 9 solutions of the NCERT maths textbook for NCERT Solutions for Maths Chapter 9 are available below by SimplyAcad. This will ensure support to the students throughout their learning process and gain insights of the chapter through minor details missed out before. The solutions are prepared by keeping the updated syllabus in mind as prescribed by NCERT. Our subject expert team has reviewed the solutions of all the exercises in the textbook to ensure clarity for students.
Overview of the Exercises of NCERT Solutions for Circle Class 9 Maths Chapter 9
- Exercise 9.1: The first chapter focuses on the understanding of the basics, including definition and properties of a circle. Students are asked to find the radius of the circle here.
- Exercise 9.2: The second exercise deals with tangents and its properties of a circle. Students are asked to calculate the length of the tangents, the angle between the radius and tangent, and the distance of point from the centre of the circle.
- Exercise 9.3: The third exercise deals with secants of a circle.Students are asked to find the length, angle between the secant and the tangent, and the intersecting point of two secants.
- Exercise 9.4: The fourth exercise focuses on the chords of a circle. Students are asked to find the length of the chord, the angle between two chords, and its perpendicular bisector.
- Exercise 9.5: The fifth exercise focuses on the practical applications of circles. Students are asked to find the radius of a wheel and the length of the belt, followed by calculating the distance between the two parallel tracks, and the different number of revolutions completed by the wheel.
- Exercise 9.6: The final exercise is a compilation of the above ones. It covers all the concepts learnt in the chapter and its applications to solve the questions. These problems consist of calculating the areas of the shaded portion, calculating the length of the tangents, etc.
Circles Chapter 9: NCERT Solutions for Maths Class 9 Exercise 9.1
1. Prove that equal chords of congruent circles subtend equal angles at their centres.
Solution:
Given:
- Two congruent circles with centers
- Chords
To Prove:
.
Proof:
Consider the congruent circles
and
with equal chords
. Since the circles are congruent, their radii are equal, i.e.,
Now, consider the triangles
and
:
(Radii of congruent circles)
(Radii of congruent circles)
(Given, equal chords)
By the Side-Side-Side (SSS) criterion of congruence:
Since the triangles are congruent, their corresponding angles are equal. Therefore, by CPCT (Corresponding Parts of Congruent Triangles):
Hence, equal chords of congruent circles subtend equal angles at their centres
2. Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Solution:
Consider the following diagram.
Here, it is given that ∠AOB = ∠COD, i.e., they are equal angles.
Now, we will have to prove that the line segments AB and CD are equal, i.e., AB = CD.
Proof:
In triangles AOB and COD,
∠AOB = ∠COD (As given in the question.)
OA = OC and OB = OD (These are the radii of the circle.)
So, by SAS congruency, ΔAOB ≅ ΔCOD
∴ By the rule of CPCT, we have,
AB = CD (Hence, proved.)
Circles Chapter 9: NCERT Solutions for Maths Class 9 Exercise 9.2
Question 1. Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Solution:
Given:
-
- Radius of the first circle,
- Radius of the second circle,
- Distance between the centers of the circles,
d=4cm
- Radius of the first circle,
-
Determine the Length of the Common Chord:
Let
and be the centers of the two circles, and be the common chord. The distance between
and is perpendicular to the common chord , and this perpendicular bisects at .
Let
(distance between the centers)
and . (radius of the first circle) (radius of the second circle) - Apply the Pythagorean Theorem in
Δ A Q R \Delta AQR In
In
- Equate (1) and (2):
- Substitute
into equation (1):
- Find the Length of the Common Chord
:Since
Thus, the length of the common chord
is 6 cm.
Question 3. If two equal chords of a circle intersect within the circle, prove that the line
joining the point of intersection to the centre makes equal angles with the chords.
Solution:
From the question, we know the following:
(i) AB and CD are 2 chords which are intersecting at point E.
(ii) PQ is the diameter of the circle.
(iii) AB = CD.
Now, we will have to prove that ∠BEQ = ∠CEQ
For this, the following construction has to be done.
Construction:
Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE. The constructed diagram will look as follows:
Now, consider the triangles ΔOEM and ΔOEN.
Here,
(i) OM = ON [The equal chords are always equidistant from the centre.]
