Category: CBSE Class 8

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2.

• Data Handling Class 8 Ex 5.1
• Data Handling Class 8 Ex 5.3

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 5
Chapter Name	Data Handling
Exercise	Ex 5.2
Number of Questions Solved	5
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2

Question 1.
A survey was made to find the type of music that a certain group of young people liked in a city. The adjoining pie chart shows the findings of this survey.

From this pie chart answer the following:
(i) If 20 people liked classical music, how many young people were surveyed?
(ii) Which type of music is liked by the maximum number of people?
(iii) If a cassette company were to make 1000 CD’s, how many of each type would they make?
Solution.
(i) Suppose that x young people were surveyed. Then, the number of young people who liked classical music = 10% of x
\(\frac { 10 }{ 100 } \times x=\frac { x }{ 10 } \)

According to the question,
\(\frac { x }{ 10 } =20\)
⇒ x = 20 x 10
⇒ x = 200
Hence, 200 young people were surveyed.

(ii) Light music is liked by the maximum number of people.

(iii) Total number of CD’s = 1000 Number of CD’s of Semi Classical music = 20% of 1000
⇒ \(\frac { 20 }{ 100 } \times 1000=200\)
Number of CD’s of Classical music = 10% of 1000
⇒ \(\frac { 10 }{ 100 } \times 1000=100\)
Number of CD’s of Folk music = 30% of 1000
⇒ \(\frac { 30 }{ 100 } \times 1000=300\)
Number of CD’s of Light music = 40% of 1000
⇒ \(\frac { 40 }{ 100 } \times 1000=400\)

Question 2.
A group of 360 people was asked to vote for their favorite season from the three seasons rainy, winter and summer.

(i) Which season got the most votes?
(ii) Find the central angle of each sector.
(iii) Draw a pie chart to show this information.
Solution.
(i) Winter season got the most votes.

(ii) Total votes = 90 + 120 + 150 = 360. Central angle of winter sector
\(= \frac { Number\quad of\quad people\quad who\quad vote\quad for\quad winter\quad season }{ Total\quad number\quad of\quad people } \times { 360 }^{ \circ }\)
= \(\frac { 150 }{ 360 } \times { 360 }^{ \circ }={ 150 }^{ \circ }\)
Central angle of summer sector
\(= \frac { Number\quad of\quad people\quad who\quad vote\quad for\quad summer\quad season }{ Total\quad number\quad of\quad people } \times { 360 }^{ \circ }\)
= \(\frac { 90 }{ 360 } \times { 360 }^{ \circ }={ 90 }^{ \circ }\)
Central angle of rainy sector
\(= \frac { Number\quad of\quad people\quad who\quad vote\quad for\quad rainy\quad season }{ Total\quad number\quad of\quad people } \times { 360 }^{ \circ }\)
= \(\frac { 120}{ 360 } \times { 360 }^{ \circ }={ 120 }^{ \circ }\)

(iii)

Question 3.
Draw a pie chart showing the following information. The table shows the colors preferred by a group of people.

Find the proportion of each sector. For example, Blue is \(\frac { 18 }{ 36 } =\frac { 1 }{ 2 } \) ; Green is \(\frac { 9 }{ 36 } =\frac { 1 }{ 4 } \) ; and so on. Use this to find the corresponding angles.
Solution.

Question 4.
The adjoining pie chart gives the marks scored in an examination by a student in Hindi, English, Mathematics, Social Science and Science. If the total marks obtained by the students were 540, answer the following questions.

(i) In which subject did the student score 105 marks?
(Hint: For 540 marks, the central angle = 360°. So, for 105 marks, what is the central angle ?)
(ii) How many more marks were obtained by the student in Mathematics than in Hindi?
(iii) Examine whether the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.
(Hint; Just study the central angles).
Solution.
(i) Total marks = 540
∵ Central angle corresponding to 540 = 360°
∴ Central angle corresponding to 105
\(\frac { { 360 }^{ \circ } }{ 540 } \times \left( 105 \right) ={ 70 }^{ \circ }\)
Since the sector having central angle 70° is corresponding to Hindi, therefore, the student scored 105 marks in Hindi.

