## RS Aggarwal Solutions for Class 9 Maths Exercise 10B PDF

RS Aggarwal solutions contain answers to all the questions which are based on the current CBSE syllabus. Each question is solved based on the evaluation process which is followed by the exam board. The students can use the RS Aggarwal Solutions to prepare for the exams more methodically. The solutions are extremely accurate and detailed in simple language to help students gain more marks. Solving the exercise wise problems and examples in RS Aggarwal textbook improve the problem solving abilities in students. RS Aggarwal Solutions for Class 9 Maths Chapter 10 Quadrilaterals Exercise 10B are provided here.

## RS Aggarwal Solutions for Class 9 Chapter 10: Quadrilaterals Exercise 10B Download PDF

## Access RS Aggarwal Solutions for Class 9 Chapter 10: Quadrilaterals Exercise 10B

## Exercise 10(B)

**1. In the adjoining figure, ABCD is a parallelogram in which âˆ A = 72 ^{o}. Calculate âˆ B, âˆ C and âˆ D.**

**Solution:**

We know that in a parallelogram opposite angles are equal

So we get

âˆ A = âˆ C = 72^{o}

We that that in a parallelogram the sum of all the angles is 360^{o}

It can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values in the above equation

72^{o} + âˆ B + 72^{o} + âˆ D = 360^{o}

We know that âˆ A = âˆ C and âˆ B = âˆ D

So we can write it as

2 âˆ B + 144^{o} = 360^{o}

On further calculation

2 âˆ B = 360^{o} – 144^{o}

By subtraction

2 âˆ B = 216^{o}

By division

âˆ B = 108^{o}

Therefore, âˆ B = 108^{o}, âˆ C = 72^{o} and âˆ D = 108^{o}.

**2. In the adjoining figure, ABCD is a parallelogram in which âˆ DAB = 80 ^{o} and âˆ DBC = 60^{o}. Calculate âˆ CDB and âˆ ADB.**

**Solution:**

It is given that ABCD is a parallelogram in which âˆ DAB = 80^{o} and âˆ DBC = 60^{o}

We know that opposite angles are equal in parallelogram

So we get

âˆ C = âˆ A = 80^{o}

From the figure we know that AD || BC and BD is a transversal

We know that âˆ ADB and âˆ DBC are alternate angles

So we get

âˆ ADB = âˆ DBC = 60^{o}

Consider â–³ ABD

Using the sum property of triangle

âˆ A + âˆ ADB + âˆ ABD = 180^{o}

By substituting values in the above equation

80^{o} + 60^{o} + âˆ ABD = 180^{o}

On further calculation

âˆ ABD = 180^{o} – 80^{o} – 60^{o}

By subtraction

âˆ ABD = 180^{o} – 140^{o}

So we get

âˆ ABD = 40^{o}

It can be written as

âˆ ABC = âˆ ABD + âˆ DBC

By substituting values we get

âˆ ABC = 40^{o} + 60^{o}

By addition

âˆ ABC = 100^{o}

We know that the opposite angles are equal in a parallelogram

âˆ ADC = âˆ ABC = 100^{o}

We get

âˆ ADC = âˆ CDB + âˆ ADB

On further calculation

âˆ CDB = âˆ ADC – âˆ ADB

By substituting values

âˆ CDB = 100^{o} – 60^{o}

By subtraction

âˆ CDB = 40^{o}

Therefore, âˆ ADB = 60^{o} and âˆ CDB = 40^{o}.

**3. In the adjoining figure, M is the midpoint of side BC of a parallelogram ABCD such that âˆ BAM = âˆ DAM. Prove that AD = 2CD.**

**Solution:**

It is given that ABCD is a parallelogram

So we know that AD || BC

From the figure we know that âˆ DAM and âˆ AMB are alternate angles

So we get

âˆ DAM = âˆ AMB

We know that âˆ BAM = âˆ DAM

It can be written as

âˆ BAM = âˆ AMB

From the figure we know that the sides opposite to equal angles are equal

So we get

BM = AB

We know that the opposite sides of a parallelogram are equal

AB = CD

So we can write it as

BM = AB = CD â€¦â€¦. (1)

We know that M is the midpoint of the line BC

So we get

BM = Â½ BC

We know that BC = AD

We get

BM = Â½ AD

Based on equation (1)

CD = Â½ AD

By cross multiplication

AD = 2CD.

Therefore, it is proved that AD = 2CD.

**4. In the adjoining figure, ABCD is a parallelogram in which âˆ A = 60 ^{o}. If the bisectors of âˆ A and âˆ B meet DC at P, prove that**

**(i) âˆ APB = 90 ^{o},**

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

**(iii) DC = 2AD**

**Solution:**

(i) We know that opposite angles are equal in a parallelogram.

