RS Aggarwal Class 7 Solutions Chapter 23 Probability Ex 23

RS Aggarwal Class 7 Solutions Chapter 23 Probability Ex 23

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 23 Probability Ex 23.

Question 1.
Solution:
(i) Here, total number of trials = 300
Number of heads got = 136.
P(E) = \(\frac { 136 }{ 300 }\) = \(\frac { 34 }{ 75 }\)
(ii) Total number of trials = 300
Number of tails got = 164
P(E) = \(\frac { 164 }{ 300 }\) = \(\frac { 41 }{ 75 }\)

Question 2.
Solution:
Number times, the two coins were tossed = 200
Number of times got two heads = 58
Number of times got one head = 83
and number of times got no head = 59
(i) Probability of getting 2 heads : P(E) = \(\frac { 58 }{ 200 }\) = \(\frac { 29 }{ 100 }\)
(ii) Probability of getting one head : P(E) = \(\frac { 83 }{ 200 }\)
(iii) Probability of getting no head : P(E) = \(\frac { 59 }{ 200 }\)

Question 3.
Solution:
Number of times, a dice was thrown = 100
(i) Number of times got 3 = 18
Probability will be
P(E) = \(\frac { 18 }{ 100 }\) = \(\frac { 9 }{ 50 }\)
(ii) Number of times got 6 = 9
Probability will be
P(E) = \(\frac { 9 }{ 100 }\)
iii) Number of times got 4 = 15
Probability will be
P(E) = \(\frac { 15 }{ 100 }\) = \(\frac { 3 }{ 20 }\)
(iv) Number of times got 1 = 21
Probability will be
P(E) = \(\frac { 21 }{ 100 }\)

Question 4.
Solution:
Total number of ladies = 100
Number of ladies also like coffee = 36.
Number of ladies who dislike coffee = 64
(i) Probability of lady who like coffee :
P(E) = \(\frac { 36 }{ 100 }\) = \(\frac { 9 }{ 25 }\)
(ii) Probability of lady who dislikes coffee:
P(E) = \(\frac { 64 }{ 100 }\) = \(\frac { 16 }{ 25 }\)

 

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RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22

RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 22 Bar Graphs Ex 22.

Question 1.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph representing x-axis and y-axis.
(ii) Along OX, mark subjects and along y-axis, mark, number of marks
(iii) Take one division = 10 marks.
Now we shall draw bar graph as shown.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 1

Question 2.
Solution:
(i) Draw a horizontal line OX and a vertical line OY on the graph paper. These two lines represent x-axis and y-axis respectively.
(ii) Along OX, write sports and along OY, number of students choosing on division equal to 10 students.
(iii) Now draw the bars of various heights according to the no. of students as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 2

Question 3.
Solution:
(i) Draw a horizontal line OX and a vertical line OY. These represent x-axis and y-axis respectively on the graph paper.
(ii) Along OX, write years and along OY, no. of students choosing one division = 200 students.
(iii) Draw bars of various heights according to the number of students given.
This is the required bar graph as shown.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 3

Question 4.
Solution:
(i) Draw a horizontal line OX and a vertical line OY which represent x-axis and y-axis respectively on the graph.
(ii) Along OX, write years and along OY, no. of scooters
(iii) Choose 1 division = 300
(iv) Now draw bars of different heights according to given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 4

Question 5.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along OX write countries and along OY, take Birth rate per thousand.
(iii) Choose 1 division = 10.
(iv) Now draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 5

Question 6.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis on the graph paper.
(ii) Along x-axis write states and along y-axis population in lakhs.
(iii) Choose one division = 200 (Lakhs)
(iv) Draw bars of different heights according to the data given as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 6

Question 7.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis Interest in thousand cores rupees.
(iii) Choose one division = 20 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 7

Question 8.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write city and along y-axis the distance (in km).
(iii) Choose one division = 200 km.
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 8