(ii) Central angle corresponding to the sector of Mathematics = 90°
∴ Marks obtained by the student in Mathematics
\(\frac { { 90 }^{ \circ } }{ { 360 }^{ \circ } } \times 540=135\).
Marks obtained by the student in Hindi = 105.
Hence, the student obtained 135 – 105 = 30 marks more in Mathematics than in Hindi.

(iii) Sum of the central angles for Social Science and Mathematics
= 65° + 90° = 155°
Sum of the central angles for Science and Hindi
= 80° + 70° = 150°
Since the marks obtained are proportional to the central angles corresponding to various subjects and 155° > 150°, therefore the sum of the marks obtained in Social Science and Mathematics is more than that in Science and Hindi.

Question 5.
The number of students in a hostel, speaking different languages is given below. Display the data in a pie chart.

Solution.

 

We hope the NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.2
• Practical Geometry Class 8 Ex 4.3
• Practical Geometry Class 8 Ex 4.4

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 4
Chapter Name	Practical Geometry
Exercise	Ex 4.5
Number of Questions Solved	1
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5

Question 1.
Draw the following:
1. The square READ with RE = 5.1 cm.
2. A rhombus whose diagonals are 5.2 cm and 6.4 cm long.
3. A rectangle with adjacent sides of lengths 5 cm and 4 cm.
4. A parallelogram OKAY where OK = 5.5 cm and KA = 4.2 cm. Is it unique?
Solution.
1. Steps of Construction

1. Draw RE = 5.1 cm.
2. At R, draw a ray RX such that ∠ERX
   
3. From ray RX, cut RD = 5.1 cm.
4. At E, draw a ray EY such that ∠REY = 90°.
5. From ray EY, cut EA = 5.1 cm.
6. Join AD.

Then, READ is the required square.

2. Steps of Construction
[We know that the diagonals of a rhombus bisect each other at right angles. So in rhombus ABCD, the diagonals AC and BD will bisect each other at right angles.]

1. Draw AC = 5.2 cm.
2. Construct its perpendicular bisector. Let it intersect AC at O.
   
3. Cut off \(\frac { 6.4 }{ 2 } \)= 3.2 cm lengths on either side of the bisector drawn in step 2, we get B and D.
4. Join AB, BC, CD, and DA.

Then, ABCD is the required rhombus.

3. Steps of Construction
[We know that each angle of a rectangle is 90°. So, in rectangle PQRS,
∠P=∠Q=∠R=∠S= 90°.
Also, opposite sides of a rectangle are parallel.
So, in rectangle PQRS,
PQ || SR and PS || QR]

1. Draw PQ = 5 cm.
2. At Q, draw a ray QX such that ∠PQX = 90°.
   
3. From ray QX, cut QR = 4 cm.
4. At P, draw a ray PY parallel to QR.
5. At R, draw a ray RZ parallel to QP to meet the ray drawn in step 4 at S.

Then, PQRS is the required rectangle.

4. Steps of Construction
[We know that in a parallelogram, opposite sides are parallel and equal. So,
OK = YA and OK || YA;
KA = OY and KA || OY]

1. Draw OK = 5.5 cm.
   
2. At K, draw a ray KX at any suitable angle from OK.
3. From ray KX, cut KA = 4.2 cm.
4. A, draw a ray AT parallel to KO.
5. At O, draw a ray OZ parallel to KA to cut the ray drawn in step 4 at Y.

Then, OKAY is the required parallelogram.
This is not unique.
Note: We can construct countless parallelograms with these dimensions by varying ∠OKA

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.2
• Practical Geometry Class 8 Ex 4.3
• Practical Geometry Class 8 Ex 4.5

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 4
Chapter Name	Practical Geometry
Exercise	Ex 4.4
Number of Questions Solved	1
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4

Question 1.
Construct the following quadrilaterals:
(i) Quadrilateral DEAR
DE = 4 cm
EA = 5 cm
AR = 4.5 cm
∠E = 60°
∠A = 90°

(ii) Quadrilateral TRUE
TR = 3.5 cm
RU = 3 cm
UE = 4 cm
∠R = 75°
∠U= 120°
Solution.
(i) Steps of Construction

1. Draw DE = 4 cm.
2. At E, draw ray EX such that ∠DEX = 60°.
3. From ray EX, cut EA = 5 cm.
   
4. At A, draw ray AY such that ∠EAY = 90°.
5. Cut AR = 4.5 cm from ray AY.
6. Join RD.

Then, DEAR is the required quadrilateral.