So we get

âˆ C = âˆ A = 60^{o}

We know that the sum of all the angles in a parallelogram is 360^{o}

It can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

So we get

âˆ B + âˆ D = 360^{o} â€“ (âˆ A + âˆ C)

By substituting values in the above equation we get

âˆ B + âˆ D = 360^{o} â€“ (60^{o} + 60^{o})

On further calculation we get

âˆ B + âˆ D = 360^{o} â€“ 120^{o}

By subtraction

âˆ B + âˆ D = 240^{o}

We know that âˆ B = âˆ B

So the above equation becomes

âˆ B + âˆ B = 240^{o}

2 âˆ B = 240^{o}

By division

âˆ B = âˆ D = 120^{o}

We know that AB || DP and AP is a transversal

From the figure we know that âˆ APD and âˆ PAD are alternate angles

âˆ APD = âˆ PAD = 60^{o}/2

âˆ APD = âˆ PAD = 30^{o} â€¦â€¦. (1)

We know that AB || PC and BP is a transversal

So we get

âˆ ABP = âˆ CPB = âˆ B/2

i.e. âˆ ABP = âˆ CPB = 120^{o}/2

We get âˆ ABP = âˆ CPB = 60^{o} â€¦â€¦.. (2)

We know that DPC is a straight line

It can be written as

âˆ APD + âˆ APB + âˆ CPB = 180^{o}

By substituting the values we get

30^{o} + âˆ APB + 60^{o} = 180^{o}

On further calculation

âˆ APB = 180^{o} – 30^{o} – 60^{o}

By subtraction

âˆ APB = 180^{o} – 90^{o}

âˆ APB = 90^{o}

Therefore, it is proved that âˆ APB = 90^{o}

(ii) From equation (1) we know that

âˆ APD = 30^{o}

We know that

âˆ DAP = 60^{o}/2

By division

âˆ DAP = 30^{o}

So we get

âˆ APD = âˆ DAP â€¦â€¦â€¦ (3)

From the figure we know that the sides of an isosceles triangle are equal

So we get

DP = AD

We know that

âˆ CPB = 60^{o} and âˆ C = 60^{o}

By sum property of triangle

We get

âˆ C + âˆ CPB + âˆ PBC = 180^{o}

By substituting the values in the above equation

60^{o} + 60^{o} + âˆ PBC = 180^{o}

On further calculation we get

âˆ PBC = 180^{o} – 60^{o} – 60^{o}

By subtraction

âˆ PBC = 180^{o} – 120^{o}

âˆ PBC = 60^{o}

We know that all the sides of an equilateral triangle are equal

So we get

PB = PC = BC â€¦â€¦â€¦. (4)

Therefore, it is proved that AD = AP and PB = PC = BC.

(iii) We know that âˆ DPA = âˆ PAD based on equation (3)

We also know that all the sides are equal in an isosceles triangle

DP = AD

From the figure we know that the opposite sides are equal

So we get

DP = BC

Considering equation (4)

DP = PC

From the figure we know that DP = PC and P is the midpoint of the line DC

So we get

DP = Â½ DC

By cross multiplication we get

DC = 2AD

Therefore, it is proved that DC = 2AD.

**5. In the adjoining figure, ABCD is a parallelogram in which âˆ BAO = 35 ^{o}, âˆ DAO = 40^{o} and âˆ COD = 105^{o}. Calculate **

**(i) âˆ ABO,**

**(ii) âˆ ODC,**

**(iii) âˆ ACB**

**(iv) âˆ CBD.**

**Solution:**

(i) From the figure we know that âˆ AOB and âˆ COD are vertically opposite angles

So we get

âˆ AOB = âˆ COD = 105^{o}

Consider â–³ AOB

By sum property of a triangle

âˆ OAB + âˆ AOB + âˆ ABO = 180^{o}

By substituting the values in above equation

35^{o} + 105^{o} + âˆ ABO = 180^{o}

On further calculation

âˆ ABO = 180^{o} – 35^{o} – 105^{o}

By subtraction

âˆ ABO = 180^{o} – 140^{o}

âˆ ABO = 40^{o}

(ii) We know that AB || DC and BD is a transversal

From the figure we know that âˆ ABD and âˆ CDB are alternate angles

It can be written as

âˆ CDO = âˆ CDB = âˆ ABD = âˆ ABO = 40^{o}

So we get

âˆ ODC = 40^{o}

(iii) We know that AB || CD and AC is a transversal

From the figure we know that âˆ ACB and âˆ DAC are alternate opposite angles

So we get

âˆ ACB = âˆ DAC = 40^{o}

(iv) We know that âˆ B can be written as

âˆ B = âˆ CBD + âˆ ABO

So we get

âˆ CBD = âˆ B – âˆ ABO

In a parallelogram we know that the sum of all the angles is 360^{o}

So we get

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

It can be written as

2 âˆ A + 2 âˆ B = 360^{o}

By substituting values in the above equation

2 (40^{o} + 35^{o}) + 2 âˆ B = 360^{o}

On further calculation

2 (75^{o}) + 2 âˆ B = 360^{o}

So we get

150^{o} + 2 âˆ B = 360^{o}

2 âˆ B = 360^{o} – 150^{o}

By subtraction

2 âˆ B = 210^{o}

By division

âˆ B = 105^{o}

So we get

âˆ CBD = âˆ B – âˆ ABO

By substituting values

âˆ CBD = 105^{o} – 40^{o}

By subtraction

âˆ CBD = 65^{o}

**6. In a parallelogram ABCD, if âˆ A = (2x + 25) ^{o} and âˆ B = (3x – 5) ^{o}, find the value of x and the measure of each angle of the parallelogram.**