Question 9.
Solution:
(i) Draw a horizontal line OX and a vertical line OY represent x-axis and y-axis respectively on the graph paper.
(i) Along x-axis write countries and along y-axis life expectancy (in years)
(ii) Choose one division = 10 years
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 9

Question 10.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write years and along y-axis imports in thousand crores rupees.
(iii) Choose one division = 50 thousand crore rupees.
(iv) Draw bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 10

Question 11.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write months and along y-axis average rainfall in cm.
(iii) Choose one small division = 5cm.
(iv) Draw different bars of different heights according to the given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 11

Question 12.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis write Brand and along y-axis, percentage of buyers.
(iii) Choose one division = 5% of buyers.
(iv) Draw bars of different heights according to the given data, as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 12

Question 13.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write week and along y-axis. Rate per 10gm in rupees.
(iii) Choose and division = 1000
(iv) Draw bars of different heights according to the data given as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 13

Question 14.
Solution:
(i) Draw a horizontal line OX and a vertical line OY representing x-axis and y-axis respectively on the graph paper.
(ii) Along x-axis, write mode of transport and on the y-axis is number of students.
(iii) Choose one division =100 students
(iv) Draw bars of different heights according to given data as shown on the graph.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 14

Question 15.
Solution:
(i) The bar graph shows the marks obtained by a student in different subjects.
(ii) The student is very good in Mathematics.
(iii) The student is very poor in Hindi.
(iv) Average marks
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 15

Question 16.
Solution:
(i) The bar graph shows the number of members in each of the 85 families.
(ii) 40 families have 3 number each.
(iii) Number of people living alone is nil.
(iv) The families having 3 members each is most common.

Question 17.
Solution:
(i) The highest peak is Mount Everest whose heighest is 8800 m.
(ii) The required ratio between the highest peak and the next heighest peak = 8800 : 8200 or 44 : 41
(iii) Arranging the heights of peaks in descending order are : 8800 m, 8200 m, 8000 m, 7500 m, 6000 m

Question 18.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale : 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are :
Mon. 350 and 200; Tues. 400 and 450; Wed. 500 and 300; Thurs. 450 and 250; Fri. 550 and 100 and Sat. 450 and 50.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 16
(ii) The number of readers in the library was maximum on TUESDAY.
(iii) Total Number of magazine readers in a week = 200 + 450 + 300 + 250 + 100 + 50 = 1350
Mean Number of readers per day = \(\frac { 1350 }{ 6 }\) = 225

Question 19.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 4
(d) The heights of various pairs of bars in terms of the number of small divisions are:
VI 95 and 92; VII 90 and 85; VIII 82 and 78; IX 75 and 69; X 68 and 62.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 17
(ii) For class VII, total Number of Students = 90
Number of students present = 85
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 18

Question 20.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
January 2 and 1.5 ; February 3.25 and 3; March 4 and 3.5; April 4.5 and 6; May 7.75 and 5.5; June 8 and 6.5.
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 19
(ii) June
(iii) January

Question 21.
Solution:
(i) We can draw the double bar graph in the following steps :
(a) On a graph paper, draw a horizontal line OX and a vertical line OY, representing the x-axis and the y’-axis respectively.
(b) Along OX, write the names of the subjects taken at appropriate uniform gaps.
(c) Choose the scale: 1 division = 10
(d) The heights of various pairs of bars in terms of the number of small divisions are:
Town A 640000 and 750000; Town B 830000 and 920000; Town C 460000 and 630000; Town D 290000 and 320000
(e) On the x-axis, draw pairs of bars of equal width and of heights shown in step 4 at the points marked in step 2.
RS Aggarwal Class 7 Solutions Chapter 22 Bar Graphs Ex 22 20
(ii) Town B
(iii) Town D

 

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C

RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21C.