(ii) OE = OE [It is the common side.]
(iii) ∠OME = ∠ONE [These are the perpendiculars.]
So, by RHS congruency criterion, ΔOEM ≅ ΔOEN.
Hence, by the CPCT rule, ∠MEO = ∠NEO
∴ ∠BEQ = ∠CEQ (Hence, proved)
Question 4. If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD (see Fig).
Solution:
The given image is as follows:
First, draw a line segment from O to AD, such that OM ⊥ AD.
So, now OM is bisecting AD since OM ⊥ AD.
Therefore, AM = MD — (i)
Also, since OM ⊥ BC, OM bisects BC.
Therefore, BM = MC — (ii)
From equation (i) and equation (ii),
AM-BM = MD-MC
∴ AB = CD
Question 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?
Solution:
Let the positions of Reshma, Salma and Mandip be represented as A, B and C, respectively.
From the question, we know that AB = BC = 6 cm
So, the radius of the circle, i.e., OA = 5cm
Now, draw a perpendicular BM ⊥ AC.
Since AB = BC, ABC can be considered an isosceles triangle. M is the midpoint of AC. BM is the perpendicular bisector of AC, and thus it passes through the centre of the circle.
Now, let AM = y and OM = x
So, BM will be = (5-x).
By applying the Pythagorean theorem in ΔOAM, we get
⇒ — (i)
Again, by applying the Pythagorean theorem in ΔAMB,
⇒ — (ii)
Subtracting equation (i) from equation (ii), we get
36-25 =
Now, solving this equation, we get the value of x as
x =
Substituting the value of x in equation (i), we get
⇒
Solving it, we get the value of y as
y =
Thus,
AC = 2×AM
= 2×y
= 2× m
AC = 9.6 m
So, the distance between Reshma and Mandip is 9.6 m.
Question 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in
his hands to talk each other. Find the length of the string of each phone.
Solution:
First, draw a diagram according to the given statements. The diagram will look as follows:
Here, the positions of Ankur, Syed and David are represented as A, B and C, respectively. Since they are sitting at equal distances, the triangle ABC will form an equilateral triangle.
AD ⊥ BC is drawn. Now, AD is the median of ΔABC, and it passes through the centre O.
Also, O is the centroid of the ΔABC. OA is the radius of the triangle.
OA = 2/3 AD
Let the side of a triangle be ‘a’ metres, then BD = m.
Applying Pythagoras’ theorem in ΔABD,
OA = AD
20 m =
a =
So, the length of the string of the toy is .
Circles Chapter 9: NCERT Solutions for Maths Class 9 Exercise 9.3
Circles Class 9
Question 1. In Fig. 9.23, 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:
∠AOB=60∘
∠BOC=30∘
Here, ∠AOC=∠AOB+∠BOC
∠AOC=60∘+30∘
∠AOC=90∘
Since arc ABC makes an angle of 90∘ at the centre of the circle.
∠ADC=12∠AOC
∠ADC=12(90∘)
∠ADC=45∘
Hence, ∠ADC=45∘
Question 2.A chord AB of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the major arc and also at a point on the minor arc.
In \(\Delta OAB\),
AB = OA = OB = Radius
\(\therefore \Delta OAB\) is an equilateral triangle.
Therefore, each interior angle of this triangle will be of \(60^{\circ}\).
\(\Rightarrow \angle AOB =60^{\circ}\).
We know that angle subtended by an arc at the centre is double the angle subtended by it any point on the remaining part of the circle.
\(\angle ACB =\frac{1}{2}\angle AOB =\frac{1}{2}(60^{\circ})= 30^{\circ}\)
In cyclic quadrilateral ACBD,
\(\angle ACB + \angle ADB = 180^{\circ}\) (Opposite angles in a cyclic quadrilateral are supplementary)
\(\therefore \angle ADB = 180^{\circ}-30^{\circ}=150^{\circ}\)
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are \(30^{\circ}\) and \(150^{\circ}\) respectively
Question 3.In Fig. 9.24, ∠ PQR = 100°, where P, Q and R are point
Solution:
Consider PR as a chord of the circle.
Take any point S on the major arc of the circle.