**Solution:**

We know that the opposite angles are equal in a parallelogram

Consider parallelogram ABCD

So we get

âˆ A = âˆ C = (2x + 25) ^{o}

âˆ B = âˆ D = (3x – 5) ^{o}

We know that the sum of all the angles of a parallelogram is 360^{o}

So it can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values in the above equation

(2x + 25) + (3x â€“ 5) + (2x + 25) + (3x â€“ 5) = 360^{o}

By addition we get

10x + 40^{o} = 360^{o}

By subtraction

10x = 360^{o} – 40^{o}

So we get

10x = 320^{o}

By division we get

x = 32^{o}

Now substituting the value of x

âˆ A = âˆ C = (2x + 25) ^{o} = (2(32) + 25) ^{o}

âˆ A = âˆ C = (64 + 25) ^{o}

By addition

âˆ A = âˆ C = 89^{o}

âˆ B = âˆ D = (3x – 5) ^{o} = (3(32) – 5) ^{o}

âˆ B = âˆ D = (96 – 5) ^{o}

By subtraction

âˆ B = âˆ D = 91^{o}

Therefore, x = 32^{o}, âˆ A = âˆ C = 89^{o} and âˆ B = âˆ D = 91^{o}.

**7. If an angle of a parallelogram is four fifths of its adjacent angle, find the angles of the parallelogram.**

**Solution:**

Consider ABCD as a parallelogram

If âˆ A = x^{o}

We know that âˆ B is adjacent to A which can be written as 4/5 x^{o}

Opposite angles are equal in a parallelogram

So we get

âˆ A = âˆ C = x^{o} and âˆ B = âˆ D = 4/5 x^{o}

We know that the sum of all the angles of a parallelogram is 360^{o}

It can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values in the above equation

x + (4/5) x + x + (4/5) x = 360^{o}

By addition we get

2x + (8/5) x = 360^{o}

By taking the LCM as 5

(18/5) x = 360^{o}

By cross multiplication

x = (360 Ã— 5)/18

On further calculation

x = 100^{o}

By substituting the value of x

So we get

âˆ A = âˆ C = x = 100^{o}

âˆ B = âˆ D = 4/5 x^{o} = (4/5) (100^{o}) = 80^{o}

Therefore, âˆ A = âˆ C = x = 100^{o} and âˆ B = âˆ D = 80^{o}.

**8. Find the measure of each angle of a parallelogram, if one of its angles is 30 ^{o} less than twice the smallest angle.**

**Solution:**

Consider ABCD as a parallelogram

Let us take âˆ A as the smallest angle

So we get

âˆ B = 2 âˆ A – 30^{o}

We know that the opposite angles are equal in a parallelogram

âˆ A = âˆ C and âˆ B = âˆ D = 2 âˆ A – 30^{o}

We know that the sum of all the angles of a parallelogram is 360^{o}

It can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values in the above equation

âˆ A + (2 âˆ A – 30^{o}) + âˆ A + (2 âˆ A – 30^{o}) = 360^{o}

On further calculation

âˆ A + 2 âˆ A – 30^{o} + âˆ A + 2 âˆ A – 30^{o} = 360^{o}

So we get

6 âˆ A – 60^{o} = 360^{o}

By addition

6 âˆ A = 360^{o} + 60^{o}

6 âˆ A = 420^{o}

By division

âˆ A = 70^{o}

By substituting the value of âˆ A

âˆ A = âˆ C = 70^{o}

âˆ B = âˆ D = 2 âˆ A – 30^{o} = 2 (70^{o}) – 30^{o}

âˆ B = âˆ D = 110^{o}

Therefore, âˆ A = âˆ C = 70^{o} and âˆ B = âˆ D = 110^{o}.

**9. ABCD is a parallelogram in which AB = 9.5 cm and its perimeter is 30 cm. Find the length of each side of the parallelogram.**

**Solution:**

We know that the perimeter of parallelogram ABCD can be written as

Perimeter = AB + BC + CD + DA

We know that opposite sides of parallelogram are equal

AB = CD and BC = DA

By substituting the values

Perimeter = 9.5 + BC + 9.5 + BC

It is given that perimeter = 30 cm

So we get

30 = 19 + 2BC

It can be written as

2BC = 30 â€“ 19

By subtraction

2BC = 11

By division we get

BC = 5.5 cm

Therefore, AB = 9.5 cm, BC = 5.5 cm, CD = 9.5 cm and DA = 5.5 cm.

