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

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.

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

Answer:

(i)

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

From the figure, ∠POQ = 120o

Sum of angles linear pair is equal to 180o

So, ∠POQ + ∠PON = 180o

120o + ∠PON = 180o

∠PON = 180o – 120o

∠PON = 60o

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

∠POQ = ∠MNO

120o = 120o … [because corresponding angels are equal]

y = 120o

z = y

120o = 120o

Hence, x = 60o, y = 120o and z = 120o

(ii)

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

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

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

So, z = 40o

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

So, ∠NMO = a = 30o

Consider the triangle PQO,

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

∠P + ∠Q + ∠O = 180o

x + 100o + 30o = 180o

x + 130o = 180o

x = 180o – 130o

x = 50o

Then, exterior angle ∠OQP = y + z

100o = y + 40o

By transposing we get,

y = 100o – 40o

y = 60o

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

(iii) 

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

∠SPR = ∠PRQ

35o = 35o … [because alternate angles are equal]

Now consider the triangle PQR,

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

∠RPQ + ∠PQR + ∠PRQ = 180o

z + 120o + 35o = 180o

z + 155o = 180o

z = 180o – 155o

z = 25o

Then, ∠QPR = ∠PRQ

Z = x

25o = 25o … [because alternate angles are equal]

in parallelogram opposite angles are equal.

So, ∠S = ∠Q

y = 120o

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

(iv)

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

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

∠SPR = ∠PRQ

67o = 67o … [because alternate angles are equal]

Now, consider the triangle PQR

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

∠RPQ + ∠PQR + ∠PRQ = 180o

x + 70o + 67o = 180o

x + 137o = 180o

x = 180– 137o

x = 43o

Then, ∠PSR = ∠PQR

We know that, in parallelogram opposite angles are equal.

Z = 70o

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

y = 70o + 67o

y = 137o

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

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:

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.

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:

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.

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 = 180o

4y + 5y = 180o

9y = 180o

y = 180o/9

y = 20o

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

In parallelogram opposite angles are equal,

So, ∠R = ∠P = 80o

∠S = ∠Q = 100o

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.

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

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

(i) ∠A + ∠C= 180o

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 = 80o, ∠D = 70o

From the above dimension not able to form parallelogram.

Because ∠B ≠ ∠D

(iv) ∠B + ∠C= 180o

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.

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
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.
∠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
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.
∠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).

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.

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°

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

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

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

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

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.

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

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

Thanks

 Share with your friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.