In this chapter, solutions are available in PDF format to make students stress free during the exam preparation. The answers are mainly prepared with the aim of helping the students who are weak in Mathematics to clear the exam. To score good marks in the exam, the students mainly need to solve the exercise wise problems on a daily basis. This chapter explains the types of quadrilaterals and various theorems on parallelograms.

The aim of preparing the solutions by faculties at BYJUâ€™S is to help students solve the problems of each exercise and understand the methods used. The solutions are based on the evaluation process followed in the CBSE board. RS Aggarwal Solutions for Class 9 Chapter 10 Quadrilaterals are provided here.

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

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

## Exercise 10(A)

**1. Three angles of a quadrilateral are 75 ^{o}, 90^{o} and 75^{o}. Find the measure of the fourth angle. **

**Solution:**

Consider the fourth angle as x^{o}

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

So we can write it as

x^{o} + 75^{o} + 90^{o} + 75^{o} = 360^{o}

On further calculation we get

x^{o} = 360^{o} – 75^{o} – 90^{o} – 75^{o}

By subtraction

x^{o} = 360^{o} – 240^{o}

x^{o} = 120^{o}

Therefore, the measure of the fourth angle is 120^{o}.

**2. The angles of a quadrilateral are in the ratio 2: 4: 5: 7. Find the angles.**

**Solution:**

Consider the angles of a quadrilateral as 2x, 4x, 5x and 7x

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

So we can write it as

2x + 4x + 5x + 7x = 360^{o}

By addition

18x = 360^{o}

By division

x^{o} = 20^{o}

Now by substituting the value of x^{o}

2x = 2(20^{o}) = 40^{o}

4x = 4(20^{o}) = 80^{o}

5x = 5(20^{o}) = 100^{o}

7x = 7(20^{o}) = 140^{o}

Therefore, the angles are 40^{o}, 80^{o}, 100^{o} and 140^{o}.

**3. In the adjoining figure, ABCD is a trapezium in which AB || DC. If âˆ A = 55 ^{o} and âˆ B = 70^{o}, find âˆ C and âˆ D.**

**Solution:**

It is given that AB || DC

From the figure we know that âˆ A and âˆ D are consecutive interior angles

So it can written as

âˆ A + âˆ D = 180^{o}

To find âˆ D

âˆ D = 180^{o} – âˆ A

By substituting the value in the above equation

âˆ D = 180^{o} – 55^{o}

By subtraction

âˆ D = 125^{o}

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

So we can write it as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values in the above equation

55^{o} + 70^{o} + âˆ C + 125^{o} = 360^{o}

On further calculation

âˆ C = 360^{o} – 55^{o} – 70^{o} – 125^{o}

By subtraction

âˆ C = 360^{o} – 250^{o}

âˆ C = 110^{o}

Therefore, âˆ C = 110^{o} and âˆ D = 125^{o}.

**4. In the adjoining figure, ABCD is a square and â–³ EDC is an equilateral triangle. Prove that **

**(i) AE = BE,**

**(ii) âˆ DAE = 15 ^{o}**

**Solution:**

(i) From the figure we know that â–³ EDC is an equilateral triangle.

So we get âˆ EDC = âˆ ECD = 60^{o}

We know that ABCD is a square

So we get âˆ CDA = âˆ DCB = 90^{o}

Consider â–³ EDA

We get

âˆ EDA = âˆ EDC + âˆ CDA

By substituting the values in the above equation

âˆ EDA = 60^{o} + 90^{o}

So we get

âˆ EDA = 150^{o} â€¦â€¦. (1)

Consider â–³ ECB

We get

âˆ ECB = âˆ ECD + âˆ DCB

By substituting the values in the above equation

âˆ ECB = 60^{o} + 90^{o}

So we get

âˆ ECB = 150^{o}

So we get âˆ EDA = âˆ ECB â€¦â€¦ (2)

Consider â–³ EDA and â–³ ECB

From the figure we know that ED and EC are the sides of equilateral triangle

So we get

ED = EC

We also know that the sides of square are equal

DA = CB

By SAS congruence criterion

â–³ EDA â‰… â–³ ECB

AE = BE (c. p. c. t)

