# ML Aggarwal Quadrilaterals Shapes Exe-13.2 Class 8 ICSE Ch-13 Maths Solutions

ML Aggarwal Quadrilaterals Shapes Exe-13.2 Class 8 ICSE Ch-13 Maths Solutions. We Provide Step by Step Answer of Exe-13.2 Questions for Quadrilaterals Shapes as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-8.

**ML Aggarwal Quadrilaterals Shapes Exe-13.2 Class 8 ICSE Maths Solutions**

Board | ICSE |

Publications | Avichal Publishig Company (APC) |

Subject | Maths |

Class | 8th |

Chapter-13 | Quadrilaterals Shapes |

Writer | ML Aggarwal |

Book Name | Understanding |

Topics | Solution of Exe-13.2 Questions |

Edition | 2023-2024 |

**Quadrilaterals Shapes Exe-13.2**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-237

**Question 1. In the given figure, ABCD is a parallelogram. Complete each statement along with the definition or property used.**

(i) AD = ………..

(ii) DC = ………..

(iii) ∠DCB = ………..

(iv) ∠ADC = ………..

(v) ∠DAB = ………..

(vi) OC = ………..

(vii) OB = ………..

(viii) m∠DAB + m∠CDA = ………..

**Answer:**

In paralleloram ABCD

(i) AD = 6 cm (Opposite sides of parallelogram)

(ii) DC = 9 cm (Opposite sides of parallelogram)

(iii) ∠DCB = 60° (∵ ∠DCB + ∠CBA = 180°)

(iv) ∠ADC = ∠ABC = 120°

(v) ∠DAB = ∠DCB = 60°

(vi) OC = AO = 7 cm

(vii) OB = OD = 5 cm

(viii) m∠DAB + m∠CDA = 180°

**Quadrilaterals Shapes Exe-13.2**

ML Aggarwal Class 8 ICSE Maths Solutions

Page-238

**Question 2. Consider the following parallelograms. Find the values of x, y, z in each.**

**Answer**:

**(i)**

**From the figure, ∠POQ = 120 ^{o}**

Sum of angles linear pair is equal to 180^{o}

So, ∠POQ + ∠PON = 180^{o}

120^{o} + ∠PON = 180^{o}

∠PON = 180^{o} – 120^{o}

∠PON = 60^{o}

∠M = ∠O = 60^{o} … [because opposite angles of parallelogram are equal]

∠POQ = ∠MNO

120^{o} = 120^{o} … [because corresponding angels are equal]

y = 120^{o}

z = y

120^{o} = 120^{o}

Hence, x = 60^{o}, y = 120^{o} and z = 120^{o}

**(ii)**

**From the figure, it is given that ∠PQO = 100 ^{o}, ∠OMN = 30^{o}, ∠PMO =40^{o}.**

Then, ∠NOM = ∠OMP … [because alternate angles are equal]

So, z = 40^{o}

Now, ∠NMO = ∠POM … [because alternate angles are equal]

So, ∠NMO = a = 30^{o}

Consider the triangle PQO,

Sum of measures of interior angles of triangle is equal to 180^{o}.

∠P + ∠Q + ∠O = 180^{o}

x + 100^{o} + 30^{o} = 180^{o}

x + 130^{o} = 180^{o}

x = 180^{o} – 130^{o}

x = 50^{o}

Then, exterior angle ∠OQP = y + z

100^{o} = y + 40^{o}

By transposing we get,

y = 100^{o} – 40^{o}

y = 60^{o}

Hence, the value of x = 50^{o}, y = 60^{o} and z = 40^{o}.

**(iii) **

∠SPR = ∠PRQ

35^{o} = 35^{o} … [because alternate angles are equal]

Now consider the triangle PQR,

Sum of measures of interior angles of triangle is equal to 180^{o}.

∠RPQ + ∠PQR + ∠PRQ = 180^{o}

z + 120^{o} + 35^{o} = 180^{o}

z + 155^{o} = 180^{o}

z = 180^{o} – 155^{o}

z = 25^{o}

Then, ∠QPR = ∠PRQ

Z = x

25^{o} = 25^{o} … [because alternate angles are equal]

in parallelogram opposite angles are equal.

So, ∠S = ∠Q

y = 120^{o}

Hence, value of x = 25^{o}, y = 120^{o} and ∠z = 25^{o}.

**(iv)**

From the figure, it is given that ∠SPR = 67^{o} and ∠PQR = 70^{o}

∠SPR = ∠PRQ

67^{o} = 67^{o} … [because alternate angles are equal]

Now, consider the triangle PQR

sum of measures of interior angles of triangle is equal to 180^{o}.