**10. In each of the figures given below, ABCD is a rhombus. Find the value of x and y in each case.**

**(i)**

**(ii)**

**(iii)**

**Solution:**

(i) We know that all the sides are equal in a rhombus

Consider â–³ ABC

We know that AB = BC

It can be written as

âˆ CAB = âˆ ACB = x^{o}

By sum property of a triangle

We get

âˆ CAB + âˆ ABC + âˆ ACB = 180^{o}

By substituting the values in above equation

x + 110^{o} + x = 180^{o}

By addition we get

2x + 110^{o }= 180^{o}

On further calculation

2x = 180^{o} – 110^{o}

By subtraction

2x = 70^{o}

By division

x = 35^{o}

Therefore, x = 35^{o} and y = 35^{o}.

(ii) We know that all the sides are equal in a rhombus

Consider â–³ ABD

We get

AB = AD and

âˆ ABD = âˆ ADB

It can be written as

x = y â€¦â€¦. (1)

Consider â–³ ABC

We get

AB = BC and

âˆ CAB = âˆ ACB

We know that

âˆ ACB = 40^{o}

By using the sum property of a triangle

âˆ B + âˆ CAB + âˆ ACB = 180^{o}

By substituting values in the above equation

âˆ B + 40^{o} + 40^{o} = 180^{o}

On further calculation

âˆ B = 180^{o} – 40^{o} – 40^{o}

By subtraction

âˆ B = 180^{o} – 80^{o}

So we get

âˆ B = 100^{o}

âˆ DBC can be written as

âˆ DBC = âˆ B – x^{o}

By substituting the values

âˆ DBC = 100^{o} – x^{o}

From the figure we know that âˆ DBC and âˆ ADB are alternate angles

âˆ DBC = âˆ ADB = y^{o}

By substituting the value of âˆ DBC

100^{o} – x^{o} = y^{o}

Consider the equation (1) we know that x = y

100^{o} – x^{o} = x^{o}

On further calculation

2x^{o} = 100^{o}

By division

x^{o} = 50^{o}

Therefore, x = y = 50^{o}.

(iii) From the figure we know that

âˆ A = âˆ C = 62^{o}

Consider â–³ BCD

We get

BC = DC

It can be written as

âˆ CDB = âˆ DBC = y^{o}

Using the sum property of triangle

âˆ BDC + âˆ DBC + âˆ BCD = 180^{o}

By substituting the values

y + y + 62^{o} = 180^{o}

On further calculation

2y = 180^{o} – 62^{o}

By subtraction

2y = 118^{o}

By division

y = 59^{o}

We know that the diagonals of a rhombus are perpendicular to each other

Consider â–³ COD as a right angle triangle

âˆ DOC = 90^{o}

âˆ ODC = y = 59^{o}

It can be written as

âˆ DCO + âˆ ODC = 90^{o}

To find âˆ DCO

âˆ DCO = 90^{o} – âˆ ODC

By substituting the values

âˆ DCO = 90^{o} – 59^{o}

âˆ DCO = x = 31^{o}

Therefore, x = 31^{o} and y = 59^{o}.

**11. The lengths of the diagonals of a rhombus are 24cm and 18 cm respectively. Find the length of each side of the rhombus.**

**Solution:**

We know that ABCD is a rhombus

It is given that AC = 24cm and BD = 18cm

In a rhombus we know that the diagonals bisect each other at right angles

Consider â–³ AOB

We know that

âˆ AOB = 90^{o}

To find AO and BO

We know that

AO = Â½ AC

By substituting AC

AO = Â½ (24)

AO = 12 cm

BO = Â½ BD

By substituting BD

BO = Â½ (18)

BO = 9 cm

Based on the Pythagoras Theorem

AB^{2} = AO^{2} + OB^{2}

By substituting values

AB^{2} = 12^{2} + 9^{2}

So we get

AB^{2} = 144 + 81

By addition

AB^{2} = 225

AB = âˆš 225

So we get

AB = 15 cm

Therefore, the length of each side of the rhombus is 15 cm.

