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

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

**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.**

**In the given figure ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠ DBC = 60° and ∠ BAC = 40°.**

**Find (i) ∠ BCD, (ii) ∠ CAD.**

**Solution:**

In cyclic quad. ABCD, ∠ DBC = 60° and ∠BAC = 40°

∴∠ CAD and ∠ CBD are in the same segment of the circle.

∴∠ CAD = ∠ CBD or ∠ DBC

=> ∠ CAD = 60°

∴∠BAD = ∠BAC + ∠CAD

= 40° + 60° = 100°

But in cyclic quad. ABCD,

∠BAD + ∠BCD = 180°

(Sum of opposite angles)

=> 100° + ∠BCD = 180°

=> ∠BCD = 180° – 100°

∴ ∠ BCD = 80°

Hence (i) ∠BCD = 80° and

(ii) ∠CAD = 60° Ans.

**Question 2.**

**In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ∠ PSR = 150°, find ∠ RPQ**

**Solution:**

In the figure, POQ is diameter, PQRS is a cyclic quad, and ∠ PSR =150° In cyclic quad. PQRS.

∠ PSR + ∠PQR = 180°

(Sum of opposite angles)

=> 150° + ∠PQR = 180°

=> ∠PQR = 180°- 150° = 30°

=> ∠PQR =180° – 150° = 30°

Now in ∆ PQR,

∴∠ PRQ = 90° (Angle in a semicircle)

∴∠ RPQ + ∠PQR = 90°

=> ∠RPQ + 30° = 90°

=> ∠RPQ = 90° – 30° = 60° Ans.

**Question 3.**

**In the given figure, ABCD is a cyclic quadrilateral in which AB || DC.**

**If ∠BAD = 100°, find**

**(i) ∠BCD (ii) ∠ADC (iii) ∠ABC**

**Solution:**

In cyclic quad. ABCD,

AB || DC and ∠BAD = 100°

∠ ADC = ∠BAD =180°

(co-interior angles)

=> ∠ ADC + 100° = 180°

=> ∠ADC = 180° – 100° = 80°

∴ ABCD is a cyclic quadrilateral.

∴ ∠BAD + ∠BCD = 180°

=> 100° + ∠ BCD = 180°

=> ∠BCD = 180° – 100°

=> ∠BCD = 80°

Similarly ∠ABC + ∠ADC = 180°

=> ∠ABC + 80° = 180°

=> ∠ABC = 180° – 80° = 100°

Hence (i) ∠BCD = 80° (ii) ∠ADC = 80° and (iii) ∠ABC = 100° Ans.

**Question 4.**

**In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is extended to P, find ∠ PBC.**

**Solution:**

O is the centre of the circle and arc ABC subtends an angle of 130° at the centre i.e. ∠AOC = 130°. AB is produced to P

Reflex ∠AOC = 360° – 130° = 230°

Now, arc AC subtends reflex ∠ AOC at the centre and ∠ ABC at the remaining out of the circle.

**Question 5.**

**In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If ∠ABC = 92° and ∠FAE = 20°, find ∠BCD**

**Solution:**

In the figure, ABCD is a cyclic quadrilateral in which BA is produced to F and AE is drawn parallel to CD.

∠ABC = 92° and ∠FAE = 20°

ABCD is a cyclic quadrilateral.

**Question 6.**

**In the given figure, BD = DC and ∠CBD = 30°, find m(∠BAC).**

**Solution:**

In the figure, BD = DC and ∠CBD = 30°

In ∆ BCD,

BD = DC (given)

∠ BCD = ∠ CBD

(Angles opposite to equal sides)

= 30°

But ∠BCD + ∠CBD + ∠BDC = 180° (Angles of a triangle)

=> 30°+ 30°+ ∠BDC = 180°

=> 60°+ ∠BDC = 180°

=> ∠ BDC =180° – 60° = 120°

But ABDC is a cyclic quadrilateral

∠BAC + ∠BDC = 180°

=> ∠BAC + 120°= 180°

=> ∠ BAC = 180° – 120° = 60°

Hence ∠ BAC = 60° Ans.

**Question 7.**

**In the given figure, O is the centre of the given circle and measure of arc ABC is 100°. Determine ∠ ADC and ∠ ABC**

**Solution:**

(i) Arc ABC subtends ∠ AOC at the centre , and ∠ ADC at the remaining part of the circle.

∠ AOC = 2 ∠ ADC

**Question 8.**

**In die given figure, ∆ ABC is equilateral. Find (i) ∠BDC, (ii) ∠BEC.**

**Solution:**

In the figure, ABC is an equilateral triangle inscribed is a circle

Each angle is of 60°.

∠ BAC = ∠ BDC

(Angles in the same segment)

∠BDC = 60°

BECD is a cyclic quadrilateral.