Other Exercises

Question 1.
Solution:
(i) Arranging in ascending order :
4, 6, 7, 8, 8, 8, 8, 10, 11, 15
We see that 8 occurs maximum times
Mode = 8
(ii) Arranging in ascending order :
18, 21, 23, 27, 27, 27, 27, 27, 36, 39, 40
We see that 27 occurs maximum times
Mode = 27

Question 2.
Solution:
Arranging in ascending order :
28, 31, 32, 32, 32, 32, 34, 36, 38, 40, 41.
We see that 32 occurs maximum times
Mode = 32 years

Question 3.
Solution:
We prepare the table as given below:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 1
Here, number of terms = 45, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 45 + 1 }{ 2 }\) = \(\frac { 46 }{ 2 }\) th term
= 23th term = 450
Now, mode = 3(median) – 2(mean)
= 3 x 450 – 2 x 470
= 1350 – 940
= 410

Question 4.
Solution:
We prepare the table as given below:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 2
Here, number of terms (N) = 41, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 41 + 1 }{ 2 }\) th term
= \(\frac { 42 }{ 2 }\) = 21 th term = 22 {value of 18 to 29 = 22}
Mode = 3 (median) – 2 (mean)
= 3 x 22 – 2 x 21.92 = 66 – 43.84 = 22.16

Question 5.
Solution:
We prepare the table as given below:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 3
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21C 4

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B

RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21B

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B.

Other Exercises

Question 1.
Solution:
(i) Arranging in ascending order.
2, 2, 3, 5, 7, 9, 9, 10, 11
Here number of terms = 9 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 9 + 1 }{ 2 }\) th term
= 5th term = 7
Hence median = 7
(ii) Arranging in ascending order,
6, 8, 9, 15, 16, 18, 21, 22, 25
Here, number of terms (n) = 9 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 9 + 1 }{ 2 }\) th term
= 5th term = 16
(iii) Arranging in ascending order,
6, 8, 9, 13, 15, 16, 18, 20, 21, 22, 25 Here number of terms (n) = 11 which is odd
Median= \(\frac { n +1 }{ 2 }\) = \(\frac { 11 + 1 }{ 2 }\) th term
= 6th term = 16

Question 2.
Solution:
(i) Arranging in ascending order,
9, 10, 17, 19, 21, 22, 32, 35
Here, number of terms = 8 which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 1
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 2

Question 3.
Solution:
First 15 odd numbers are 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.
Here, number of terms (n) = 15 which is odd
Median = \(\frac { n + 1 }{ 2 }\) = \(\frac { 15 + 1 }{ 2 }\) th term
= 8th term = 15

Question 4.
Solution:
First 10 even numbers are 2, 4, 6, 8, 10, 12, 14, 16, 18, 20
Here, number of terms = 10 which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 3

Question 5.
Solution:
First 50 whole numbers
0, 1, 2, 3, 4, …, 49
Here, number of terms = 50, which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 4

Question 6.
Solution:
Arranging in ascending order,
17, 17, 19, 19, 20, 21, 22, 23, 24, 25, 26, 29, 31, 35, 40 .
Here, number of terms = 15 which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term = \(\frac { 15 + 1 }{ 2 }\) = \(\frac { 16 }{ 2 }\) th
= 8th term = 23

Question 7.
Solution:
Arranging is ascending order,
31, 34, 36, 37, 40, 43, 46, 50, 52, 53
Here, number of terms = 10, which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 5

Question 8.
Solution:
Preparing the cumulative frequency table
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 6
Here, number of terms (N) = 41, which is odd
Median = \(\frac { n + 1 }{ 2 }\) th term
= \(\frac { 41 + 1 }{ 2 }\) = \(\frac { 42 }{ 2 }\) th = 21 th term = 50kg (value of 20 to 28 = 50)
Hence median = 50kg

Question 9.
Solution:
Arranging in order and preparing the cumulative frequency table.
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 7
Here, number of terms (N) = 37 which is odd.
Median = \(\frac { n + 1 }{ 2 }\) th term
= \(\frac { 37 + 1 }{ 2 }\) = \(\frac { 38 }{ 2 }\) th term
= 19th term = 22 (Value of 18 to 21 = 22)
Hence median = 22

Question 10.
Solution:
Arranging in order and then preparing its cumulative frequency table :
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 8
Here, number of terms (N) = 50, which is even
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21B 9
Hence median = 154.5 cm

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RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A

RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data (Mean, Median and Mode) Ex 21A

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A.