**12. Each side of a rhombus is 10cm long and one of its diagonals measures 16cm. Find the length of the other diagonal and hence find the area of the rhombus.**

**Solution:**

We know that the diagonals of a rhombus bisect at right angles

We get

AO = OC = Â½ AC

We know that AC = 16 cm

AO = OC = Â½ (16) = 8 cm

Consider â–³ AOB

Using the Pythagoras Theorem

AB^{2} = AO^{2} + OB^{2}

By substituting values

10^{2} = 8^{2} + OB^{2}

On further calculation

OB^{2} = 100 â€“ 64

By subtraction

OB^{2} = 36

So we get

OB = âˆš 36

OB = 6 cm

We know that the length of other diagonal

BD = 2 Ã— OB

By substituting the value

BD = 2 Ã— 6

BD = 12 cm

From the figure the area of

â–³ ABC = Â½ Ã— AC Ã— OB

By substituting the values

â–³ ABC = Â½ Ã— 16 Ã— 6

By multiplication

â–³ ABC = 48 cm^{2}

â–³ ACD = Â½ Ã— AC Ã— OD

By substituting the values

â–³ ACD = Â½ Ã— 16 Ã— 6

By multiplication

â–³ ACD = 48 cm^{2}

So we can calculate the area of rhombus as

Area of rhombus = area of â–³ ABC + area of â–³ ACD

We get

Area of rhombus = 48 + 48 = 96 cm^{2}

Therefore, the length of other diagonal is 12 cm and the area of rhombus is 96 cm^{2}.

**13. In each of the figures given below, ABCD is a rectangle. Find the values of x and y in each case. **

**(i)**

**(ii)**

**Solution:**

(i) From the figure we know that the diagonals are equal and bisect at point O.

Consider â–³ AOB

We get AO = OB

We know that the base angles are equal

âˆ OAB = âˆ OBA = 35^{o}

By using the sum property of triangle

âˆ AOB + âˆ OAB + âˆ OBA = 180^{o}

By substituting the values we get

âˆ AOB + 35^{o }+ 35^{o }= 180^{o}

On further calculation

âˆ AOB = 180^{o} – 35^{o }– 35^{o}

By subtraction

âˆ AOB = 180^{o} – 70^{o}

âˆ AOB = 110^{o}

From the figure we know that the vertically opposite angles are equal

âˆ DOC = âˆ AOB = y = 110^{o}

In â–³ ABC

We know that âˆ ABC = 90^{o}

Consider â–³ OBC

We know that

âˆ OBC = x^{o} = âˆ ABC – âˆ OBA

By substituting the values

âˆ OBC = 90^{o} – 35^{o}

By subtraction

âˆ OBC = 55^{o}

Therefore, x = 55^{o} and y = 110^{o}.

(ii) From the figure we know that the diagonals of a rectangle are equal and bisect each other.

Consider â–³ AOB

We get

OA = OB

We know that the base angles are equal

âˆ OAB = âˆ OBA

By using the sum property of triangle

âˆ AOB + âˆ OAB + âˆ OBA = 180^{o}

By substituting the values

110^{o} + âˆ OAB + âˆ OBA = 180^{o}

We know that âˆ OAB = âˆ OBA

So we get

2 âˆ OAB = 180^{o} – 110^{o}

By subtraction

2 âˆ OAB = 70^{o}

By division

âˆ OAB = 35^{o}

We know that AB || CD and AC is a transversal

From the figure we know that âˆ DCA and âˆ CAB are alternate angles

âˆ DCA = âˆ CAB = y^{o} = 35^{o}

Consider â–³ ABC

We know that

âˆ ACB + âˆ CAB = 90^{o}

So we get

âˆ ACB = 90^{o }– âˆ CAB

By substituting the values in above equation

âˆ ACB = 90^{o }– 35^{o}

By subtraction

âˆ ACB = x = 55^{o}

Therefore, x = 55^{o} and y = 35^{o}.

**14. In a rhombus ABCD, the altitude from D to the side B bisects AB. Find the angles of the rhombus.**

**Solution:**

From the figure consider DP bisects the side AB at point P

Construct a line at BD

Consider â–³ AMD and â–³ BMD

We know that M is the mid-point of AB

AM = BM

From the figure we know that

âˆ AMD = âˆ BMD = 90^{o}

MD is common i.e. MD = MD

By SAS congruence criterion

â–³ AMD â‰… â–³ BMD

AD = BD (c. p. c. t)

We know that the sides of a rhombus are equal

So we get

AD = AB

It can be written as

AD = AB = BD

We know that â–³ ADB is an equilateral triangle

So we get

âˆ A = 60^{o}

From the figure we know that the opposite angles are equal

âˆ C = âˆ A = 60^{o}

We know that

âˆ B + âˆ A = 180^{o}

On further calculation

âˆ B = 180^{o} – âˆ A

âˆ B = 180^{o} – 60^{o}

By subtraction

âˆ B = 120^{o}

We know that âˆ D = âˆ B = 120^{o}

Therefore, the angles are âˆ C = âˆ A = 60^{o} and âˆ D = âˆ B = 120^{o}.

