Important questions with solutions for class 9 Maths Chapter 10 (Circles) are provided as per **NCERT** book and **CBSE** syllabus 2019-2020. These questions have been provided by our subject experts to help students score good marks in their final **exam 2020**. Also, there are some extra questions given at the below section to have a better practice and revision.

The chapter circles will include concepts of circumscribed and inscribed and questions based on them. Practice all chapters of class 9 Maths important questions to score excellent marks in this subject.

**Also Check:**

- Important 2 Marks Questions for CBSE 9th Maths
- Important 3 Marks Questions for CBSE 9th Maths
- Important 4 Marks Questions for CBSE 9th Maths

## Important Questions & Solutions For Class 9 Maths Chapter 10 (Circles)

**Q.1.Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90Â° â€“ (Â½)A, 90Â° â€“ (Â½)B and 90Â° â€“ (Â½)C.**

Solution:

Consider the following diagram:

Here, ABC is inscribed in a circle with center O and the bisectors of âˆ A, âˆ B and âˆ C intersect the circumcircle at D, E and F respectively.

Now, join DE, EF and FD

As angles in the same segment are equal, so,

âˆ FDA = âˆ FCA â€”â€”â€”â€”-(i)

âˆ FDA = âˆ EBA â€”â€”â€”â€”-(i)

Adding equations (i) and (ii) we have,

âˆ FDA + âˆ EDA = âˆ FCA + âˆ EBA

Or, âˆ FDE = âˆ FCA + âˆ EBA = (Â½)âˆ C + (Â½)âˆ B

We know, âˆ A + âˆ B + âˆ C = 180Â°

So, âˆ FDE = (Â½)[âˆ C + âˆ B] = (Â½)[180Â° â€“ âˆ A]

â‡’ âˆ FDE = [90 â€“ (âˆ A/2)]

In a similar way,

âˆ FED = [90 â€“ (âˆ B/2)]

And,

âˆ EFD = [90 â€“ (âˆ C/2)]

**Q.2.In any triangle ABC, if the angle bisector of âˆ A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.**

Solution:

Consider this diagram:

Here, join BE and CE.

Now, since AE is the bisector of âˆ BAC,

âˆ BAE = âˆ CAE

Also,

âˆ´ arc BE = arc EC

This implies chord BE = chord EC

Now, for triangles Î”BDE and Î”CDE,

DE = DE (It is the common side)

BD = CD (It is given in the question)

BE = CE (Already proved)

So, by SSS congruency, Î”BDE â‰Œ Î”CDE.

Thus, âˆ´âˆ BDE = âˆ CDE

We know, âˆ BDE = âˆ CDE = 180Â°

Or, âˆ BDE = âˆ CDE = 90Â°

âˆ´ DE âŠ¥BC (hence proved).

**Q.3: Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.**

Solution:

To prove: A circle drawn with Q as centre, will pass through A, B and O (i.e. QA = QB = QO)

Since all sides of a rhombus are equal,

AB = DC

Now, multiply (Â½) on both sides

(Â½)AB = (Â½)DC

So, AQ = DP

â‡’ BQ = DP

Since Q is the midpoint of AB,

AQ= BQ

Similarly,

RA = SB

Again, as PQ is drawn parallel to AD,

RA = QO

Now, as AQ = BQ and RA = QO we have,

QA = QB = QO (hence proved).

**Q.4: Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6, find the radius of the circle.**

Solution:

Here, OM âŠ¥ AB and ON âŠ¥ CD. is drawn and OB and OD are joined.

As we know, AB bisects BM as the perpendicular from the centre bisects the chord.

Since AB = 5 so,

BM = AB/2

Similarly, ND = CD/2 = 11/2

Now, let ON be x.

So, OM = 6âˆ’ x.

Consider Î”MOB,

OB^{2} = OM^{2} + MB^{2}

Or,

OB^{2} = 36 + x^{2} – 12x + 25/4 â€¦â€¦(1)

Consider Î”NOD,

OD^{2} = ON^{2} + ND^{2}

Or,

OD^{2} = x^{2}+121/4 â€¦â€¦â€¦.(2)

We know, OB = OD (radii)

From eq. (1) and eq. (2) we have;

36 + x^{2} -12x + 25/4 = x^{2} + 121/4

12x = 36 + 25/4 – 121/4

12x = (144 + 25 -121)/4

12x = 48/4 = 12

x = 1

Now, from eq. (2) we have,

OD2 = 11 + (121/4)

Or OD = (5/2) Ã— âˆš5

**Q.5: If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lies on the third side.**

Solution:

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

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

Proof:

As we know, angle in the semi-circle are equal

So, âˆ ADB = âˆ ADC = 90Â°

Hence, âˆ ADB + âˆ ADC = 180Â°

âˆ´ âˆ BDC is a straight line.

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

**Q.6: If the non-parallel sides of a trapezium are equal, prove that it is cyclic.**

Solution:

Contruction-Consider a trapezium ABCD with AB||CD and BC = AD.

Draw AM âŠ¥CD and BN âŠ¥ CD

In âˆ†AMD and âˆ†BNC;

AD = BC (Given)

âˆ AMD = âˆ BNC (90Â°)

AM =BN (perpendiculars between parallel lines)

âˆ†AMD = âˆ†BNC (By RHS congruency)

âˆ†ADC = âˆ†BCD (By CPCT rule) â€¦â€¦.(i)

âˆ BAD and âˆ ADC are on the same side of transversal AD.