PQRS is a cyclic quadrilateral.
∠PQR + ∠PSR = 180° (Opposite angles of a cyclic quadrilateral)
⇒ ∠PSR = 180° − 100° = 80°
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
∴ ∠POR = 2∠PSR = 2 (80°) = 160°
In ΔPOR,
OP = OR (Radii of the same circle)
∴ ∠OPR = ∠ORP (Angles opposite to equal sides of a triangle)
∠OPR + ∠ORP + ∠POR = 180° (Angle sum property of a triangle)
2 ∠OPR + 160° = 180°
2 ∠OPR = 180° − 160° = 20º
∠OPR = 10°
Question 4.In Fig. 9.25, ∠ ABC = 69°, ∠ ACB = 31°, find ∠ BDC.
Solution:
Given that ∠ABC=69∘ and ∠ACB=31∘.
Using angle sum property in △ABC, we have
∠BAC+∠ABC+∠ACB=180∘.
i.e., ∠BAC+69∘+31∘=180∘
⟹∠BAC=180∘−100∘=80∘
Now, since the angles in the same segment are equal,
∠BAC=∠BDC.
∴∠BDC=80∘
Question 5.In Fig. 9.26, 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 : Given,
\(\angle BEC = 130^\circ\). \(\angle ECD = 20^\circ\)
\(\angle BEC + \angle DEC =180^\circ\) [Linear pair]
\(130^\circ + \angle DEC =180^\circ\)
\(\angle DEC = 180^\circ – 130^\circ = 50^\circ\)
Now, in \(\Delta DEC\)
\(\angle DEC + \angle ECD + \angle CDE =180^\circ\) [Angle sum property of a triangle]
\( 50^\circ + 20^\circ + \angle CDE =180^\circ\)
\(\angle CDE = 180^\circ -70^\circ = 110^\circ\)
\(\angle CDE = \angle BAC\) [Angles in the same segment]
\(\therefore \angle BAC = 110^\circ\)
Question 6.ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \(\angle DBC = 70^{\circ}, \angle BAC \text{ is }30^{\circ}\), find \(\angle BCD\) . Further, if AB = BC, find \(\angle ECD\).
Solution :
Solution: Given
\(\angle DBC = 70^{\circ}, \angle BAC=30^{\circ}\)
For chord CD,
\(\angle CBD = \angle CAD \text{ (Angles in the same segment)} \angle CAD =70^{\circ}\)
\(\angle BAD = \angle BAC + \angle CAD = 30^{\circ} + 70^{\circ} = 100^{\circ}\)
(Opposite angles of a cyclic quadrilateral)
\(\angle BCD + 100^{\circ} = 180^{\circ}\)
In \(\Delta ABC\),
AB = BC (Given)
\(\Rightarrow \angle BCA = \angle CAB\) (Angles opposite to equal sides of a triangle)
\(\therefore \angle BCA = 30^{\circ}\)
\(\angle BCA + \angle ACD = 80^{\circ}\)
\(30^{\circ} + \angle ACD = 80^{\circ}\)
\(\Rightarrow \angle ACD = 50^{\circ}\)
\(\therefore \angle ECD = 50^{\circ}\)
Question 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:
Let ABCD be a cyclic quadrilateral having diagonals BD and AC, intersecting each other at point O.
We know that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
\(\angle BAD=\frac{1}{2}\angle BOD =\frac{180}{2}=90^{\circ}\)
(Opposite angles of a cyclic quadrilateral)
\(\angle BCD = 180^{\circ} – 90^{\circ}= 90^{\circ}\)
We know that angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
\(\angle ADC =\frac{1}{2}\angle AOC=\frac{1}{2} (180^{\circ})= 90^{\circ}\)
\(\angle ADC + \angle ABC = 180^{\circ}\) (Opposite angles of a cyclic quadrilateral)
\(90^{\circ} + \angle ABC = 180^{\circ}\)
\(\angle ABC = 90^{\circ}\)
Each interior angle of a cyclic quadrilateral is of \(90^{\circ} \). Hence, it is a rectangle.
Question 8.If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
Solution:
Consider a trapezium ABCD with AB | |CD and BC = AD.