**15. In the adjoining figure, ABCD is a square. A line segment CX cuts AB at X and the diagonal BD at O such that âˆ COD = 80 ^{o} and âˆ OXA = x^{o}. Find the value of x.**

**Solution:**

In â–³ ABD

We know that AB = AD

From the figure we know that the base angles are equal

âˆ ADB = âˆ ABD

We know that âˆ A = 90^{o}

It can be written as

âˆ ADB + âˆ ABD = 90^{o}

We know that âˆ ADB = âˆ ABD

So we get

2 âˆ ADB = 90^{o}

By division

âˆ ADB = 45^{o}

Consider â–³ OXB

From the figure we know that âˆ XOB and âˆ DOC are vertically opposite angles

âˆ XOB = âˆ DOC = 80^{o}

We also know that

âˆ ABD = âˆ XBD = 45^{o}

We can write it as

Exterior âˆ AXO = âˆ XOB + âˆ XBD

By substituting the values

x^{o} = 80^{o} + 45^{o}

By addition

x^{o} = 125^{o}

Therefore, the value of x is 125^{o}.

**16. In a rhombus ABCD show that diagonal AC bisects âˆ A as well as âˆ C and diagonal BD bisects âˆ B as well as âˆ D.**

**Solution:**

Consider â–³ ABC and â–³ ADC

We know that the sides of rhombus are equal

AB = AD and BC = CD

AC is common i.e. AC = AC

By SSS congruence criterion

â–³ ABC â‰… â–³ ADC

We know that

âˆ BAC = âˆ DAC and âˆ BCA = âˆ DCA (c. p. c. t)

Therefore, AC bisects âˆ A as well as âˆ C.

Consider â–³ BAD and â–³ BCD

We know that the sides of rhombus are equal

AB = BC and AD = CD

BD is common i.e. BD = BD

By SSS congruence criterion

â–³ BAD â‰… â–³ BCD

We know that

âˆ ABD = âˆ CBD and âˆ ADB = âˆ CDB (c. p. c. t)

Therefore, BD bisects âˆ B as well as âˆ D.

**17. In a parallelogram ABCD, points M and N have been taken on opposite sides AB and CD respectively such that AM = CN. Show that AC and MN bisect each other.**

**Solution:**

Consider â–³ AMO and â–³ CNO

We know that AB || CD

From the figure we know that âˆ MAO and âˆ NCO are alternate angles

It is given that AM = CN

We know that âˆ AOM and âˆ CON are vertically opposite angles

âˆ AOM = âˆ CON

By ASA congruence criterion

â–³ AMO â‰… â–³ CNO

So we get

AO = CO and MO = NO (c. p. c. t)

Therefore, it is proved that AC and MN bisect each other.

**18. In the adjoining figure, ABCD is a parallelogram. If P and Q are points on AD and BC respectively such that AP = 1/3 AD and CQ = 1/3 BC, prove that AQCP is a parallelogram.**

**Solution:**

Consider â–³ ABQ and â–³ CDP

We know that the opposite sides of a parallelogram are equal

AB = CD

So we get âˆ B = âˆ D

We know that

DP = AD â€“ PA

i.e. DP = 2/3 AD

BQ = BC â€“ CQ

i.e. BQ = BC â€“ 1/3 BC

BQ = (3-1)/3 BC

We know that AD = BC

So we get

BQ = 2/3 BC = 2/3 AD

We get BQ = DP

By SAS congruence criterion

â–³ ABQ â‰… â–³ CDP

AQ = CP (c. p. c. t)

We know that

PA = 1/3 AD

We know that AD = BC

CQ = 1/3 BC = 1/3 AD

So we get

PA = CQ

âˆ QAB = âˆ PCD (c. p. c. t)â€¦ (1)

We know that

âˆ QAP = âˆ A – âˆ QAB

Consider equation (1)

âˆ A = âˆ C

âˆ QAP = âˆ C – âˆ PCD

From the figure we know that the alternate interior angles are equal

âˆ QAP = âˆ PCQ

So we know that AQ and CP are two parallel lines.

Therefore, it is proved that PAQC is a parallelogram.

**19. In the adjoining figure, ABCD is a parallelogram whose diagonals intersect each other at O. A line segment EOF is drawn to meet AB at E and DC at F. Prove that OE = OF.**

**Solution:**

We know that ABCD is a parallelogram whose diagonals intersect each other at O

Consider â–³ AOE and â–³ COF

We know that âˆ CAE and âˆ DCA are alternate angles

From the figure we know that the diagonals are equal and bisect each other

AO = CO

We know that âˆ AOE and âˆ COF are vertically opposite angles

âˆ AOE =âˆ COF

By ASA congruence criterion

â–³ AOE â‰… â–³ COF

OE = OF (c. p. c. t)

Therefore, it is proved that OE = OF.

