## RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B.

**Other Exercises**

- RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9A
- RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9B
- RS Aggarwal Solutions Class 9 Chapter 9 Quadrilaterals and Parallelograms Ex 9C

**Question 1.**

**Solution:**

In parallelogram ABCD.

∠ A = 72°

But ∠ A = ∠ C (opposite angle of a ||gm)

∴ ∠ C = 72°

∴ AD || BC

∴ ∠ A + ∠ B = 180° (co-interior angles)

=> 72° + ∠B = 180°

=> ∠B = 180° – 72°

=> ∠B = 108°

But ∠ B = ∠ D (opposite angles of a ||gm)

∴ ∠D = 108°

Hence ∠D = 108°, ∠ C = 72° and ∠ D = 108° Ans.

**Question 2.**

**Solution:**

In || gm ABCD, BD is its diagonal

and ∠DAB = 80° and ∠DBC = 60°

∴AB || DC

∴∠DAB + ∠ADC – 180°

(co-interior angles)

=> 80° + ∠ADC = 180°

=> ∠ ADC = 180° – 80°

=> ∠ ADC = 100°

But ∠ ADB = ∠ DBC (Alternate angles)

∴∠ ADB = 60°

But ∠ ADB + ∠ CDB = 100°

(∴∠ ADC = 100°)

60° + ∠CDB = 100°

=> ∠CDB = 100° – 60° = 40°

Hence ∠CDB = 40° and ∠ ADB = 60° Ans.

**Question 3.**

**Solution:**

Given : In ||gm ABCD,

∠ A = 60° Bisectors of ∠ A and ∠ B meet DC at P.

To Prove : (i) ∠APB = 90°

(ii) AD = DP and PB = PC = BC

(iii) DC = 2AD

Proof: ∴ AD || B (opposite sides of a ||gm)

∴∠ A + ∠ B = 180° (co-interior angles)

But AP and BP are the bisectors of ∠A and ∠B

∴\(\frac { 1 }{ 2 } \)∠A + \(\frac { 1 }{ 2 } \) ∠B = 90°

=> ∠PAB + ∠PBA = 90°

But in ∆ APB,

∠PAB + ∠PBA + ∠APB = 180° (angles of a triangle)

=> 90° + ∠APB = 180°

=> ∠APB = 180° – 90° = 90°

Hence ∠APB = 90°

(ii) ∠ A + ∠ D = 180° (co-interior angles)

and ∠ A – 60°

∴ ∠D = 180° – 60° = 120°

But ∠DAP = \(\frac { 1 }{ 2 } \) ∠A = \(\frac { 1 }{ 2 } \) x 60° = 30°

∴∠DPA = 180° – (∠DAP + ∠D)

= 180° – (30° + 120°)

= 180° – 150° = 30°

∠DAP = ∠DPA (each = 30°)

Hence AD = DP (sides opposite to equal angles)

In ∆ BCP,

∠ C = 60° (opposite to ∠ A)

∠CBP = \(\frac { 1 }{ 2 } \) ∠ B = \(\frac { 1 }{ 2 } \) x 120° = 60°

But ∠CPB + ∠CBP + ∠C = 180°

(Angles of a triangle)

=> ∠CPB + 60° + 60° = 180°

=> ∠CPB + 120° = 180°

=> ∠CPB = 180° – 120° = 60°

∆ CBP is an equilateral triangle and BC = CP = BP

=> PB – PC = BC

(iii) DC = DP + PC

= AD + BC

(∴ DP = AD and PC = BC proved)

= AD + AD (∴ AD = BC opposite sides of a ||gm)

= 2AD

Hence DC = 2AD.

Hence proved.

**Question 4.**

**Solution:**

In ||gm ABCD,

AC and BD are joined

∠BAO = 35°, ∠ DAO = 40°

∠COD = 105°

∴ ∠AOB = ∠COD

(vertically opposite angles)

∴∠AOB = 105°

(i) Now in ∆ AOB,

∠ABO + ∠AOB + ∠OAB = 180°

(angles of a triangle)

=> ∠ABO + 105° + 35° = 180°

=> ∠ABO + 140° = 180°

=> ∠ABO = 180° – 140°

∠ ABO = 40°

(ii) ∴ AB || DC

∴ ∠ ABO = ∠ ODC (alternate angles)

∴ ∠ ODC = 40°

(iii) ∴ AD || BC

∴ ∠ ACB = ∠ DAO or ∠ DAC

(alternate angles)

= 40°

(iv) ∴ ∠ A + ∠ B = 180° (co-interior angles)

=> (40° + 35°) + ∠B = 180°

=> ∠B = 180° – 75° = 105°

=> ∠ CBD + ∠ABO = 105°

=> ∠CBD + 40° = 105°

=> ∠CBD = 105° – 40° = 65°

Hence ∠ CBD = 65° Ans.

**Question 5.**

**Solution:**

In ||gm ABCD

( ∠A = (2x + 25)° and ∠ B = (3x – 5)°

∴AD || BC (opposite sides of parallelogram)

∴∠ A + ∠B = 180° (co-interior angles)

=> 2x + 25° + 3x – 5° = 180°

=> 5x + 20° = 180°

=> 5x = 180° – 20°

=> 5x = 160° => x = \(\frac { { 160 }^{ o } }{ 5 } \) = 32°

∴x = 32°

Now ∠A = 2x + 25° = 2 x 32° + 25°

= 64° + 25° = 89°

∠B = 3x – 5 = 3 x 32° – 5°

= 96° – 5° = 91°

∠ C = ∠ A (∴ opposite angles of ||gm)

= 89°

Similarly ∠B = ∠D

∠D = 91°

Hence ∠ A = 89°, ∠ B = 91°, ∠ C = 89°, ∠D = 91° Ans.