(ii) Steps of Construction

1. Draw TR = 3.5 cm.
2. At R, draw ray RX such that ∠TRX = 75°.
   
3. Cut RU = 3 cm from ray RX.
4. At U, draw ray UY such that ∠RUY = 120°.
5. Cut UE = 4 cm from ray UY.
6. Join ET.

Then, TRUE is the required quadrilateral.

 

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.2
• Practical Geometry Class 8 Ex 4.4
• Practical Geometry Class 8 Ex 4.5

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 4
Chapter Name	Practical Geometry
Exercise	Ex 4.3
Number of Questions Solved	1
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3

Question 1.
Construct the following quadrilaterals :
(i) Quadrilateral MORE
MO = 6 cm
OR = 4.5 cm
∠M = 60°
∠O = 105°
∠R = 105°

(ii) Quadrilateral PLAN
PL = 4 cm
LA = 6.5 cm
∠P = 90°
∠A = 110°
∠N – 85°.

(iii) Parallelogram HEAR
HE = 5 cm
EA = 6 cm
∠R = 85°

(iv) Rectangle OKAY
OK = 7 cm
KA = 5 cm.
Solution.
(i) Steps of Construction

1. Draw MO = 6 cm.
2. At 0, draw ray OX such that Z MOX = 105°.
   
3. Cut OR = 4.5 cm from ray OX.
4. At M, draw ray MY such that ∠OMY = 60°.
5. At R, draw ray RZ such that ∠ORZ = 105°.
6. Let the rays MY and RZ meet at E.

Then, MORE is the required quadrilateral.

(ii) Steps of Construction

1. Draw PL = 4 cm.
2. At L, draw ray LX such that ∠PLX = 75°.
   
   By Angle-sum property of a quadrilateral,
   ∠P + ∠A + ∠N + ∠L = 360°
   ⇒ 90° + 110° + 85° + Z L = 360°
   ⇒ 285° + ∠ L = 360°
   ⇒ ∠L = 360° – 285°
   ⇒ ∠L = 75°.
3. Cut LA = 6.5 cm from ray LX.
4. At A, draw ray AY such that ∠LAY = 110°.
5. At P, draw ray PZ such that ∠LPZ = 90°.
   Let the rays AY and PZ meet at N.

Then, PLAN is the required quadrilateral.

(iii) Steps of Construction

1. Draw HE = 5 cm.
2. At E, draw ray EX such that ∠HEX = 85°
   ∴ Opposite angles of a parallelogram are equal.
   ∵ ∠E = ∠R = 85°
3. Cut EA = 6 cm from the ray EX.
   
4. With A as centre and radius AR = 5 cm, draw an arc.
5. With H as centre and radius HR = 6 cm; draw another arc to intersect the arc drawn in step 4 at R.
   ∴ opposite sides of a parallelogram are equal in length
   ∵ AR = EH = 5 cm
   and HR = EA = 6 cm
6. Join AR and HR.

Then, HEAR is the required parallelogram.

(iv) Steps of Construction
[We know that each angle of a rectangle
is 90°.
∴ ∠O=∠K=∠A=∠Y= 90°.
Also, opposite sides of a rectangle are equal in length.
∴ OY = KA = 5 cm and AY = KO = 7 cm]

1. Draw OK = 7 cm.
2. At K, draw ray KX such that ∠OKX = 90°.
3. Cut KA – 5 cm from ray KX.
   
4. Taking A as centre and radius AY = 7 cm, draw an arc.
5. Taking O as centre and radius OY = 5 cm, draw another arc to intersect the arc drawn in step 4 at Y.
6. Join AY and OY.

Then OKAY is the required rectangle.

 

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2.