Other Exercises

Question 1.
Solution:
(i) Data : A collection of numerical figures giving some particular type of information is called data
(ii) Raw data : Data obtained in the original form is called raw data.
(iii) Array : Arranging the numerical figures of a data in ascending or descending order is called an array.
(iv) Tabulation of data : Arranging the data in a systematic form in the form of a table is called tabulation of the data.
(v) Observations : Each numerical figure in a data is called an observation.
(vi) Frequency of an observation : The number of times a particular observation occurs is called its frequency.
(vii) Statistics : It is the subject that deals with the collection presentation analysis and interpretation of numerical data.

Question 2.
Solution:
Arranging the given data in ascending order is as given below :
1, 1, 2, 2, 2, 2, 2, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6 and 6 its frequency table will be as under:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 1

Question 3.
Solution:
Arranging the given data in ascending order,
260, 260, 300, 300, 300, 300, 360, 360, 360, 360, 360, 360, 400, 400, 400.
and its frequency table will be as under.
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 2

Question 4.
Solution:
Arranging the given data in ascending order we find
5, 5, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 10, 10 and its frequency table will be as under.
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 3

Question 5.
Solution:
(i) Data means information in the form of numerical figures.
(ii) Data obtained in the original form is called raw data.
(iii) Arranging the numerical figures in ascending or descending order is called an array.
(iv) The number of times a particular observation occurs is called its frequency.
(v) Arranging the data in the form of a table is called tabulation of data.

Question 6.
Solution:
First five natural numbers are 1, 2, 3, 4, 5
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 4

Question 7.
Solution:
First six odd natural numbers are 1, 3, 5, 7, 9, 11
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 5

Question 8.
Solution:
First seven even natural numbers are 2, 4, 6, 8, 10, 12, 14
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 6

Question 9.
Solution:
First five prime numbers are 2, 3, 5, 7, 11
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 7

Question 10.
Solution:
First six multiples of 5 are 5, 10, 15, 20, 25, 30
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 8

Question 11.
Solution:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 9

Question 12.
Solution:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 10
Mean = Rs. 159

Question 13.
Solution:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 11
Mean = Rs. 318

Question 14.
Solution:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 12

Question 15.
Solution:
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 13
RS Aggarwal Class 7 Solutions Chapter 21 Collection and Organisation of Data Ex 21A 14

 

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper.

Other Exercises

Question 1.
Solution:
We know that all the angles of a rectangle are 90° and the diagonal divides the rectangle into two right angled triangles.
So, one side of the triangle will be 48 m and the diagonal, which is 50 m, will be the hypotenuse.
According to Pythagoras theorem :
(Hypotenuse)² = (Base)² + (Perpendicular)²
Perpendicular
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 1
Other side of the rectangular plot = 14m
Area of the rectangular plot = 48 m x 14 m = 672 m²
Hence, the area of a rectangular plot is 672 m².

Question 2.
Solution:
Length = 9 m; Breadth = 8 m
Height = 6.5 m
Area of the four walls = {2 (l + b) x h} sq. units
= {2 (9 + 8) x 6.5} m² = {34 x 6.5) m² = 221 m²
Area of one door = (2 x 1 .5) m² = 3m²
Area of one window = (1.5 x 1) m² = 1.5 m²
Area of four windows = (4 x 1.5) m² = 6 m²
Total area of one door and four windows = (3 + 6) m² = 9 m²
Area to be painted = (221 – 9) m² = 212 m²
Rate of painting = ₹ 50 per m²
Total cost of painting = ₹ (212 x 50) = ₹ 10,600