**20. The angle between two altitudes of a parallelogram through the vertex of an obtuse angle of the parallelogram is 60 ^{o}. Find the angles of the parallelogram.**

**Solution:**

From the figure

âˆ DCM = âˆ DCN + âˆ MCN

By substituting values in the above equation

90^{o} = âˆ DCN + 60^{o}

On further calculation

âˆ DCN = 90^{o} – 60^{o}

By subtraction

âˆ DCN = 30^{o}

Consider â–³ DCN

Using the sum property

âˆ DNC + âˆ DCN + âˆ D = 180^{o}

By substituting values in the above equation

90^{o} + 30^{o} + âˆ D = 180^{o}

On further calculation

âˆ D = 180^{o} – 90^{o} – 30^{o}

By subtraction

âˆ D = 180^{o} – 120^{o}

âˆ D = 60^{o}

We know that the opposite angles of parallelogram are equal

âˆ B = âˆ D = 60^{o}

It can be written as

âˆ A + âˆ B = 180^{o}

By substituting values

âˆ A + 60^{o} = 180^{o}

By subtraction

âˆ A = 180^{o} – 60^{o}

âˆ A =120^{o}

So we get âˆ A = âˆ C =120^{o}

Therefore, âˆ B = âˆ D = 60^{o} and âˆ A = âˆ C =120^{o}.

**21. ABCD is a rectangle in which diagonal AC bisects âˆ A as well as âˆ C. Show that**

**(i) ABCD is a square,**

**(ii) Diagonal BD bisects âˆ B as well as âˆ D.**

**Solution:**

(i) We know that ABCD is a rectangle in which diagonal AC bisects âˆ A as well as âˆ C.

So we get

âˆ BAC = âˆ DAC â€¦â€¦.. (1)

âˆ BCA = âˆ DCA â€¦â€¦.. (2)

We know that every rectangle is a parallelogram

So we get AB || DC and AC is a transversal

From the figure we know that âˆ BAC and âˆ DCA are alternate angles

âˆ BAC = âˆ DCA

By considering equation (1)

We get

âˆ DAC = âˆ DCA

Consider â–³ ADC

We know that the opposite sides of equal angles are equal

AD = CD

Since ABCD is a rectangle

We get AB = BC and CD = AD

So we get AB = BC = CD = AD

Therefore, it is proved that ABCD is a square.

(ii) Consider â–³ BAD and â–³ BCD

We know that AB = CD and AD = BC

BD is common i.e. BD = BD

By SSS congruence criterion

â–³ BAD â‰… â–³ BCD

We know that

âˆ ABD = âˆ CBD and âˆ ADB = âˆ CDB (c. p. c. t)

Therefore, diagonal BD bisects âˆ B as well as âˆ D.

**22. In the adjoining figure, ABCD is a parallelogram in which AB is produced to E so that BE = AB. Prove that ED bisects BC.**

**Solution:**

Consider â–³ OCD and â–³ OBE

From the figure we know that âˆ DOC and âˆ EOB are vertically opposite angles

âˆ DOC = âˆ EOB

We know that AB || CD and BC is a transversal

âˆ OCD and âˆ OBE are alternate angles

âˆ OCD = âˆ OBE

From the figure we know that AB = CD and BE = AB

So we can write DC = BE

By AAS congruence criterion

â–³ OCD â‰… â–³ OBE

OC = OB (c. p. c. t)

Therefore, it is proved that ED bisects BC.

**23. In the adjoining figure, ABCD is a parallelogram and E is the midpoint of side BC. If DE and AB when produced to meet at F, prove that AF = 2AB.**

**Solution:**

Consider â–³ DEC and â–³ FEB

From the figure we know that âˆ DEC and âˆ FEB are vertically opposite angles

âˆ DEC = âˆ FEB

âˆ DCE and âˆ FBE are alternate angles

âˆ DCE = âˆ FBE

It is given that CE = EB

By AAS congruence criterion

â–³ DEC â‰… â–³ FEB

DC = FB (c. p. c. t)

From the figure

AF = AB + BF

We know that BF = DC and AB = DC

So we get

AF = AB + DC

AF = AB + AB

By addition

AF = 2AB

Therefore, it is proved that AF = 2AB.

**24. Two parallel lines l and m are intersected by a transversal t. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle.**

**Solution:**

We know that l || m and t is a transversal

From the figure we know that âˆ APR and âˆ PRD are alternate angles

âˆ APR = âˆ PRD

We can write it as

Â½ âˆ APR = Â½ âˆ PRD

We know that PS and RQ are the bisectors of âˆ APR and âˆ PRD

So we get

âˆ SPR = âˆ PRQ

Hence, PR intersects PS and RQ at points P and R respectively

We get

PS || RQ

In the same way SR || PQ

Therefore, PQRS is a parallelogram

We know that the interior angles are supplementary

âˆ BPR + âˆ PRD = 180^{o}

From the figure we know that PQ and RQ are the bisectors of âˆ BPR and âˆ PRD

We can write it as

2 âˆ QPR + 2 âˆ QRP = 180^{o}

Dividing the equation by 2

âˆ QPR + âˆ QRP = 90^{o} â€¦â€¦ (1)

Consider â–³ PQR

Using the sum property of triangle

âˆ PQR + âˆ QPR + âˆ QRP = 180^{o}

By substituting equation (1)