• Practical Geometry Class 8 Ex 4.1
• Practical Geometry Class 8 Ex 4.3
• Practical Geometry Class 8 Ex 4.4
• Practical Geometry Class 8 Ex 4.5

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 4
Chapter Name	Practical Geometry
Exercise	Ex 4.2
Number of Questions Solved	1
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2

Question 1.
Construct the following quadrilaterals:
(i) Quadrilateral LIFT
LI = 4 cm
IF = 3 cm
TL = 2.5 cm
LF = 4.5 cm
IT = 4 cm

(ii) Quadrilateral GOLD
OL = 7.5 cm
GL = 6 cm
GD = 6 cm
LD = 5 cm
OD = 10 cm

(iii) Rhombus BEND
BN – 5.6 cm
DE = 6.5 cm
Solution.
(i) Steps of Construction

1. Draw LI = 4 cm.
2. With L as center and radius LT = 2.5 cm, draw an arc.
   
3. With I as center and radius, IT = 4 cm, draw another arc to intersect the arc drawn in step 2 at T.
4. With I as center and radius IF = 3 cm, draw an arc.
5. With L as center and radius LF = 4.5 cm, draw another arc to intersect the arc drawn in step 4 at F.
6. Join IF, FT, TL, LF and IT.

Then, LIFT is the required quadrilateral.

(ii) Steps of Construction

1. Draw LD = 5 cm.
2. With L as center and radius LG = 6 cm, draw an arc.
   
3. With D as center and radius DG = 6 cm, draw another arc to intersect the arc drawn in step 2 at G.
4. With L as center and radius LO = 7.5 cm, draw an arc.
5. With D as center and radius DO = 10 cm, draw another arc to intersect the arc drawn in step 4 at O.
6. Join DG, GO, OL, LG and DO.

Then GOLD is the required quadrilateral.

(iii) Steps of Construction

1. Draw DE = 6.5 cm.
2. Draw perpendicular bisector PQ of DE so as to intersect DE at M. Then M is the mid-point of DE.
3. With M as centre and radius
   \(=\frac { 1 }{ 2 } \times \left( 5.6 \right) =2.8 cm\)
   opposite sides of DE to intersect MP at N and MQ at B.
   
4. Join DN, NE, EB, and BD.

Then, BEND is the required rhombus.

 

We hope the NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4.

• Understanding Quadrilaterals Class 8 Ex 3.1
• Understanding Quadrilaterals Class 8 Ex 3.2
• Understanding Quadrilaterals Class 8 Ex 3.3

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 3
Chapter Name	Understanding Quadrilaterals
Exercise	Ex 3.4
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4

Question 1.
State whether True or False :
(a) All rectangles are squares
(b) All rhombuses are parallelograms
(c) All squares are rhombuses and also rectangles
(d) All squares are not parallelograms
(e) All kites are rhombuses
(f) All rhombuses are kites
(g) All parallelograms are trapeziums
(h) All squares are trapeziums.
Solution.
(b), (c), (f), (g), (h) are true;
others are false.

Question 2.
Identify all the quadrilaterals that have.
(a) four sides of equal length
(b) four right angles
Solution.
(a) Rhombus; square
(b) Square; rectangle

Question 3.
Explain how a square is
(i) a quadrilateral
(ii)a parallelogram
(iii) a rhombus
(iv) a rectangle.
Solution.
(i) a quadrilateral
A square is 4 sided, so it is a quadrilateral.

(ii) a parallelogram
A square has its opposite sides parallel; so it is a parallelogram.

(iii) a rhombus
A square is a parallelogram with all the 4 sides equal, so it is a rhombus.

(iv) a rectangle
A square is a parallelogram with each angle a right angle; so it is a rectangle.

Question 4.
Name the quadrilaterals whose diagonals :
(i) bisect each other
(ii) are perpendicular bisectors of each other
(iii) are equal.
Solution.
(i) bisect each other
The names of the quadrilaterals whose diagonals bisect each other are parallelogram; rhombus; square; rectangle.

(ii) are perpendicular bisectors of each other
The names of the quadrilaterals whose diagonals are perpendicular bisectors of each other are rhombus; square.

(iii) are equal
The names of the quadrilaterals whose diagonals are equal are square; rectangle.

Question 5.
Explain why a rectangle is a convex quadrilateral.
Solution.
A rectangle is a convex quadrilateral because both of its diagonals lie wholly in its interior.

Question 6.
ABC is a right-angled triangle and O is the mid-point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you).