∠BDC + ∠BEC = 180°

(opposite angles of cyclic quad.)

=> 60°+ ∠BEC = 180°

=> ∠BEC = 180° – 60°= 120°

Hence ∠BDC = 60° and ∠BEC = 120° Ans.

**Question 9.**

**In the adjoining figure, ABCD is a cyclic quadrilateral in which ∠ BCD =100° and ∠ABD = 50°. Find ∠ADB.**

**Solution:**

ABCD is a cyclic quadrilateral.

∠BCD + ∠BAD = 180°

(opposite angles of a cyclic quad.)

=> 100°+ ∠BAD = 180°

so ∠BAD = 180° – 100° = 80°

Now in ∆ ABD,

∠BAD + ∠ABD + ∠ADB = 180° (Angles of a triangle)

=> 80° + 50° + ∠ADB = 180°

=> 130°+ ∠ADB = 180°

=> ∠ADB = 180° – 130° = 50°

Hence, ∠ADB = 50° Ans.

**Question 10.**

**In the given figure, O is the centre of a circle and ∠ BOD = 150°. Find the values of x and y.**

**Solution:**

Arc BAD subtends ∠ BOD at the centre and ∠BCD at the remaining part of the circle.

**Question 11.**

**In the given figure, O is the centre of the circle and ∠DAB = 50° . Calculate the values of x and y.**

**Solution:**

In ∆ OAB,

OA = OB (radii of the same circle)

∠OAB = ∠OBA = 50°

and Ext ∠BOD = ∠OAB + ∠OBA

=>x° = 50° + 50° – 100°

ABCD is a cyclic quadrilateral

∠BAD + ∠BCD = 180°

(opposite angles of a cyclic quad.)

=> 50°+ y° = 180°

=> y° = 180° – 50° = 130°

Hence x = 100° and y = 130° Ans.

**Question 12.**

**In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If ∠CBF = 1300 and ∠CDE = x°, find the value of x.**

**Solution:**

Sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively.

∠CBF = 130°, ∠CDE = x.

∠CBF + ∠CBA = 180° (Linear pair)

=> 130°+ ∠CBA = 180°

=> ∠CBA = 180° – 130° = 50°

But Ext. ∠ CDE = Interior opp. ∠ CBA (In cyclic quad. ABCD)

=> x = 50° Ans.

**Question 13.**

**In the given figure, AB is a diameter of a circle with centre O and DO || CB.**

**If ∠ BCD = 120°, calculate**

**(i) ∠BAD (ii) ∠ABD**

**(iii) ∠CBD (iv) ∠ADC**

**Also, show that ∆ AOD is an equilateral triangle.**

**Solution:**

In a circle with centre O AB is its diameter and DO || CB is drawn. ∠BCD = 120°

To Find : (i) ∠BAD (ii) ABD

(iii) ∠CBD (iv) ∠ADC

(v) Show that ∆ AOD is an equilateral triangle.

(i) ABCD is a cyclic quadrilateral.

∠BCD + ∠BAD = 180°

120° + ∠BAD = 180°

**Question 14.**

**Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 6cm, BP = 2cm and PD = 2.5 cm, find CD.**

**Solution:**

AB = 6cm, BP = 2cm, DP = 2.5cm

Let CD = xcm

Two chords AB and CD

**Question 15.**

**In the given figure, O is the centre of a circle. If ∠ AOD = 140° and ∠CAB = 50°, calculate**

**(i) ∠EDB, (ii) ∠EBD.**

**Solution:**

O is the centre of the circle

∠ AOD = 140° and ∠CAB = 50°

BD is joined.

(i) ABDC is a cyclic quadrilateral.

**Question 16.**

**In the given figure, ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet in E. Prove that ∆ EBC ~ ∆ EDA.**

**Solution:**

Given : ABCD is a cyclic quadrilateral whose sides AB and DC are produced to meet each other at E.

To Prove : ∆ EBC ~ ∆ EDA

Proof : In ∆ EBC and ∆ EDA

∠ E = ∠ E (common)

∠ECB = ∠EAD

{Exterior angle of a cyclic quad, is equal to its interior opposite angle}

and ∠ EBC = ∠EDA

∆ EBC ~ ∆ EDA (AAS axiom)

Hence proved

**Question 17.**

**In the given figure, ∆ ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.**

**Solution:**

Solution Given : In an isosceles ∆ ABC, AB = AC

A circle is drawn x in such a way that it passes through B and C and intersects AB and AC at D and E respectively.

DE is joined.

To Prove : DE || BC

Proof : In ∆ ABC,

AB = AC (given)

∠ B = ∠ C (angles opposite to equal sides)

But ∠ ADE = ∠ C (Ext. angle of a cyclic quad, is equal E to its interior opposite angle)

∠ADE = ∠B

But, these are corresponding angles

DE || BC.