âˆ BAD + âˆ ADC = 180Â° â€¦â€¦(ii)

âˆ BAD + âˆ BCD = 180Â° (by equation (i))

Since, the opposite angles are supplementary, therefore, ABCD is a cyclic quadrilateral.

**Q.7: 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.

https://cdn1.byjus.com/wp-content/uploads/2020/03/Class-9-Maths-Chapter-10-Circles-01.png

Consider the chord CD,

As we know, 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°

As we know, the opposite angles of a cyclic quadrilateral sums up to 180 degrees.

So,

âˆ BCD + âˆ BAD = 180Â°

Since, âˆ BAD = 100Â°

So, âˆ BCD = 80Â°

Now consider the Î”ABC.

Here, it is given that AB = BC

Also, âˆ BCA = âˆ CAB (Angles opposite to equal sides of a triangle)

âˆ BCA = 30Â°

also, âˆ BCD = 80Â°

âˆ BCA + âˆ ACD = 80Â°

So, âˆ ACD = 50Â° and,

âˆ ECD = 50Â°

**Q.8: In Figure, âˆ ABC = 69Â°, âˆ ACB = 31Â°, find âˆ BDC.**

Solution:

As we know, angles in the segment of the circle are equal so,

âˆ BAC = âˆ BDC

Now in the In Î”ABC, 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° â€“ 69Â° â€“ 31Â°

So, âˆ BAC = 80Â°

**Q.9: In Figure, 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 = 2 Ã— âˆ PQR

We know the values of angle PQR as 100Â°

So, âˆ POR = 2 Ã— 100Â° = 200Â°

âˆ´ âˆ POR = 360Â° â€“ 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°

â‡’ âˆ OPR + âˆ OPR = 180Â° â€“ 160Â°

As âˆ OPR = âˆ ORP

â‡’ 2âˆ OPR = 20Â°

Thus, âˆ OPR = 10Â°

**Q.10: 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 âŠ¥ BC is drawn. Now, AD is median of Î”ABC and it passes through the centre O.

Also, O is the centroid of the Î”ABC. OA is the radius of the triangle.

OA = 2/3 AD

Let the side of a triangle a metres then BD = a/2 m.

Applying Pythagoras theorem in Î”ABD,

AB^{2} = BD^{2} + AD^{2}

â‡’ AD^{2} = AB^{2} â€“ BD^{2}

â‡’ AD^{2} = a^{2} â€“ (a/2)^{2}

â‡’ AD^{2} = 3a2/4

â‡’ AD = âˆš3a/2

OA = 2/3 AD

â‡’ 20 m = 2/3 Ã— âˆš3a/2

â‡’ a = 20âˆš3 m

So, the length of the string of the toy is 20âˆš3 m.

**Q.11: 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 have the following conditions:

(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 âˆ BEQ = âˆ CEQ

For this, the following construction has to be done:

Construction:

Draw two perpendiculars are drawn as OM âŠ¥ AB and ON âŠ¥ CD. Now, join OE. The constructed diagram will look as follows:

Now, consider the triangles Î”OEM and Î”OEN.

Here,

(i) OM = ON [Since the equal chords are always equidistant from the centre]

(ii) OE = OE [It is the common side]

(iii) âˆ OME = âˆ ONE [These are the perpendiculars]

So, by RHS similarity criterion, Î”OEM â‰… Î”OEN.

Hence, by CPCT rule, âˆ MEO = âˆ NEO

âˆ´ âˆ BEQ = âˆ CEQ (Hence proved).

**Q.12: If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.**

Solution:

It is given that two circles intersect each other at P and Q.

To prove:

OOâ€™ is a perpendicular bisector of PQ.

Proof:

Triangle Î”POOâ€™ and Î”QOOâ€™ are similar by SSS congruency since

OP = OQ and Oâ€™P = OQ (Since they are also the radii)

OOâ€™ = OOâ€™ (It is the common side)

So, It can be said that Î”POOâ€™ â‰… Î”QOOâ€™

âˆ´ âˆ POOâ€™ = âˆ QOOâ€™ â€” (i)

Even triangles Î”POR and Î”QOR are similar by SAS congruency as

OP = OQ (Radii)

âˆ POR = âˆ QOR (As âˆ POOâ€™ = âˆ QOOâ€™)

OR = OR (Common arm)

So, Î”POR â‰… Î”QOR

âˆ´ âˆ PRO = âˆ QRO

Also, As we know,

âˆ PRO + âˆ QRO = 180Â°

Hence, âˆ PRO = âˆ QRO = 180Â°/2 = 90Â°

So, OOâ€™ is the perpendicular bisector of PQ.

**Q.13: 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 âˆ AOB = âˆ 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,

âˆ AOB = âˆ COD (as given in the question)

OA = OC and OB = OD ((these are the radii of the circle)

So, by SAS congruency, Î”AOB â‰… Î”COD.

âˆ´ By the rule of CPCT, AB = CD. (Hence proved).

### Extra Questions (CBSE) For Class 9 Maths Chapter 10

- Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
- 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.
- If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
- 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
- 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?
- 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.Â
- 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.
- In Figure, âˆ ABC = 69Â°, âˆ ACB = 31Â°, find âˆ BDC.Â
- Prove that a cyclic parallelogram is a rectangle.
- Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that âˆ ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
- ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE, = AD.
- AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.