Solution.
Construction: Produce BO to D such that BO = OD. Join AD and CD.
Proof. AO = OC ∵ O is the mid-point of AC
BO = OD By construction
∴ Diagonals of quadrilateral ABCD bisect each other.
∴ Quadrilateral ABCD is a parallelogram.
Now, ∠ABC = 90° given
∴ ABCD is a rectangle.
Since the diagonals of a rectangle bisect each other, therefore,
O is the mid-point of AC and BD both. But AC = BD
∵ Diagonals of a rectangle are equal
∴ OA = OC =\(\frac { 1 }{ 2 } \)AC =\(\frac { 1 }{ 2 } \)BD = OB
⇒ OA = OB = OC.

We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.4, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3.

• Understanding Quadrilaterals Class 8 Ex 3.1
• Understanding Quadrilaterals Class 8 Ex 3.2
• Understanding Quadrilaterals Class 8 Ex 3.4

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 3
Chapter Name	Understanding Quadrilaterals
Exercise	Ex 3.3
Number of Questions Solved	12
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3

Question 1.
Given a parallelogram ABCD. Complete each statement along with the definition with the definition or property used.

(i) AD = ………
(ii) ∠DCB = …………….
(iii) OC = ……………….
(iv) m∠DAB + m∠CDA = …………..
Solution.
(i) AD = BC
Opposite sides of a parallelogram are equal

(ii) ∠DCB = ∠DAB
Opposite angles of a parallelogram are equal

(iii) OC = OA
∵ Diagonals of a parallelogram bisect each other

(iv) m∠DAB + m∠CDA = 180°
∵ Adjacent angles of a parallelogram are supplementary.

Question 2.
Consider the following parallelo¬grams. Find the values of the unknowns x, y, z.

Solution.
(i) y = 100°
Opposite angles of a parallelogram are equal
x + 100° = 180°
Adjacent angles in a parallelogram are supplementary
⇒ x = 180° – 100°
⇒ x = 80°
⇒ z – x = 80°
Opposite angles of a parallelogram are of equal measure

(ii) x + 50° = 180°
Adjacent angles in a parallelogram are supplementary
⇒ x = 180° – 50° = 130°
⇒ y = x = 130°
The opposite angles of a parallelogram are of equal measure
180° – z = 50°
Opposite angles of a parallelogram are of equal measure
⇒ z = 180° – 50° = 130°

(iii) x = 90°
Vertically opposite angles are equal
x + y + 30° = 180°
By angle sum property of a triangle
⇒ 90° + y + 30° = 180°
⇒ 120° + y = 180°
⇒ y = 180° – 120° = 60° z + 30° + 90° – 180°
By angle sum property of a triangle
z = 60°

(iv) y = 80°
Opposite angles of a parallelogram are of equal measure
x + 80° = 180°
Adjacent angles in a parallelogram are supplementary
⇒ x = 180° – 80°
⇒ x = 100°
⇒ 180°-2+ 80°= 180°
Linear pair property and adjacent angles in a parallelogram are supplementary.
z = 80°

(v) y = 112°
Opposite angles of a parallelogram are equal
x + y + 40° = 180°
By angle sum property of a triangle
⇒ x + 112° + 40° = 180°
⇒ x + 152° = 180°
⇒ x = 180°- 152°
⇒ x = 28°
z = x = 28°.
Alternate interior angles

Question 3.
Can a quadrilateral ABCD be a parallelogram if
(i) ∠D + ∠B = 180° ?
(ii) AB = DC = 8 cm, AD = 4 cm and BC = 4.4 cm
(iii) ∠A = 70° and ∠C = 65°?
Solution.
(i) Can be, but need not be
(ii) No: in a parallelogram, opposite sides are equal; but here, AD ≠ BC.
(iii) No: in a parallelogram, opposite angles are of equal measure; but here ∠A ≠ ∠C.

Question 4.
Draw a rough figure of a quadrilateral that is not a parallelogram but has exactly two opposite angles of equal measure.
Solution.
A kite, for example

Question 5.
The measures of two adjacent angles of a parallelogram are in the ratio 3:2. Find the measure of each of the angles of the parallelogram.
Solution.
Let the two adjacent angles be 3x° and 2x°.
Then,
3x° + 2x° = 180°
∴ Sum of the two adjacent angles of a parallelogram is 180°
⇒ 5x° = 180°
⇒ \({ x }^{ \circ }=\frac { { 180 }^{ \circ } }{ 5 } \)
⇒ x° = 36°
⇒ 3x° = 3 x 36° = 108°
and
2x° = 2 x 36° = 72°.
Since, the opposite angles of a parallelogram are of equal measure, therefore the measures of the angles of the parallelogram are 72°, 108°, 72°, and 108°.