Hence proved.

**Question 18.**

**ABC is an isosceles triangle in which AB = AC. If D and E are midpoints of AB and AC respectively, prove that the points D, B, C, E are concyclic.**

**Solution:**

Given : ∆ ABC is an isosceles triangle in which AB = AC.

D and E are midpoints of AB and AC respectively.

DE is joined.

To Prove : D, B, C, E are concyclic.

Proof: D and E are midpoints of sides AB and AC respectively.

DE || BC

In ∆ ABC, AB = AC

∠B = ∠C

But ∠ ADE = ∠ B (alternate angles)

∠ADE =∠C

But ∠ADE is exterior angle of quad. DBCE which is equal to its interior opposite angle C.

DBCE is a cyclic quadrilateral.

Hence D, B, C, E are con cyclic.

Hence proved.

**Question 19.**

**Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.**

**Solution:**

Given : ABCD is a cyclic quadrilateral whose perpendicular bisectors l, m, n, p of the side are drawn

To prove : l, m, n and p are concurrent.

Proof : The sides AB, BC, CD and DA are the chords of the circle passing through the vertices’s of quad. A, B, C and D. and perpendicular bisectors of a chord always passes through the centre of the circle.

l,m, n and p which are the perpendicular bisectors of the sides of cyclic quadrilateral will pass through O, the same point Hence, l, m, n and p are concurrent.

Hence proved.

**Question 20.**

**Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.**

**Solution:**

Given : ABCD is a rhombus and four circles are drawn on the sides AB, BC, CD and DA as diameters. Diagonal AC and BD intersect each other at O.

**Question 21.**

**ABCD is a rectangle. Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.**

**Solution:**

Given: ABCD is a rectangle whose diagonals AC and BD intersect each other at O.

To prove : O is the centre of the circle passing through A, B, C and D

**Question 22.**

**Give a geometrical constructions for finding the fourth point lying on a circle passing through three given points without finding the centre of the circle. Justify the construction.**

**Solution:**

Construction.

(i) Let A, B and C are three points

(ii) With A as centre and BC as radius draw an arc

(iii) With centre C, and radius AB, draw another arc which intersects the first arc at D.

D is the required point.

Join BD and CD, AC and BA and CB

BC = BC (common)

AC = BD (const.)

AB = DC

∴ ∆ ABC ≅ ∆ DBC (SSS axiom)

∴ ∠BAC ≅ ∠BDC (c.p.c.t.)

But these are angles on the same sides of BC

Hence these are angles in the same segment of a circle

A, B, C, D are concyclic Hence D lies on the circle passing througtvA, B and C.

Hence proved.

**Question 23.**

**In a cyclic quadrilateral ABCD if (∠ B – ∠ D) = 60°, Show that the smaller of the two is 60°**

**Solution:**

Given : ABCD is a cylic quadrilateral (∠B – ∠D) = 60°

To prove : The small angle of the quad, is 60°

**Question 24.**

**In the given figure, ABCD is a quadrilateral in which AD = BC, ∠ ADC = ∠ BCD. Show that the points A, B, C, D lie on a circle.**

**Solution:**

Given : ABCD is a quadrilateral in which AD = BC and ∠ ADC = ∠BCD

To prove : A, B, C and D lie on a circle

**Question 25.**

**In the given figure, ∠BAD = 75°, ∠ DCF = x° and ∠DEF = y°. Find the values of x and y.**

**Solution:**

Given : In the figure, two circles intersect each other at D and C

∠BAD = 75°, ∠DCF = x° and ∠DEF = y°

**Question 26.**

**The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.**

**Solution:**

Given : ABCD is a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angle.

**Question 27.**

**In the given figure, chords AB and CD of a circle are produced to meet at E. Prove that ∆ EDB and ∆ EAC are similar.**

**Solution:**

In a circle, two chords AB and CD intersect each other at E when produced.

AD and BC are joined.

**Question 28.**

**In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that ∆ AEB is isosceles.**

**Solution:**

Given : Two parallel chords AB and CD of a circle BD and AC are joined and produced to meet at E.

**Question 29.**

**In the given figure, AB is a diameter of a circle with centre O. If ADE and CBE are straight lines, meeting at E such that ∠ BAD = 35° and ∠ BED = 25°, find (i) ∠ DBC (ii) ∠ DCB (iii) ∠ BDC.**

**Solution:**

Given : In a circle with centre O, AB is its diameter. ADE and CBE are lines meeting at E such that ∠BAD = 35° and ∠BED = 25°.

To Find : (i) ∠DBC (ii) ∠DCB (iii) ∠BDC

Solution. Join BD and AC,

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

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