Question 3.
Solution:
Given that the diagonal of a square is 64 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 2

Question 4.
Solution:
Let ABCD be the square lawn
and PQRS be the outer boundary of the square path
Let one side of the lawn (AB) be x m
Area of the square lawn = x²
Length PQ = (x m + 2 m + 2 m) =(x + 4) m
Area of PQRS = (x + 4)² = (x² + 8x + 16) m²
Now, Area of the path = Area of PQRS – Area of the square lawn
⇒ 136 = x² + 8x + 16x – x²
⇒ 136 = 8x + 16
⇒ 136 – 16 = 8x
⇒ 120 = 8x
⇒ x = 15
Side of the laws = 15 m
Area of the lawn = (Side)² = (15 m)² = 225 m²

Question 5.
Solution:
Let ABCD be the rectangular park
EFGH and IJKL are the two rectangular roads with width 2 m.
Length of the rectangular park AD = 30 cm
Breadth of the rectangular park CD = 20 cm
Area of the road EFGH = 30 m x 2 m = 60 m²
Area of the road IJKL = 20 m x 2m = 40 m²
Clearly, area of MNOP is common to the two roads.
Area of MNOP = 2m x 2m = 4m²
Area of the roads = Area(EFGH) + Area (IJKL) – Area (MNOP)
= (60 + 40) m² – 4 m² = 96 m²

Question 6.
Solution:
Let ABCD be the rhombus whose diagonals intersect at O.
Then, AB = 13 cm
AC = 24 cm
The diagonals of a rhombus bisect each other at right angles.
Therefore, ∆AOB is a right-angled triangle, right angled at O, such that:
OA = \(\frac { 1 }{ 2 }\) AC = 12 cm
AB = 13 cm
By Pythagoras theorem :
(AB)² = (OA)² + (OB)²
⇒ (13)² = (12)² + (OB)²
⇒ (OB)² = (13)² – (12)²
⇒ (OB)2 = 169 – 144 = 25
⇒ (OB)² = (5)²
⇒ OB = 5 cm
BD = 2 x OB = 2 x 5 cm = 10 cm
Area of the rhombus ABCD = \(\frac { 1 }{ 2 }\) x AC x BD cm²
= \(\frac { 1 }{ 2 }\) x 24 x 10
= 120 cm²

Question 7.
Solution:
Let the base of the parallelogram be x m.
The, the altitude of the parallelogram will be 2x m.
It is given that the area of the parallelogram is 338 m².
Area of a parallelogram = Base x Altitude
⇒ 338 = x x 2x
⇒ 338 = 2x²
⇒ x² = 169 m²
⇒ x = 13 m
Base = x m = 13 m
Altitude = 2x m = (2 x 13) m = 26 m

Question 8.
Solution:
Consider ∆ABC Here, ∠B = 90°
AB = 24 cm
AC = 25 cm
Now, AB² + BC² = AC²
BC² = AC² – AB² = (25² – 24²) =(625 – 576) = 49
BC = (√49) cm = 7 cm
Area of ∆ABC = \(\frac { 1 }{ 2 }\) x BC x AB Sq.units
= \(\frac { 1 }{ 2 }\) x 7 x 24 cm² = 84 cm²
Hence, area of the right angled triangle is 84 cm².

Question 9.
Solution:
Radius of the wheel = 35 cm
Circumference of the wheel = 2πr
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 3

Question 10.
Solution:
Let the radius of the circle be r cm
Area = (πr²) cm²
πr² = 616
⇒ \(\frac { 22 }{ 7 }\) x r x r = 616
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 4
Hence, the radius of (he given circle is 14 cm.