Question 6.
Two adjacent angles of a parallelogram have equal measure. Find the measure of each of the angles of the parallelogram.
Solution.
Let the two adjacent angles of a parallelogram be x° each.
Then,
x° + x° = 180°
∴ Sum of the two adjacent angles of a parallelogram is 180°.
⇒ 2x° = 180°
⇒ \({ x }^{ \circ }=\frac { { 180 }^{ \circ } }{ 2 } \)
⇒ x° = 90°.
Since the opposite angles of a parallelogram are of equal measure, therefore the measure of each of the angles of the parallelogram is 90°, i.e., each angle of the parallelogram is a right angle.

Question 7.
The adjacent figure HOPE is a parallelogram. Find the angle measures x, y and z. State the properties you use to find them.
Solution.
x = 180° – 70° = 110°
Linear pair property and the opposite angles of a parallelogram are of equal measure.
∵ HOPE is a || gm
∴ HE || OP
and HP is a transversal
∴ y = 40°
alternate interior angles
40° + z + x = 180°
The adjacent angles in a parallelogram are supplementary

⇒ 40° + z + 110° = 180°
⇒ z + 150° = 180°
⇒ z = 180° – 150°
⇒ z = 30°.

Question 8.
The following figures GUNS and RUNS are parallelograms. Find x and y. (Lengths are in cm)
(i)

(ii)

Solution.
(i)
For Figure GUNS
Since the opposite sides of a parallelogram are of equal length, therefore,
⇒ 3x = 18
⇒ \(x=\frac { 18 }{ 3 } =6\)
and, 3y – 1 = 26
⇒ 3y = 26 + 1
⇒ 3y = 27
\(y=\frac { 27 }{ 3 } =9\)
Hence, x = 6; y = 9.

(ii)
For Figure RUNS
Since the diagonals of a parallelogram bisect each other, therefore,
⇒ x + y = 16 …(1)
and, y + 7 = 20 …(2)
From (2),
⇒ y – 20 – 7 = 13
Putting y = 13 in (1), we get
⇒ x + 13 = 16 ⇒ x = 16 – 13 = 3.
Hence, x = 3; y = 13.

Question 9.
In the below figure both RISK and CLUE are parallelograms. Find the value of x.

Solution.

Question 10.
Explain how this figure is a trapezium. Which of its two sides is parallel?
Solution.

∵ ∠KLM + ∠NML = 80° + 100° = 180°
∴ KL || NM
∵ The sum of consecutive interior angles is 180°
∴ Figure KLMN is a trapezium.
Its two sides \(\overline { KL } \) and \(\overline { NM } \) are parallel.

Question 11.
Find m∠C in the figure, if \(\overline { AB } \) || \(\overline { DC } \).

Solution.
∵ \(\overline { AB } \) || \(\overline { DC } \)
∴ m∠C + m∠B = 180°
∵ The sum of consecutive interior angles is 180°
m∠C+ 120° = 180°
⇒ m∠C = 180° – 120° = 60°.

Question 12.
Find the measure of ∠P and ∠S, if \(\overline { SP } \) || \(\overline { RQ } \) in the figure. (If you find mZ R, is there more than one method to find m∠P ?)
Solution.

∵ \(\overline { SP } \) || \(\overline { RQ } \)
∴ m∠P+m∠Q = 180°
∵ The sum of consecutive interior angles is 180°
⇒ m∠P + 130° = 180°
⇒ m∠P = 180° – 130°
⇒ m∠P = 50°
Again, m∠R + m∠S = 180°
∵ The sum of consecutive interior angles is 180°
⇒ 90° + m Z S = 180°
⇒ m∠S = 180° – 90° = 90°
Yes; there is one more method of finding m∠P if m∠R is given and that is by using the angle sum property of a quadrilateral.
We have,
m∠P + m∠Q + m∠R + m∠S = 360°
⇒ m∠P + 130° + 90° + 90° = 360°
⇒ m∠P + 310° = 360°
⇒ m∠P = 360° – 310° = 50°.