(ii) Consider â–³ EDA

We know that

ED = DA

From the figure we know that the base angles are equal

âˆ DEA = âˆ DAE

Based on equation (1) we get âˆ EDA = 150^{o}

By angle sum property

âˆ EDA + âˆ DAE + âˆ DEA = 180^{o}

By substituting the values we get

150^{o} + âˆ DAE + âˆ DEA = 180^{o}

We know that âˆ DEA = âˆ DAE

So we get

150^{o} + âˆ DAE + âˆ DAE = 180^{o}

On further calculation

2 âˆ DAE = 180^{o} – 150^{o}

By subtraction

2 âˆ DAE = 30^{o}

By division

âˆ DAE = 15^{o}

**5. In the adjoining figure, BM âŠ¥ AC and DN âŠ¥ AC. If BM = DN, prove that AC bisects BD.**

**Solution:**

It is given that BM âŠ¥ AC and DN âŠ¥ AC

From the figure we know that âˆ DON and âˆ MOB are vertically opposite angles

âˆ DON = âˆ MOB

Since DN and MB are perpendiculars

We know that

âˆ DNO = âˆ BMO = 90^{o}

It is given that BM = DN

By AAS congruence criterion

â–³ DNO â‰… â–³ BMO

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

Therefore, it is proved that AC bisects BD.

**6. In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that **

**(i) AC bisects âˆ A and âˆ C,**

**(ii) BE = DE,**

**(iii) âˆ ABC = âˆ ADC**

**Solution:**

(i) Consider â–³ ABC and â–³ ADC

It is given that AB = AD and BC = DC

AC is common i.e. AC = AC

By SSS congruence criterion

â–³ ABC â‰… â–³ ADC â€¦â€¦â€¦ (1)

âˆ BAC = âˆ DAC (c. p. c. t)

So we get

âˆ BAE = âˆ DAE

We know that AC bisects the âˆ BAD i.e. âˆ A

So we get

âˆ BCA = âˆ DCA (c. p. c. t)

It can be written as

âˆ BCE = âˆ DCE

So we know that AC bisects âˆ BCD i.e. âˆ C

(ii) Consider â–³ ABE and â–³ ADE

It is given that AB = AD

AE is common i.e. AE = AE

By SAS congruence criterion

â–³ ABE â‰… âˆ ADE

BE = DE (c. p. c. t)

(iii) We know that â–³ ABC â‰… â–³ ADC

Therefore, by c. p. c. t âˆ ABC = âˆ ADC

**7. In the given figure, ABCD is a square and âˆ PQR = 90 ^{o}. If PB = QC = DR, prove that **

**(i) QB = RC,**

**(ii) PQ = QR,**

**(iii) âˆ QPR = 45 ^{o}**

**Solution:**

(i) We know that the line segment QB can be written as

QB = BC â€“ QC

Since ABCD is a square we know that BC = DC and QC = DR

So we get

QB = CD â€“ DR

From the figure we get

QB = RC

(ii) Consider â–³ PBQ and â–³ QCR

It is given that PB = QC

Since ABCD is a square we get

âˆ PBQ = âˆ QCR = 90^{o}

By SAS congruence criterion

â–³ PBQ â‰… â–³ QCR

PQ = QR (c. p. c. t)

(iii) We know that PQ = QR

Consider â–³ PQR

From the figure we know that âˆ QPR and âˆ QRP are base angles of isosceles triangle

âˆ QPR = âˆ QRP

We know that the sum of all the angles in a triangle is 180^{o}

âˆ QPR + âˆ QRP + âˆ PRQ = 180^{o}

By substituting the values in the above equation

âˆ QPR + âˆ QRP + 90^{o} = 180^{o}

On further calculation

âˆ QPR + âˆ QRP = 180^{o} – 90^{o}

By subtraction

âˆ QPR + âˆ QRP = 90^{o}

We know that âˆ QPR = âˆ QRP

So we get

âˆ QPR + âˆ QPR = 90^{o}

By addition

2 âˆ QPR = 90^{o}

By division

âˆ QPR = 45^{o}

Therefore, it is proved that âˆ QPR = 45^{o}.

**8. If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.**

**Solution:**

It is given that O is a point within a quadrilateral ABCD

Consider â–³ ACO

We know that in a triangle the sum of any two sides is greater than the third side

We get

OA + OC > AC â€¦.. (1)

In the same way

Consider â–³ BOD

We get

OB + OD > BD â€¦.. (2)

By adding both the equations we get

OA + OC + OB + OD > AC + BD

Therefore, it is proved that OA + OC + OB + OD > AC + BD.