Mark (✓) against the correct answer in each of the following:
Question 11.
Solution:
(a) 14 cm
Let the radius of the circle be r cm
Then, its area will be (πr²) cm²
πr² = 154
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 5

Question 12.
Solution:
(b) 154 cm²
Let the radius of the circle be r cm.
Circumference = (2πr) cm
(2πr) = 44
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 6

Question 13.
Solution:
(c) 98 cm²
Given that the diagonal of a square is 14 cm
Area of a square
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 7

Question 14.
Solution:
(b) 10 cm
Given that the area of the square is 50 cm²
We know:
Area of a square
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 8

Question 15.
Solution:
(a) 192 m²
Let the length of the rectangular park be 4x.
Breadth = 3x
Perimeter of the park = 2 (l + b) = 56 m (given)
⇒ 56 = 2 (4x + 3x)
⇒ 56 = 14x
⇒ x = 4
Length = 4x = (4 x 4) = 16 m
Breadth = 3x = (3 x 4) = 12 m
Area of the rectangular park = 16 m x 12 m= 192 m²

Question 16.
Solution:
(a) 84 cm²
Let a = 13 cm, b = 14 cm and c = 15 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 9

Question 17.
Solution:
(a) 16√3 cm²
Given that each side of an equilateral triangle is 8 cm
Area of the equilateral triangle
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration CCE Test Paper 10

Question 18.
Solution:
(b) 91 cm²
Base = 14 cm
Height = 6.5 cm
Area of the parallelogram = Base x Height
= (14 x 6.5) cm² = 91 cm²

Question 19.
Solution:
(b) 135 cm²
Area of the rhombus = \(\frac { 1 }{ 2 }\) x (Product of the diagonals)
= \(\frac { 1 }{ 2 }\) x 18 x 15 = 135 cm²
Hence, the area of the rhombus is 135 cm².

Question 20.
Solution:
(i) If d1, and d2 be the diagonals of a rhombus, then its area is \(\frac { 1 }{ 2 }\) d1d2 sq. units.
(ii) If l, b and h be the length, breadth and height respectively of a room, then area of its 4 walls = [2h (l + b)] sq. units.
(iii) 1 hectare = (1000) m². (since 1 hecta metre = 100 m)
1 hectare = (100 x 100) m²
(iv) 1 acre = 100 m².
(v) If each side of a triangle is a cm, then its area = \(\frac {\surd 3 }{ 4 }\) a² cm².

Question 21.
Solution:
(i) False
Area of a triangle = \(\frac { 1 }{ 2 }\) x Base x Height
(ii) True
(iii) False
Area of a circle = πr²
(iv) True

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G.

Other Exercises

Objective Questions
Mark (✓) against the correct answer in each of the following :
Question 1.
Solution:
(c)
Length of rectangle AB = 16 cm
and diagonal BD = 20 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 1
But, in right ∆ABD
BD² = AB² + AD²
⇒ (20)² = (16)² + AD²
⇒ 400 = 256 + AD²
⇒ AD² = 400 – 256 = 144 = (12)²
⇒ AD = 12 cm
Area = l x b = 16 x 12 = 192 cm²

Question 2.
Solution:
(b)
Diagonal of square = 12 cm
Let side = 9
diagonal = √2 a
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 2

Question 3.
Solution:
(b)
Area = 200 cm²
side = √200 =√2 x 10
and diagonal = √2 a = √2 x √2 x 10 = 20

Question 4.
Solution:
(a)
Area of square = 0.5 hectare = 0.5 x 10000 = 5000 m²
= √10000 = 100 m

Question 5.
Solution:
(c)
Perimeter of rectangle = 240m
l + b = \(\frac { 240 }{ 2 }\) = 120 m
Let breadth = x, then length = 3x .
3x + x = 120
⇒ 4x = 120
⇒ x = 30
Length = 3x = 3 x 30 = 90 m

Question 6.
Solution:
Answer = (d)
Let original side of square = x
area = x²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 3
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 4

Question 7.
Solution:
(b)
Let side of square = a
Then its diagonal = √2 a
Now, area of square = a²
and area of square on diagonal = (√2 a)² = 2a²
Ratio = a² : 2a² = 1 : 2