 

We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.3, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2.

• Understanding Quadrilaterals Class 8 Ex 3.1
• Understanding Quadrilaterals Class 8 Ex 3.3
• Understanding Quadrilaterals Class 8 Ex 3.4

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 3
Chapter Name	Understanding Quadrilaterals
Exercise	Ex 3.2
Number of Questions Solved	6
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2

Question 1.
Find x in the following figures.

Solution.

Question 2.
Find the measure of each exterior angle of a regular polygon of
(i) 9 sides
(ii) 15 sides
Solution.
(i) 9 sides
Measure of each exterior angle=\(\frac { { 360 }^{ \circ } }{ 9 } ={ 40 }^{ \circ }\)

(ii) 15 slides
Measure of each exterior angle=\(\frac { { 360 }^{ \circ } }{ 15 } ={ 24 }^{ \circ }\)

Question 3.
How many sides does a regular polygon have if the measure of an exterior angle is 24°?
Solution.
Let the number of sides be n. Then, n(24°)=360°.
⇒ \(n=\frac { { 360 }^{ \circ } }{ 24 } =15\)
Hence, the number of sides is 15.

Question 4.
How many sides does a regular polygon have if each of its interior angles is 165°?
Solution.
∵ Each interior angle=165°
∴ Each exterior angle
= 180°-165°=15°
linear pair property
Let the number of sides be n. Then,
n(15°)=360°
\(n=\frac { { 360 }^{ \circ } }{ { 15 }^{ \circ } } =24\)
Hence, the number of sides is 24.

Question 5.
(a) Is it possible to have a regular polygon with a measure of each exterior angle at 22°?
(b) Can it be an interior angle of a regular polygon? Why?
Solution.
(a) No ; (since 22 is not a factor of 360).
(b) No ; (because each exterior angle is 180° – 22° = 158°, which is not a factor of 360°).

Question 6.
(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?
Solution.
(a) The equilateral triangle is a regular polygon of 3 sides has the minimum measure of an interior angle = 60°.
(b) By (a), we can see that the maximum exterior angle possible for a regular polygon is 180° – 60° = 120°.

 

We hope the NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 3 Understanding Quadrilaterals Ex 3.2, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6.

• Linear Equations in One Variable Class 8 Ex 2.1
• Linear Equations in One Variable Class 8 Ex 2.2
• Linear Equations in One Variable Class 8 Ex 2.3
• Linear Equations in One Variable Class 8 Ex 2.4
• Linear Equations in One Variable Class 8 Ex 2.5

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 2
Chapter Name	Linear Equations in One Variable
Exercise	Ex 2.6
Number of Questions Solved	3
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6

Question 1.
Solve the following equations:

Solution.
1. \(\frac { 8x-3 }{ 3x } =2\)

2. \(\frac { 9x }{ 7-6x } =15\)

3. \(\frac { z }{ z+15 } =\frac { 4 }{ 9 }\)

4. \(\frac { 3y+4 }{ 2-6y } =\frac { -2 }{ 5 }\)

5. \(\frac { 7y+4 }{ y+2 } =\frac { -4 }{ 3 }\)

Question 2.
The ages of Hari and Harry are in the ratio 5: 7. Four years from now the ratio of their ages will be 3 :4. Find their present ages.
Solution.
Let the present ages of Hari and Harry be 5x years and 7x years respectively.

Question 3.
The denominator of a rational number is greater than its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, the number obtained is \(\frac { 3 }{ 2 } \). Find the rational number.
Solution.

We hope the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.6, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5.

• Linear Equations in One Variable Class 8 Ex 2.1
• Linear Equations in One Variable Class 8 Ex 2.2
• Linear Equations in One Variable Class 8 Ex 2.3
• Linear Equations in One Variable Class 8 Ex 2.4
• Linear Equations in One Variable Class 8 Ex 2.6

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 2
Chapter Name	Linear Equations in One Variable
Exercise	Ex 2.5
Number of Questions Solved	2
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5