**9. In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that**

**(i) AB + BC + CD + DA > 2AC**

**(ii) AB + BC + CD > DA**

**(iii) AB + BC + CD + DA > AC + BD.**

**Solution:**

(i) Consider â–³ ABC

We know that

AB + BC > AC â€¦â€¦ (1)

Consider â–³ ACD

We know that

AD + CD > AC â€¦â€¦. (2)

By adding both the equations we get

AB + BC + AD + CD > AC + AC

So we get

AB + BC + AD + CD > 2AC â€¦â€¦.. (3)

Therefore, it is proved that AB + BC + AD + CD > 2AC.

(ii) Consider â–³ ABC

We know that

AB + BC > AC

Add CD both sides of the equation

AB + BC + CD > AC + CD â€¦â€¦ (4)

Consider â–³ ACD

We know that

AC + CD > DA â€¦â€¦. (5)

By substituting (5) in (4) we get

AB + BC + CD > DA â€¦â€¦.. (6)

(iii) Consider â–³ ABD and â–³ BDC

We know that

AB + DA > BD â€¦â€¦. (7)

So we get

BC + CD > BD â€¦â€¦. (8)

By adding (7) and (8) we get

AB + DA + BC + CD > BD + BD

On further calculation

AB + DA + BC + CD > 2BD â€¦â€¦. (9)

By adding equations (9) and (3) we get

AB + DA + BC + CD + AB + BC + AD + CD > 2BD + 2AC

So we get

2 (AB + BC + CD + DA) > 2 (BD + AC)

Dividing by 2

AB + BC + CD + DA > BD + AC

Therefore, it is proved that AB + BC + CD + DA > BD + AC.

**10. Prove that the sum of all the angles of a quadrilateral is 360 ^{o}.**

**Solution:**

Construct a quadrilateral ABCD

Consider â–³ ABC

We can write it as

âˆ CAB + âˆ B + âˆ BCA = 180^{o} â€¦â€¦.. (1)

Consider â–³ ACD

We can write it as

âˆ DAC + âˆ ACD + âˆ D = 180^{o} â€¦â€¦â€¦ (2)

Adding equations (1) and (2) we get

âˆ CAB + âˆ B + âˆ BCA + âˆ DAC + âˆ ACD + âˆ D = 180^{o} + 180^{o}

So we get

âˆ CAB + âˆ DAC + âˆ B + âˆ BCA + âˆ ACD + âˆ D = 360^{o}

So it can be written as

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

## 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.

## Exercise 10(C)

**1. P, Q, R and S are respectively the midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD. Show that**

**(i) PQ || AC and PQ = Â½ AC**

**(ii) PQ || SR**

**(iii) PQRS is a parallelogram.**

**Solution:**

(i) Consider â–³ ABC

We know that P and Q are the mid points of AB and BC.

From the figure we know that PQ || AC

So we get PQ = Â½ AC â€¦â€¦ (1)

Therefore, it is proved that PQ || AC and PQ = Â½ AC

(ii) Consider â–³ ADC

We know that R and S are the mid points of CD and AD

From the figure we know that SR || AC

So we get SR = Â½ AC â€¦â€¦.. (2)

Considering equations (1) and (2)

We get

PQ = SR and PQ || SR

Therefore, it is proved that PQ || SR.

(iii) We know that one pair of opposite sides are equal and parallel in quadrilateral PQRS.

Therefore, PQRS is a parallelogram.

**2. A square is inscribed in an isosceles in an isosceles right triangle so that the square and the triangle have one angle common. Show that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.**

**Solution:**

Consider â–³ ABC as an isosceles right triangle which is right angled at B

So we get

AB = BC

Consider a square PBSR inscribed in â–³ ABC with âˆ B as common

So we get

PB = BS = SR = RP

We can write it as

AB â€“ PB = BC â€“ BS

So we get

AP = CS â€¦â€¦ (1)

Consider â–³ APR and â–³ CSR

We know that

âˆ APR = âˆ CSR = 90^{o}

From the figure we know that the sides of a square are equal

PR = SR

By SAS congruence criterion

â–³ APR â‰… â–³ CSR

AR = CR (c. p. c. t)

So we know that point R bisects the hypotenuse AC.

Therefore, it is proved that the vertex of the square opposite the vertex of the common angle bisects the hypotenuse.

**3. In the adjoining figure, ABCD is a parallelogram in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.**

**Solution:**

From the figure we know that AD, EF and BC are three line segment and DC and AB are the two transversals.