Construction:
Draw \(AM \perp CD\) and \(BN \perp CD\).
In \(\Delta AMD and \Delta BNC\)
AD = BC (Given)
\(\angle AMD = \angle BNC\) (By construction, each angle is \(90^{\circ}\))
AM = BN (Perpendicular distance between two parallel lines is same)
\(\therefore \Delta AMD \cong \Delta BNC\) (RHS congruence rule)
\(\Rightarrow \angle ADC = \angle BCD (CPCT) … (1)\)
\(\angle BAD ~and~ \angle ADC\) are on the same side of transversal AD.
\(\angle BAD +\angle ADC = 180^{\circ} … (2)\)
\(\angle BAD + \angle BCD = 180^{\circ}\) [Using equation (1)]
This equation shows that the opposite angles are supplementary.
Therefore, ABCD is a cyclic quadrilateral.
Question 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. 9.27). Prove that ∠ ACP = ∠ QCD.
Solution:
Construction:
Join chords AP and DQ.
For chord AP,
\(\angle PBA = \angle ACP\) (Angles in the same segment) … (1)
For chord DQ,
\(\angle DBQ = \angle QCD\) (Angles in the same segment) … (2)
ABD and PBQ are line segments intersecting at B.
\(\therefore \angle PBA = \angle DBQ\) (Vertically opposite angles) … (3)
From equations (1), (2), and (3), we obtain
\(\angle ACP = \angle QCD\)
Question 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:
Consider a \(\Delta ABC\).
Two circles are drawn while taking AB and AC as the diameter.
Let they intersect each other at D.
Assume that the point of intersection ‘D’ doesn’t lie on BC and join AD.
\(\angle ADB = 90^{\circ}\) (Angle subtended by semi-circle)
\(\angle BDC =\angle ADB + \angle ADC = 90^{\circ} + 90^{\circ} =180\)
Therefore, BDC is a straight line and hence, our assumption was wrong.
Thus, Point D lies on third side BC of \(\Delta ABC\).
Question 11.ABC and ADC are two right triangles with common hypotenuse AC. Prove that
∠ CAD = ∠ CBD.
Solution:
In \(\Delta ABC\),
\(\angle ABC + \angle BCA + \angle CAB = 180^{\circ}\) (Angle sum property of a triangle)
\(\angle 90^{\circ} + \angle BCA + \angle CAB = 180^{\circ}\)
\(\therefore \angle BCA + \angle CAB = 90^{\circ} … (1)\)
In \(\Delta ADC\),
\(\angle CDA + \angle ACD + \angle DAC = 180^{\circ}\) (Angle sum property of a triangle)
\(\angle 90^{\circ} + \angle ACD + \angle DAC = 180^{\circ}\)
\(\therefore \angle ACD + \angle DAC = 90^{\circ} … (2)\)
Adding equations (1) and (2), we obtain
\((\angle BCA + \angle ACD) + (\angle CAB + \angle DAC)= 180^{\circ}\)
\(\angle BCD + \angle DAB = 180^{\circ} … (3)\)
However, it is given that \(\angle B + \angle D = 90^{\circ} + 90^{\circ} = 180^{\circ} … (4)\)
From equations (3) and (4), it can be observed that the sum of the measures of opposite angles of quadrilateral ABCD is \(180^{\circ}\).
Therefore, it is a cyclic quadrilateral.
Consider chord CD.\(\angle CAD = \angle CBD\) (Angles in the same segment)
Hence proved.
Question 12.Prove that cyclic parallelogram is rectangle.
Solution:
Let ABCD be a cyclic parallelogram.
\(\angle A + \angle C = 180^{\circ}\) (Opposite angles of a cyclic quadrilateral)
We know that opposite angles of a parallelogram are equal.
\(\therefore \angle A = \angle C~ and ~\angle B = \angle D\)
\(\angle A + \angle C = 180^{\circ}\)
Then \(\angle A + \angle A = 180^{\circ}\)
\(\therefore 2 \angle A = 180^{\circ}\)
\(\Rightarrow \angle A = \angle C= 90^{\circ}\)
Parallelogram ABCD has one of its interior angles as \(90^{\circ}\). Therefore, it is a rectangle.
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