âˆ PQR + 90^{o} = 180^{o}

On further calculation

âˆ PQR = 180^{o} – 90^{o}

By subtraction

âˆ PQR = 90^{o}

We know that PQRS is a parallelogram

It can be written as

âˆ PQR = âˆ PSR = 90^{o}

We know that the adjacent angles in a parallelogram are supplementary

âˆ SPQ + âˆ PQR = 180^{o}

By substituting the values in above equation

âˆ SPQ + 90^{o} = 180^{o}

On further calculation

âˆ SPQ = 180^{o} – 90^{o}

By subtraction

âˆ SPQ = 90^{o}

We know that all the interior angles of quadrilateral PQRS are right angles

Therefore, it is proved that the quadrilateral formed by the bisectors of interior angles is a rectangle.

**25. K, L, M and N are points on the sides AB, BC, CD and DA respectively of a square ABCD such that AK = BL = CM = DN. Prove that KLMN is a square.**

**Solution:**

It is given that AK = BL = CM = DN

ABCD is a square

So we get

BK = CL = DM = AN â€¦â€¦ (1)

Consider â–³ AKN and â–³ BLK

It is given AK = BL

From the figure we know that âˆ A = âˆ B = 90^{o}

Using equation (1)

AN = BK

By SAS congruence criterion

â–³ AKN â‰… â–³ BLK

We get

âˆ AKN = âˆ BLK and âˆ ANK = âˆ BKL (c. p. c. t)

We know that

âˆ AKN + âˆ ANK = 90^{o}

âˆ BLK + âˆ BKL = 90^{o}

By adding both the equations

âˆ AKN + âˆ ANK + âˆ BLK + âˆ BKL = 90^{o} + 90^{o}

On further calculation

2 âˆ ANK + 2 âˆ BLK = 180^{o}

Dividing the equation by 2

âˆ ANK + âˆ BLK = 90^{o}

So we get

âˆ NKL = 90^{o}

In the same way

âˆ KLM = âˆ LMN = âˆ MNK = 90^{o}

Therefore, it is proved that KLMN is a square.

**26. A â–³ ABC is given. If lines are drawn through A, B, C, parallel respectively to the sides BC, CA and AB, forming â–³ PQR, as shown in the adjoining figure, show that BC = Â½ QR.**

**Solution:**

We know that AR || BC and AB || RC

From the figure we know that ABCR is a parallelogram

So we get

AR = BC â€¦â€¦ (1)

We know that AQ || BC and QB || AC

From the figure we know that AQBC is a parallelogram

So we get

QA = BC â€¦â€¦â€¦ (2)

By adding both the equations

AR + QA = BC + BC

We know that AR + QA = QR

So we get

QR = 2BC

Dividing by 2

BC = QR/2

BC = Â½ QR

Therefore, it is proved that BC = Â½ QR.

**27. In the adjoining figure, â–³ ABC is a triangle and through A, B, C lines are drawn, parallel respectively to BC, CA and AB, intersecting at P, Q and R. Prove that the perimeter of â–³ PQR is double the perimeter of â–³ ABC.**

**Solution:**

We know that AR || BC and AB || RC

From the figure we know that ABCR is a parallelogram

So we get

AR = BC â€¦â€¦.. (1)

We know that AQ || BC and QB || AC

From the figure we know that AQBC is a parallelogram

So we get

QA = BC â€¦â€¦â€¦ (2)

By adding both the equations we get

AR + QA = BC + BC

We know that AR + QA = QR

So we get

QR = 2BC

It can be written as

BC = QR/2

BC = Â½ QR

In the same way

AB = Â½ RP and AC = Â½ PQ

Perimeter of â–³ PQR = PQ + QR + RP

It can be written as

Perimeter of â–³ PQR = 2AC + 2BC + 2AB

By taking 2 as common

Perimeter of â–³ PQR = 2 (AC + BC + 2AB)

Perimeter of â–³ PQR = 2 (Perimeter of â–³ ABC)

Therefore, it is proved that the perimeter of â–³ PQR is double the perimeter of â–³ ABC.

### Access other exercise solutions of Class 9 Maths Chapter 10: Quadrilaterals

Exercise 10A Solutions 10 Questions

Exercise 10C Solutions 18 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 10 – Quadrilaterals Exercise 10B

RS Aggarwal Solutions Class 9 Maths Chapter 10 Quadrilaterals Exercise 10B consists of theorems which are based on various structures like parallelogram, rhombus, rectangle and square.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 10: Quadrilaterals Exercise 10B

- The solutions are prepared in simple language based on the understanding capacity of students.
- By referring the PDF, students can improve their logical thinking abilities, which is important in solving problems.
- RS Aggarwal solutions are well designed by the teachers with various methods used for a single problem.
- They mainly help students to complete the entire syllabus and prepare effectively for their board exams.