We know that the intercepts made by the line on AB and CD are equal

So we get

AE = EB and DF = FC

Let us prove that FE is parallel to AD by using the method of contradiction

Assume FE is not parallel to AD

Construct FR parallel to AD

Based on the Intercept theorem

We get AR = RB as DF = FC

It is given that AE = EB

AB cannot have two mid points i.e. R and E

So our assumption is not correct

We know that AD || EF || BC

By using the Intercept theorem

GP = PH

So we know that GH is a transversal and the intercepts made by AD, EF and BC on GH are equal as DF = FC.

Therefore, it is proved that GP = PH.

**4. M and N are points on opposite sides AD and BC of a parallelogram ABCD such that MN passes through the point of intersection O of its diagonals AC and BD. Show that MN is bisected at O.**

**Solution:**

Consider â–³ AOM and â–³ CON

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

âˆ MAO = âˆ OCN

We know that the diagonals of parallelogram bisect each other

AO = OC

âˆ AOM and âˆ CON are vertically opposite angles

âˆ AOM = âˆ CON

By ASA congruence criterion

â–³ AOM â‰… â–³ CON

MO = NO (c. p. c. t)

Therefore, it is proved that MN is bisected at O.

**5. In the adjoining figure, PQRS is a trapezium in which PQ || SR and M is the midpoint of PS. A line segment MN || PQ meets QR at N. Show that N is the midpoint of QR.**

**Solution:**

Construct a line to join diagonal QS.

Diagonal QS intersect the line MN at point O

It is given that PQ || SR and MN || PQ

We can write it as

PQ || MN || SR

Consider â–³ SPQ

We know that MO || PQ and M is the midpoint to the side SP

O is the midpoint of the line QS

We know that MN || SR

In â–³ QRS we know that ON || SR

O is the midpoint of the diagonal QS

Hence, based on the converse midpoint theorem we know that N is the midpoint of QR.

Therefore, it is proved that N is the midpoint of QR.

**6. In a parallelogram PQRS, PQ = 12 cm and PS = 9 cm. The bisector of âˆ P meets SR in M. PM and QR both when produced meet at T. Find the length of RT.**

**Solution:**

From the figure we know that PM is the bisector of âˆ P

So we get

âˆ QPM = âˆ SPM â€¦â€¦. (1)

We know that PQRS is a parallelogram

From the figure we know that PQ || SR and PM is a transversal âˆ QPM and âˆ PMS are alternate angles

âˆ QPM = âˆ PMS â€¦â€¦â€¦ (2)

Consider equation (1) and (2)

âˆ SPM = âˆ PMS â€¦â€¦â€¦. (3)

We know that the sides opposite to equal angles are equal

MS = PS = 9 cm

âˆ RMT and âˆ PMS are vertically opposite angles

âˆ RMT = âˆ PMS â€¦â€¦â€¦. (4)

We know that PS || QT and PT is the transversal

âˆ RTM = âˆ SPM

It can be written as

âˆ RTM = âˆ RMT

We know that the sides opposite to equal angles are equal

RT = RM

We get

RM = SR â€“ MS

By substituting the values

RM = 12 â€“ 9

RM = 3 cm

RT = RM = 3cm

Therefore, the length of RT is 3 cm.

**7. In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced to meet at E. Also, AC and PQ intersect at R. Prove that **

**(i) DQ = QE,**

**(ii) PR || AB and**

**(iii) AR = RC.**

**Solution:**

(i) In â–³ QCD and â–³ QBE

We know that âˆ DQC and âˆ BQE are vertically opposite angles

So we get

âˆ DQC = âˆ BQE

From the figure we know that Q is the midpoint of BC

It can be written as

CQ = BQ

We know that AE || DC and BC is a transversal

From the figure we know that âˆ QDC and âˆ QEB are alternate angles

âˆ QDC = âˆ QEB

By ASA congruence criterion

â–³ QCD â‰… â–³ QBE

DQ = QE (c. p. c. t)

Therefore, it is proved that DQ = QE

(ii) Based on the midpoint theorem

We know that PQ || AE

From the figure we know that AB is a part of AE

So we have PQ || AB

We know that the intercepts on AD are made by the lines AB, PQ and DC

So we get PQ || AB || DC

We know that PR is a part of PQ and is parallel to AB

So we get PR || AB || DC

Therefore, it is proved that PR || AB.

