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

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.

BoardCBSE
TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 5
Chapter NameData Handling
ExerciseEx 5.2
Number of Questions Solved5
CategoryNCERT 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.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 1
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
\(\frac { 10 }{ 100 } \times x=\frac { x }{ 10 } \)
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 2
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.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 3
(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)
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 4NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 5

Question 3.
Draw a pie chart showing the following information. The table shows the colors preferred by a group of people.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 6
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.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 7

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.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 8
(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.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 9
Solution.
NCERT Solutions for Class 8 Maths Chapter 5 Data Handling Ex 5.2 10

 

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

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.

BoardCBSE
TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 4
Chapter NamePractical Geometry
ExerciseEx 4.5
Number of Questions Solved1
CategoryNCERT 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
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 1
  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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 2
  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°.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 3
  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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.5 4
  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

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

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.

BoardCBSE
TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 4
Chapter NamePractical Geometry
ExerciseEx 4.4
Number of Questions Solved1
CategoryNCERT 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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 1
  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°.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.4 2
  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.

 

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

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.

BoardCBSE
TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 4
Chapter NamePractical Geometry
ExerciseEx 4.3
Number of Questions Solved1
CategoryNCERT 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°.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 1
  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°.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 2
    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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 3
  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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.3 4
  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.

 

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

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.

BoardCBSE
TextbookNCERT
ClassClass 8
SubjectMaths
ChapterChapter 4
Chapter NamePractical Geometry
ExerciseEx 4.2
Number of Questions Solved1
CategoryNCERT 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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 1
  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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 2
  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.
    NCERT Solutions for Class 8 Maths Chapter 4 Practical Geometry Ex 4.2 3
  4. Join DN, NE, EB, and BD.

Then, BEND is the required rhombus.

 

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