Question 8.
Solution:
(c)
If perimeters of a square and a rectangle are equal Then the area of the square will be greater than that of a rectangle
A > B

Question 9.
Solution:
(b)
Perimeter of rectangle = 480m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 5

Question 10.
Solution:
(a)
Total cost of carpet = Rs. 6000
Rate per metre = Rs. 50
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 6

Question 11.
Solution:
(a)
Sides are 13 cm, 14 cm, 15 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 7

Question 12.
Solution:
(b)
Base of triangle = 12 m
and height = 8m
Area= \(\frac { 1 }{ 2 }\) x Base x height
= \(\frac { 1 }{ 2 }\) x 12 x 8 = 48 m²

Question 13.
Solution:
(b)
Let side = a
then area = \(\frac { \surd 3 }{ 4 }\) a²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 8

Question 14.
Solution:
(c)
Side of an equilateral triangle = 8cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 9

Question 15.
Solution:
(b)
Let a be the side of an equilateral triangle
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 10

Question 16.
Solution:
(b)
One side (Base) of parallelogram = 16 cm
and altitude = 4.5 cm
Area = base x altitude = 16 x 4.5 = 72 cm²

Question 17.
Solution:
(b)
Length of diagonals of a rhombus are 24 cm and 18 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 11

Question 18.
Solution:
(c)
Let r be the radius of the circle Then
c = 2πr
2πr – r = 37
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 12
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 13

Question 19.
Solution:
(c) Perimeter of room = 18 m
and height = 3 m
Area of 4 walls = Perimeter x height = 18 x 3 = 54 m²

Question 20.
Solution:
(a) Area of floor = l x b = 14 x 9 = 126 m²
Area of carpet = 126 m²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 14

Question 21.
Solution:
(c)
Perimeter = 46 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 15

Question 22.
Solution:
(b)
Ratio in area of two squares = 9 : 1
Let area of bigger square = 9x²
and of smaller square = x²
Side of bigger square = √9x² = 3x
and perimeter = 4 x side = 4 x 3x = 12x
Side of smaller square = √x² = x
Perimeter = 4x
Now ratio in their perimeter = 12x : 4x = 3 : 1

Question 23.
Solution:
(d)
Let the diagonals of two square be 2d and d
Area of bigger square 2 (2d)² = 8d²
and of smaller = 2 (d)² = 2d²
Ratio in their area = \(\frac { 8{ d }^{ 2 } }{ 2{ d }^{ 2 } } =\frac { 4 }{ 1 }\)
= 4 : 1

Question 24.
Solution:
(c)
Side of square = 84 m
Area of square = (84)² = 7056 m²
Area of rectangle = 7056 m²
Length of rectangle = 144 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 16

Question 25.
Solution:
(d)
Side of a square = a
Area = a²
Side of equilateral triangle = a
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 17
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 18

Question 26.
Solution:
(a)
Let a be the side of a square
Area = a²
Then area of circle = a²
Let r be the radius
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 19

Question 27.
Solution:
(b)
Let each side of an equilateral triangle = a
Then area = \(\frac {\surd 3 }{ 4 }\) a²
Now radius of the circle = a
Then area = πr² = πa²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 20

Question 28.
Solution:
(c)
Area of rhombus = 36 cm²
Length of one diagonal = 6 cm
Length of second diagonal
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 21

Question 29.
Solution:
(d)
Area of a rhombus =144 cm²
Let one diagonal (d1) = a
then Second diagonal (d2) = 2a
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 22
Larger diagonal = 2a = 2 x 12 = 24 cm

Question 30.
Solution:
(c)
Area of a circle = 24.64 m²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 23

Question 31.
Solution:
(c)
Let original radius = r
Then its area = πr²
Radius of increased circle = r + 1
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 24
2r = 7 – 1 = 6
⇒ r = \(\frac { 6 }{ 2 }\) = 3
Radius of original circle = 3 cm

Question 32.
Solution:
(c)
Radius of a circular wheel (r) = 1.75 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20G 25

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F.