(iii) From the figure we know that the lines PR, AB and DC are cut by AC and AD respectively.

Based on the intercept theorem we know that AR = RC.

Therefore, it is proved that AR = RC.

**8. In the adjoining figure, AD is a median of â–³ ABC and DE || BA. Show that BE is also a median of â–³ ABC.**

**Solution:**

Consider â–³ ABC

It is given that DE || AB

We know that D is the midpoint of BC

Based on the midpoint theorem we know that

E is the midpoint of AC

So we know that BE is the median of â–³ ABC drawn through B.

Therefore, it is proved that BE is also a median of â–³ ABC.

**9. In the adjoining figure, AD and BE are the medians of â–³ ABC and DF || BE. Show that CF = Â¼ AC.**

**Solution:**

Consider â–³ CBE

From the figure we know that D is the midpoint of BC and DF is parallel to BE.

By using the midpoint theorem

We know that F is the midpoint of EC.

It can be written as

CF = Â½ EC

We know that BE is the median through B

So we get

CF = Â½ [(1/2) AC]

CF = Â¼ AC

Therefore, it is proved that CF = Â¼ AC.

**10. Prove that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.**

**Solution:**

From the figure we know that F and E are the mid points of AB and AC

Based on the mid-point theorem

EF = Â½ BC

In the same way

FD = Â½ AC and ED = Â½ AB

Consider â–³ AFE and â–³ BFD

We know that AF = FB

Based on the midpoint theorem

FE = Â½ BC = BD

FD = Â½ AC = AE

By SSS congruence criterion

â–³ AFE â‰… â–³ BFD

Consider â–³ BFD and â–³ FED

We know that FE || BC

So we get FE || BD and AB || ED

Using the midpoint theorem

FB || ED

Hence, BDEF is a parallelogram

So we know that FD is a diagonal which divides the parallelogram into two congruent triangles

â–³ BFD â‰… â–³ FED

In the same way we can prove that FECD is a parallelogram

â–³ FED â‰… â–³ EDC

So we know that â–³ BFD, â–³ FDE, â–³ FED and â–³ EDC are congruent to each other.

Therefore, it is proved that the line segments joining the middle points of the sides of a triangle divide it into four congruent triangles.

**11. In the adjoining figure, D, E, F are the midpoints of the sides BC, CA and AB respectively, of â–³ ABC. Show that âˆ EDF = âˆ A, âˆ DEF = âˆ B and âˆ DFE = âˆ C.**

**Solution:**

Consider â–³ AFE and â–³ DFE

Using the midpoint theorem

AF = Â½ AB = ED

AE = Â½ AC = FD

FE is common i.e. FE = EF

By SSS congruence criterion

â–³ AFE â‰… â–³ DFE

âˆ A = âˆ FDE (c. p. c. t)

In the same way we get âˆ B = âˆ DEF and âˆ C= âˆ DFE.

Therefore, it is proved that âˆ EDF = âˆ A, âˆ DEF = âˆ B and âˆ DFE = âˆ C.

**12. Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.**

**Solution:**

Consider â–³ ABC

We know that P and Q are the midpoints of AB and BC

Based on the midpoint theorem

We know that PQ || AC and PQ = Â½ AC

Consider â–³ ADC

Based on the midpoint theorem

We know that RS || AC and RS = Â½ AC

It can be written as PQ || RS and

PQ = RS = Â½ AC â€¦â€¦. (1)

Consider â–³ BAD

We know that P and S are the midpoints of AB and AD

Based on the midpoint theorem

We know that PS || BD and PS = Â½ DB

Consider â–³ BCD

We know that RQ || BD and RQ = Â½ DB

It can be written as PS || RQ and

PS = RQ = Â½ DB â€¦â€¦â€¦ (2)

We know that the diagonals of a rectangle are equal

It can be written as

AC = BD â€¦â€¦â€¦. (3)

Comparing equations (1), (2) and (3)

We know that

PQ || RS and PS || RQ

So we get

PQ = QR = RS = SP

Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

**13. Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.**

**Solution:**

Consider â–³ ABC

We know that P and Q are the mid points of AB and BC

By using the midpoint theorem

We know that PQ || AC and PQ = Â½ AC

Consider â–³ ADC

We know that RS || AC and RS = Â½ AC

It can be written as PQ || RS and

PR = RS = Â½ AC â€¦â€¦. (1)

Consider â–³ BAD

We know that P and S are the mid points of AB and AD

Based on the midpoint theorem

We know that PS || BD and PS = Â½ DB

Consider â–³ BCD

We know that RQ || BD and RQ = Â½ DB

It can be written as PS || RQ and

PS = RQ = Â½ DB â€¦â€¦. (2)

By considering equations (1) and (2)

The diagonals intersects at right angles in a rhombus

So we get

âˆ EQF = 90^{o}

We know that RQ || DB

So we get RE || FO

In the same way SR || AC

So we get FR || OE

So we know that OERF is a parallelogram.