**Question 6.**

**Solution:**

Let ∠A and ∠B of a ||gm ABCD are adjacent angles.

∠A + ∠B = 180°

Let ∠B = x

Then ∠ A = \(\frac { 4 }{ 5 } \) x

∴ x + \(\frac { 4 }{ 5 } \) x = 180°

\(\frac { 9 }{ 5 } \) x = 180°

=>\(\frac { { 180 }^{ o }\quad X\quad 5 }{ 9 } \) = 100°

∴ ∠A = \(\frac { 4 }{ 5 } \) x 100° = 80°

and ∠B = 100°

But ∠ C = ∠ A and ∠ D = ∠ B

(opposite angles of a || gm)

∴∠C = 80°, and ∠D = 100°

Hence ∠A = 80°, ∠B = 100°, ∠C = 80° and ∠ D = 100° Ans.

**Question 7.**

**Solution:**

Let the smallest angle ∠ A and the other angle ∠ B

Let ∠ A = x

Then ∠ B = 2x – 30°

But ∠ A + ∠ B = 180° (co-interior angles)

∴x + 2x – 30° = 180°

=> 3x = 180° + 30° = 210°

=> x = \(\frac { { 210 }^{ o } }{ 3 } \) = 70°

∴ ∠ A = 70°

and ∠ B = 2x – 30° = 2 x 70° – 30°

= 140° – 30° = 110°

But ∠C = ∠ A and ∠D = ∠B

(opposite angles of a ||gm)

∠C = 70° and ∠D = 110°

Hence ∠A = 70°, ∠B = 110°, ∠C = 70° and ∠D = 110° Ans.

**Question 8.**

**Solution:**

In ||gm ABCD,

AB = 9.5 cm and perimeter = 30 cm

=> AB + BC + CD + DA = 30cm

=> AB + BC + AB + BC = 30 cm

( ∴ AB = CD and BC – DA opposite sides)

=> 2(AB + BC) = 30cm

=> AB + BC = 15cm

=> 9 5cm + BC = 15cm

∴BC = 15cm – 9.5cm = 5.5cm

Hence AB = 9.5cm, BC = 5.5cm,

CD = 9.5cm and DA = 5.5cm Ans.

**Question 9.**

**Solution:**

ABCD is a rhombus

AB = BC = CD = DA

(i)∴ AB || DC

∴ ∠ B + ∠ C = 180° (co-interior angles)

=> 110° + ∠C = 180°

=> ∠C = 180° – 110° = 70°

**Question 10.**

**Solution:**

In a rhombus,

Diagonals bisect each other at right angles

∴ AC and BC bisect each other at O at right angles.

But AC = 24 cm and BD = 18 cm

**Question 11.**

**Solution:**

Let ABCD he the rhombus whose diagonal are AC and BD which bisect each other at right angles at O.

**Question 12.**

**Solution:**

ABCD is a rectangle whose diagonals AC and BD bisect each other at O.

**Question 13.**

**Solution:**

ABCD is a square. A line CX cuts AB at X and diagonal BD at O such that

∠ COD = 80° and ∠ OXA = x°

∴∠ BOX = ∠ COD

(vertically opposite angles)

∴∠BOX = 80°

∴Diagonal BD bisects ∠ B and ∠ D

∴ ∠ABO or ∠ABD = ∠ ADO or ∠ ADB

∴ ∠OBA or ∠OBX = 45°

Now in ∆ OBX,

Ext. ∠ OXA = ∠ BOX + ∠ OBX

=>x° = 80° + 45° = 125° Ans.

**Question 14.**

**Solution:**

Given : In ||gm ABCD, AC is joined. AL ⊥ BD and CM ⊥ BD

To prove :

(i) ∆ ALD ≅ ∆ CMB

(ii) AL = CM

Proof : In ∆ ALD and ∆ BMC

AD = BC (opposite sides of ||gm)

∠L = ∠M (each 90°)

∠ ADL = ∠ CBM (Alternate angles)

∴ ∆ ALD ≅ ∆ BMC. (AAS axiom)

∴ or A ALD ≅ A CMB.

AL = CM (c.p.c.t.) Hence proved.

**Question 15.**

**Solution:**

Given : In ∆ ABCD, bisectors of ∠A and ∠B meet each other at P.

To prove : ∠APB = 90°

Proof : AD || BC

∠A + ∠B = .180° (co-interior angles)

PA and PB are the bisectors of ∠ A and ∠B

**Question 16.**

**Solution:**

In ||gm ABCD,

P and Q are the points on AD and BC respectively such that AP = \(\frac { 1 }{ 3 } \) AD and CQ = \(\frac { 1 }{ 3 } \) BC

**Question 17.**

**Solution:**

Given : In ||gm ABCD, diagonals AC and BD bisect each other at O.

A line segment EOF is drawn, which meet AB at E and DC at F.

To -prove : OE = OF

**Question 18.**

**Solution:**

Given : ABCD is a ||gm.

AB is produced to E. Such that AB = BE

DE is joined which intersects BC in O.

To prove : ED bisects BC i.e. BO = OC

**Question 19.**

**Solution:**

Given : In ||gm ABCD, E is the midpoint BC

DE is joined and produced to meet AB on producing at F.

**Question 20.**

**Solution:**

Given : ∆ ABC and lines are drawn through A, B and C parallel to respectively BC, CA and AB forming ∆ PQR.

To prove : BC = \(\frac { 1 }{ 2 } \) QR

**Question 21.**

**Solution:**

Given : In ∆ ABC, parallel lines are drawn through A, B and C respectively to the sides BC, CA and AB intersecting each other at P, Q and R.

Hope given RS Aggarwal Class 9 Solutions Chapter 9 Quadrilaterals and Parallelograms Ex 9B are helpful to complete your math homework.

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