∠RPQ + ∠PQR + ∠PRQ = 180^{o}

x + 70^{o} + 67^{o} = 180^{o}

x + 137^{o} = 180^{o}

x = 180^{o }– 137^{o}

x = 43^{o}

Then, ∠PSR = ∠PQR

We know that, in parallelogram opposite angles are equal.

Z = 70^{o}

Also we know that, exterior angle ∠SRT = ∠PSR + ∠SPR

y = 70^{o} + 67^{o}

y = 137^{o}

Hence, value of x = 43^{o}, y = 137^{o} and z = 70^{o}

**Question 3. Two adjacent sides of a parallelogram are in the ratio 5 : 7. If the perimeter of parallelogram is 72 cm, find the length of its sides.**

**Answer:**

Consider the parallelogram PQRS,

From the question it is given that, two adjacent sides of a parallelogram are in the ratio 5 : 7.

Perimeter of parallelogram = 72 cm

2(SP + RQ) = 72 cm

SP + RQ = 72/2

SP + RQ = 36 cm

Let us assume the length of side SP = 5y and RQ = 7y,

5y + 7y = 36

12y = 36

y = 36/12

y = 3

Therefore, SP = 5y = 5 × 3 = 15 cm

RQ = 7y = 7 × 3 = 21 cm

**Question 4. The measure of two adjacent angles of a parallelogram are in the ratio 4 : 5. Find the measure of each angle of the parallelogram.**

**Answer:**

Consider the parallelogram PQRS,

From the question it is given that, The measure of two adjacent angles of a parallelogram is in the ratio 4 : 5.

So, ∠P: ∠Q = 4: 5

Let us assume the ∠P = 4y and ∠Q = 5y.

∠P + ∠Q = 180^{o}

4y + 5y = 180^{o}

9y = 180^{o}

y = 180^{o}/9

y = 20^{o}

Hence, ∠P = 4y = 4 × 20^{o} = 80^{o} and ∠Q = 5y = (5 × 20^{o}) = 100^{o}

In parallelogram opposite angles are equal,

So, ∠R = ∠P = 80^{o}

∠S = ∠Q = 100^{o}

**Question 5. Can a quadrilateral ABCD be a parallelogram, give reasons in support of your answer.**

(i) ∠A + ∠C= 180°?

(ii) AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm?

(iii) ∠B = 80°, ∠D = 70°?

(iv) ∠B + ∠C= 180°?

**Answer:**

From the question it is given that, quadrilateral ABCD can be a parallelogram.

In parallelogram opposite sides are equal and opposites angles are equal.

So, AB = DC and AD = BC also ∠A = ∠C and ∠B = ∠D.

**(i) ∠A + ∠C= 180 ^{o}**

From the above condition it may be a parallelogram and may not be a parallelogram.

**(ii) AD = BC = 6 cm, AB = 5 cm, DC = 4.5 cm**

From the above dimension not able to form parallelogram.

Because AB ≠ DC

**(iii) ∠B = 80 ^{o}, ∠D = 70^{o}**

From the above dimension not able to form parallelogram.

Because ∠B ≠ ∠D

**(iv) ∠B + ∠C= 180 ^{o}**

From the above condition it may be a parallelogram and may not be a parallelogram.

**Question 6. In the following figures HOPE and ROPE are parallelograms. Find the measures of angles x, y and z. State the properties you use to find them.**

**Answer:**

**(i) In parallelogram HOPE, HO is produced to D**

∠AOP + ∠POD = 180° (Linear pair)

∴ ∠AOP + 70°= 180°

∠AOP = 180°- 70°= 110°

But ∠AOP = ∠HEP (Opposite angles of a ||gm)

∠HEP = 110°

⇒ x = 110°

∠HPO = ∠EHP (Alternate angles)

∴ y = 40°

In ∆HOP, Ext. ∠POD = y + z

⇒ 70° = y + z

⇒ 70° = 40° + z

⇒ z = 70° – 40° = 30°

∴ x= 110°, y = 40°, z = 30°

**(ii) In ||gm ROPE, RO is produced to D**

∠POD = 80°, ∠EOP = 60°

∠P = ∠POD (Alternate angles)

∴ y = 80°

∠ROE + ∠EOP + ∠POD = 180° (Angles on one side of a line)

x + 60° + 80° = 180° ⇒ x + 140° = 180°

∴ x = 180°- 140° = 40°

z = x (Alternate angles)

∴ z = 40°

Hence, x = 40°, y = 80°, z = 40°

Quadrilaterals Shapes Exe-13.2

**ML Aggarwal Class 8 ICSE Maths Solutions**

Page-239

**Question 7. In the given figure TURN and BURN are parallelograms. Find the measures of x and y (lengths are in cm).**

**Answer:**

(i) Consider the parallelogram TURN

In parallelogram opposite sides are equal.

