## RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11A

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A.

**Other Exercises**

- RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A
- RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B
- RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C

**Question 1.**

**Solution:**

Let AB be a chord of a circle with centre O. OC⊥AB and OA be the radius of the circle, then

AB = 16cm, OA = 10cm .

OC ⊥ AB.

OC bisects AB at C

AC = \(\frac { 1 }{ 2 } \) AB = \(\frac { 1 }{ 2 } \) x 16 = 8cm

**Question 2.**

**Solution:**

Let AB be the chord of the circle with centre O and OC ⊥ AB, OA be the radius of the circle,

then OC = 3cm, OA = 5cm

Now in right ∆ OAC,

OA² = AC² = OC² (Pythagoras Theorem)

**Question 3.**

**Solution:**

Let AB be the chord, OA be the radius of

the circle OC ⊥ AB

then AB = 30 cm, OC = 8cm

**Question 4.**

**Solution:**

AB and CD are parallel chords of a circle with centre O.

**Question 5.**

**Solution:**

Let AB and CD be two chords of a circle with centre O.

OA and OC are the radii of the circle. OL⊥AB and OM⊥CD.

**Question 6.**

**Solution:**

In the figure, a circle with centre O, CD is its diameter AB is a chord such that CD⊥AB.

AB = 12cm, CE = 3cm.

Join OA.

∵ COD⊥AB which intersects AB at E

**Question 7.**

**Solution:**

A circle with centre O, AB is diameter which bisects chord CD at E

i.e. CE = ED = 8cm and EB = 4cm

Join OC.

Let radius of the circle = r

**Question 8.**

**Solution:**

Given : O is the centre of a circle AB is a chord and BOC is the diameter. OD⊥AB

To prove : AC || OD and AC = 20D

Proof : OD⊥AB

∵ D is midpoint of AB

**Question 9.**

**Solution:**

Given : O is the centre of the circle two

chords AB and CD intersect each other at P inside the circle. PO bisects ∠BPD.

To prove : AB = CD.

**Question 10.**

**Solution:**

Given : PQ is the diameter of the circle with centre O which is perpendicular to one chord AB and chord AB || CD.

PQ intersects AB and CD at E and F respectively

To prove : PQ⊥CD and PQ bisects CD.

**Question 11.**

**Solution:**

Two circles with centre O and O’ intersect each other.

To prove : The two circles cannot intersect each other at more than two points.

Proof : Let if opposite, the two circles intersect each other at three points P, Q and R.

Then these three points are non-collinear. But, we know that through three non- collinear points, one and only one circle can be drawn.

∵ Our supposition is wrong

Hence two circle can not intersect each other at not more than two points.

Hence proved

**Question 12.**

**Solution:**

Given : Two circles with centres O and O’ intersect each other at A and B. AB is a common chord. OO’ is joined.

AO and AO is joined.

**Question 13.**

**Solution:**

Given : Two equal circles intersect each other at P and Q.

A straight line drawn through

P, is drawn which meets the circles at A and B respectively

To prove : QA = QB

**Question 14.**

**Solution:**

Given : A circle with centre 0. AB and CD are two chords and diameter PQ bisects AB and CD at L and M

To Prove : AB || CD.

**Question 15.**

**Solution:**

Given : Two circles with centres A and B touch each other at C internally. A, B arc joined. PQ is the perpendicular bisector of AB intersecting it at L and meeting the bigger circle at P and Q respectively and radii of the circles are 5cm and 3cm. i.e. AC = 5cm,BC = 3cm

Required : To find the lenght of PQ

**Question 16.**

**Solution:**

Given : AB is a chord of a circle with centre O. AB is produced to C such that BC = OB, CO is joined and produced to meet the circle at D.

∠ ACD = y°, ∠ AOD = x°

To prove : x = 3y

**Question 17.**

**Solution:**

Given : O is the centre of a circle AB and AC are two chords such that AB = AC

OP⊥AB and OQ⊥AC.

which intersect AB and AC at M and N

respectively. PB and QC are joined.

To prove : PB = QC.

**Question 18.**

**Solution:**

Given : In a circle with centre O, BC is its diameter. AB and CD are two chords such that AB || CD.

To prove : AB = CD

Const. Draw OL⊥AB

OM⊥CD.

Proof : In ∆ OLB and ∆ OCM,

OB = OC (radii of the same circle)

∠ OLB = ∠ OMC (each 90°)

∠ OBL = ∠ OCM (alternate angles)

**Question 19.**

**Solution:**

Equilateral ∆ ABC in inscribed in a circle in which

AB = BC = CA = 9cm.

**Question 20.**

**Solution:**

Given : AB and AC are two equal chords of a circle with centre O

To Prove : O lies on the bisector of ∠ BAC

**Question 21.**

**Solution:**

Given : OPQR is a square with centre O, a circle is drawn which intersects the square at X and Y.

To Prove : Q = QY

Const. Join OX and OY

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A are helpful to complete your math homework.

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