Question 1.
Solve the following linear equations:
1. \(\frac { x }{ 2 } -\frac { 1 }{ 5 } =\frac { x }{ 3 } +\frac { 1 }{ 4 } \)
2. \(\frac { n }{ 2 } -\frac { 3n }{ 4 } +\frac { 5n }{ 6 } =21\)
3. \(x+7-\frac { 8x }{ 3 } =\frac { 17 }{ 6 } -\frac { 5x }{ 2 } \)
4. \(\frac { x-5 }{ 3 } =\frac { x-3 }{ 5 } \)
5. \(\frac { 3t-2 }{ 4 } -\frac { 2t+3 }{ 3 } =\frac { 2 }{ 3 } -t\)
6. \(m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 } \)
Solution.
1. \(\frac { x }{ 2 } -\frac { 1 }{ 5 } =\frac { x }{ 3 } +\frac { 1 }{ 4 } \)

2. \(\frac { n }{ 2 } -\frac { 3n }{ 4 } +\frac { 5n }{ 6 } =21\)

3. \(x+7-\frac { 8x }{ 3 } =\frac { 17 }{ 6 } -\frac { 5x }{ 2 } \)

4. \(\frac { x-5 }{ 3 } =\frac { x-3 }{ 5 } \)

5. \(\frac { 3t-2 }{ 4 } -\frac { 2t+3 }{ 3 } =\frac { 2 }{ 3 } -t\)

6. \(m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 } \)
We have \(m-\frac { m-1 }{ 2 } =1-\frac { m-2 }{ 3 } \)

Question 2.
Simplify and solve the following linear equations:
7. 3(t-3)=5(2t+1)
8. 15(y-4)-2(y-9)+5 (y+6) = 0
9. 3(5z-7)-2 (9z-11)=4 (8z-13)-17
10. 0.25(4f-3) = 0.05 (10f-9).
Solution.
7. 3(t-3)=5(2t+1)

8. 15(y-4)-2(y-9)+5 (y+6) = 0

9. 3(5z-7)-2 (9z-11)=4 (8z-13)-17

10. 0.25(4f-3) = 0.05 (10f-9)

 

We hope the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 are part of NCERT Solutions for Class 8 Maths. Here we have given NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4.

• Linear Equations in One Variable Class 8 Ex 2.1
• Linear Equations in One Variable Class 8 Ex 2.2
• Linear Equations in One Variable Class 8 Ex 2.3
• Linear Equations in One Variable Class 8 Ex 2.5
• Linear Equations in One Variable Class 8 Ex 2.6

Board	CBSE
Textbook	NCERT
Class	Class 8
Subject	Maths
Chapter	Chapter 2
Chapter Name	Linear Equations in One Variable
Exercise	Ex 2.4
Number of Questions Solved	10
Category	NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4

Question 1.
Amina thinks of a number and subtracts \(\frac { 5 }{ 2 } \) from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution.
Let the number be x.
Then, according to the question,

Question 2.
A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution.
Let the numbers be x and 5x.
If 21 is added to both the numbers, then first new number = x + 21
and, second new number = 5x + 21

Question 3.
Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?
Solution.
Let the unit’s digit of the two-digit number be x.
Then, the ten’s digit of the two-digit number = 9 – x

Question 4.
One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?
Solution.
Let in the original number

Question 5.
Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?
Solution.
Let the present age of Shobo be x  year. Then, the present age of Shobo’s mother = 6x years.
Five years from now
Shobo’s age = (x + 5) years
According to the question,

Hence, their present ages are 5 years and 30 years.

Question 6.
There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate of ₹ 100 per meter, it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution.
Let the length and breadth of the plot be 11x m and 4x m respectively.
Then, the perimeter of the plot

Question 7.
Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per meter and trouser material that costs him ₹ 90 per meter. For every 2 meters of the trouser material, he buys 3 meters of the shirt material. He sells the materials at 12% and 10% profit respectively. His total sale is ₹ 36,600. How much trouser material did he buy?
Solution.
Suppose that he bought x meters of trouser material.

Question 8.
Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.
Solution.
Let the number of deer in the herd be x.

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution.
Let the present age of granddaughter be x years
Then, the present age of grandfather is 10x years
According to the question,

Question 10.
Am.an’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.
Solution.
Let the present age of Aman’s son be x years
Then, the present age of Aman

 

We hope the NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 help you. If you have any query regarding NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4, drop a comment below and we will get back to you at the earliest.