We know that the opposite angles are equal in a parallelogram

So we get

âˆ FRE = âˆ EOF = 90^{o}

So we know that PQRS is a parallelogram having âˆ R = 90^{o}

Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

**14. Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.**

**Solution:**

Consider â–³ ABC

We know that E and F are the midpoints

Based on the midpoint theorem

We know that EF || AC and EF = Â½ AC

Consider â–³ ADC

We know that H and G are the midpoints

Based on the midpoint theorem

We know that HG || AC and HG = Â½ AC

So we get EF || HG and

EF = HG = Â½ AC â€¦â€¦â€¦.. (1)

Consider â–³ BAD

We know that H and E are the midpoints

Based on the midpoint theorem

We know that HE || BD and HE = Â½ BD

Consider â–³ BCD

We know that G and F are the midpoints

Based on the midpoint theorem

We know that GF || BD and GF = Â½ BD

So we get HE || GF and

HE = GF = Â½ BD â€¦â€¦.. (2)

We know that the diagonals of a square are equal

So we get

AC = BD â€¦â€¦.. (3)

By using equations (1), (2) and (3)

So we get GF || BD and HE || GF

We have

EF = GH = GH = HE

We know that EFGH is a rhombus

From the figure we know that the diagonals of a square are equal and intersect at right angles

âˆ DOC = 90^{o}

We know that the sum of adjacent angles of parallelogram is 180^{o}

It can be written as

âˆ DOC + âˆ GKO = 180^{o}

By substituting the values

90^{o} + âˆ GKO = 180^{o}

On further calculation

âˆ GKO = 180^{o} – 90^{o}

By subtraction

âˆ GKO = 90^{o}

From the figure we know that âˆ GKO and âˆ EFG are corresponding angles

We get âˆ GKO = âˆ EFG = 90^{o}

We know that âˆ EFG = 90^{o}

Hence, EFGH is a square.

Therefore, it is proved that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square.

**15. Prove that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.**

**Solution:**

Construct lines KH, BD and GL

We know that K and H are the midpoints of AD and AB

Consider â–³ ABD

Using the mid-point theorem

We get

KH = Â½ BD

Consider â–³ CBD

Based on the midpoint theorem

GL = Â½ BD

So we get

KH = GL

Consider â–³ KOH and â–³ GOL

We know that KH = GL

From the figure we know that alternate angles are equal

So we get

âˆ OKH = âˆ GLO and âˆ OHK = âˆ OGL

By SAS congruence criterion

â–³ KOH â‰… â–³ GOL

OG = OH and OK = OL (c. p. c. t)

So we know that GH and KL bisect each other.

Therefore, it is proved that the line segments joining the midpoints of opposite sides of a quadrilateral bisect each other.

**16. The diagonals of a quadrilateral ABCD are equal. Prove that the quadrilateral formed by joining the midpoints of its sides is a rhombus.**

**Solution:**

Consider â–³ ABC

We know that P and Q are the midpoints of AB and BC

So we get PQ || AC and

PQ = Â½ AC â€¦â€¦ (1)

Consider â–³ BCD

We know that Q and R are the midpoints of BC and CD

So we get QR || BD and

QR = Â½ BD â€¦â€¦. (2)

Consider â–³ ADC

We know that S and R are the midpoints of AD and CD

So we get RS || AC and

RS = Â½ AC â€¦â€¦. (3)

Consider â–³ ABD

We know that P and S are the midpoints of AB and AD

So we get SP || BD and

SP = Â½ BD â€¦â€¦. (4)

Using all the equations

PQ || RS and QR || SP

Thus, PQRS is a parallelogram

It is given AC = BD

We can write it as

Â½ AC = Â½ BD

From the equations we get

PQ = QR = RS = SP

PQRS is a rhombus

Therefore, it is proved that the quadrilateral formed by joining the midpoints of its sides is a rhombus.