So, TU = RN

4x + 2 = 28

4x = 28 – 2

4x = 26

x = 26/4

x = 6.5 cm

and NT = RU

5y – 1 = 24

5y = 24 + 1

5y = 25

y = 25/5

y = 5

Hence, value of x = 6.5 cm and y = 5 cm.

**(ii) Consider the parallelogram BURN,**

BO = OR

x + y = 20 … [equation (i)]

UO = ON

x + 3 = 18

x = 18 – 3

x = 15

substitute the value of x in equation (i),

15 + y = 20

y = 20 – 15

y = 5

Hence, value of x = 15 and y = 5.

**Question 8. In the following figure both ABCD and PQRS are parallelograms. Find the value of x.**

**Answer:**

Two parallelograms ABCD and PQRS in which

∠A = 120° and ∠R = 50°

∠A + ∠B = 180° (Co-interior angles)

120° + ∠B = 180°

⇒ ∠B = 180°- 120° = 60°

∠P = ∠R (Opposite angles of a ||gm)

∠P = 50°

Now in ∆OPB,

∠POB + ∠P + ∠B = 180° (Angles of a triangle)

x + 50° + 60° = 180°

x + 110° = 180° ⇒ x = 180°- 110° = 70°

∴ x = 70°

**Question 9. In the given figure, ABCD, is a parallelogram and diagonals intersect at O. Find :**

(i) ∠CAD

(ii) ∠ACD

(iii) ∠ADC

**Answer:**

**(i) ∠DBC = ∠BDA = 46° (alternate angles)**

In ∆ AOD,

46° + 68° + ∠CAD = 180° (∵ ∠CAD = ∠OAD)

∠CAD = 180°- 114° = 66°

**(ii) ∠AOD + ∠COD = 180° (straight angle)**

∴ ∠COD= 180°- 68°= 112°

In ∆COD, 112° + 30° + ∠ACD = 180° (∵ ∠ACD = ∠OCD)

∠ACD = 180° – 112° – 30° = 38°

**(iii) ∠ADC = 30° + 46° = 76° (∵ ∠ADC = ∠ADO + ∠ODC)**

**(ML Aggarwal Quadrilaterals Shapes Exe-13.2 Class 8 ICSE Maths)**

**Question 10. In the given figure, ABCD is a parallelogram. Perpendiculars DN and BP are drawn on diagonal AC. Prove that:**

(i) ∆DCN ≅ ∆BAP

(ii) AN = CP

**Answer:**

In the given figure,

ABCD is a parallelogram AC is it’s one diagonal.

BP and DN are perpendiculars on AC.

To prove :

**(i) ∆DCN ≅ ∆BAP**

**(ii) AN = CP**

Proof: In ∆DCN and ∆BAP

DC=AB (Opposite sides of a ||gm)

∠N = ∠P (Each 90°)

∠DCN = ∠PAB (Alternate angle)

∴ ∆DCN ≅ ∆BAP (AAS axiom)

∴ NC = AP (c.p.c.t.)

Subtracting NP from both sides.

NC – NP = AP – NP

∴ AN = CP

**Question 11. In the given figure, ABC is a triangle. Through A, B and C lines are drawn parallel to BC, CA and AB respectively, which forms a ∆PQR. Show that**

**2(AB + BC + CA) = PQ + QR + RP.**

**Answer:**

From the figure it is given that,

Through A, B and C lines are drawn parallel to BC, CA and AB respectively.

We have to show that 2(AB + BC + CA) = PQ + QR + RP

Then, AB ∥ RC and AR ∥ CB

ABCR is a parallelogram.

So, AB = CR … [equation (i)]

CB = AR … [equation (ii)]

ABPC is a parallelogram.

AB ∥ CP and PB ∥ CA

AB = PC … [equation (iii)]

AC = PB … [equation (iv)]

ACBQ is a parallelogram

AC = BQ … [equation (v)]

AQ = BC … [equation (vi)]

By adding all the equation,

AB + AB + BC + BC + AC + AC = PB + PC + CR + AR + BQ +BC

2AB + 2BC + 2AC = PQ + QR + RP

**By taking common**

**2(AB + BC + AC) = PQ + QR + RP**

— End of Quadrilaterals Shapes **Exe-13.2 **Class 8 ICSE Maths Solutions :–

Return to : **– **ML Aggarwal Maths Solutions for ICSE Class -8

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