**17. The diagonals of a quadrilateral ABCD are perpendicular to each other. Prove that the quadrilateral formed by joining the midpoints of its sides is a rectangle.**

**Solution:**

Consider â–³ ABC

We know that P and Q are the midpoints of AB and BC

So we get PQ || AC and

PQ = Â½ AC â€¦â€¦ (1)

Consider â–³ ADC

We know that R and S are the midpoints of CD and AD

So we get RS || AC and

RS = Â½ AC â€¦â€¦.. (2)

Using the equations

We get PQ || RS and PQ = RS

We know that a pair of opposite sides are equal in a parallelogram

We know that AC and BD are the diagonals intersecting at point O

Consider â–³ ABD

We know that P and S are the midpoints of AD and AB

So we get

PS || BD

It can be written as

PN || MO

Based on equation (1)

We get

PM || NO

So we know that PM || NO and PN || MO

We know that opposite angles are equal in a parallelogram

âˆ MPN = âˆ MON

We know that

âˆ BOA = âˆ MON

So we get

âˆ MPN = âˆ BOA

We know that AC âŠ¥ BD and âˆ BOA = 90^{o}

It can be written as

âˆ MPN = 90^{o}

So we get

âˆ QPS = 90^{o}

So we know that PQRS is a parallelogram with âˆ QPS = 90^{o}

Therefore, it is proved that the quadrilateral formed by joining the midpoints of its sides is a rectangle.

**18. The midpoints of the sides AB, BC, CD and DA of a quadrilateral ABCD are joined to form a quadrilateral. If AC = BD and AC âŠ¥ BD then prove that the quadrilateral formed is a square.**

**Solution:**

Consider â–³ ABC

We know that P and Q are the midpoints of AB and BC

So we get PQ || AC and

PQ = Â½ AC â€¦â€¦ (1)

Consider â–³ BCD

We know that Q and R are the midpoints of BC and CD

So we get QR || BD and

QR = Â½ BD â€¦â€¦. (2)

Consider â–³ ACD

We know that S and R are the midpoints of AD and CD

So we get RS || AC and

RS = Â½ AC â€¦â€¦.. (3)

Consider â–³ ABD

We know that P and S are the midpoints of AB and AD

So we get SP || BD and

SP = Â½ BD â€¦â€¦. (4)

Consider all the equations

PQ || RS and QR || SP

Hence, PQRS is a parallelogram

It is given that AC = BD

It can be written as

Â½ AC = Â½ BD

So we get

PQ = QR = RS = SP

We know that AC and BD intersect at point O

So we get PS || BD

PN || MO

Based on equation (1)

We get PQ || AC

PM || NO

We know that the opposite angles are equal in a parallelogram

âˆ MPN = âˆ MON

We know that âˆ BOA = âˆ MON

So we get

âˆ MPN = âˆ BOA

We know that AC âŠ¥ BD and âˆ BOA = 90^{o}

So we get

âˆ MPN = 90^{o}

It can be written as

âˆ QPS = 90^{o}

We know that PQ = QR = RS = SP

Therefore, it is proved that PQRS is a square.

### RS Aggarwal Solutions for Class 9 Maths Chapter 10: Quadrilaterals

Chapter 10, Quadrilaterals, has 3 exercises with problems based on the types of quadrilaterals and various theorems of parallelograms. The concepts which are discussed in RS Aggarwal Solutions under chapter 10 are as follows:

- Various types of quadrilaterals
- Results on Parallelograms
- Converse of the above theorems
- Some results on rectangle, rhombus and square
- Midpoint Theorem

### RS Aggarwal Solutions Class 9 Maths Chapter 10 – Exercise list

Exercise 10A Solutions 10 Questions

Exercise 10B Solutions 27 Questions

Exercise 10C Solutions 18 Questions

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

The students can use the RS Aggarwal Solutions PDF as a vital resource to speed up their exam preparation. The solutions contain explanations in simple language to help students grasp the concepts faster. By solving the exercise wise problems regularly, the students can obtain a better hold on the topics covered in the chapter.

Quadrilaterals majorly help architects to design buildings based on the utilization of space. They are used in electronic devices like laptops, mobiles, TV etc. and in stationary items like copies, books, chart papers etc. The students, by downloading the PDF of this chapter can obtain exercise wise answers which are prepared based on their understanding level.