Other Exercises

Question 1.
Solution:
(i) Radius of the circle (r) = 21 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 1

Question 2.
Solution:
(i) Diameter of the circle = 28 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 2
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 3

Question 3.
Solution:
Circumference of a circle = 264 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 4

Question 4.
Solution:
Circumference of the circle (c) = 35.2 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 5

Question 5.
Solution:
Area of the circle = 616 cm²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 6

Question 6.
Solution:
Area of a circle = 1386 m²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 7

Question 7.
Solution:
Ratio in the radii of two circles = 4 : 5
Let radius of first circle (r1) = 4x
and radius of the second circle (r2) = 5x
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 8

Question 8.
Solution:
Length of rope (r) = 21 m
Area of the circle = πr² = \(\frac { 22 }{ 7 }\) x 21 x 21 m² = 1386 m²
The horse will graze on 1386 m² area

Question 9.
Solution:
Area of a square made of a wire = 121 cm²
Side = √Area = √121 = 11 m
Perimeter of wire = 4 x side = 4 x 11 = 44 cm
Circumference of circular wire = 44 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 9

Question 10.
Solution:
Radius of circular wire = 28 cm
Circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 28 cm = 176 cm
Perimeter of the square formed by this wire = 176 cm
Side (a) = \(\frac { 176 }{ 4 }\) = 44 cm .
Area of square so formed = a² = (44)² cm² = 1936 cm²

Question 11.
Solution:
Length of rectangular sheet (l) = 34 cm
and breadth (b) = 24 cm
Area = l x b = 34 x 24 cm² = 816 cm²
Diameter of one button = 3.5 cm
Radius (r) = \(\frac { 3.5 }{ 2 }\) cm
and area of one button = πr²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 10
Area of 64 buttons = 9.625 x 64 cm² =616 cm²
Area of remaining sheet = 816 – 616 = 200 cm²

Question 12.
Solution:
Length of ground (l) = 90 m
and breadth (b) = 32 m
Area = l x b = 90 x 32 m² = 2880 m²
Radius of circular tank (r) = 14 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 11
= 616 m²
Area of remaining portion = 2880 – 616 = 2264 m²
Rate of turfing = Rs. 50 per sq.m.
Total cost = Rs. 50 x 2264 = Rs. 113200

Question 13.
Solution:
Each side of square = 14 cm
Area of square = a² = 14 x 14 = 196 cm²
Radius of each circle at each corner of square = \(\frac { 14 }{ 2 }\) = 7 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 12
Area of shaded portion = Area of square – area of 4 quadrants
= 196 – 154 = 42 cm²

Question 14.
Solution:
Length of field = 60 m
and breadth = 40 m
Length of rope = 14 m
Area covered by the horse
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 13

Question 15.
Solution:
Diameter of largest circle (outer circle) = 21 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 14
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 15
Area of shaded portion = 346.5 – (154 + 38.5) = (346.5 – 192.5) = 154 cm²

Question 16.
Solution:
Length of plot (l) = 8m
and breadth (b) = 6m
Area of plot = l x b = 8 x 6 = 48 m²
Radius of each quadrant at the corner = 2 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20F 16

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E.

Other Exercises

Question 1.
Solution:
Radius of the circle (r) = 28 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 1
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 2

Question 2.
Solution:
(i) Diameter of circle (d) = 35 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 3

Question 3.
Solution:
Radius of a circle = 15 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 4
Circumference = 2πr = 2 x 3.14 x 15 = 94.20 cm = 94.2 cm

Question 4.
Solution:
Circumference of a circle (c) = 57.2 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 5

Question 5.
Solution:
Circumference (c) = 63.8 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 6

Question 6.
Solution:
Let c be the circumference and d be the diameter of the circle.
c = d + 30
⇒ dπ = d + 30
⇒ dπ – d = 30
⇒ d (π – 1) = 30
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 7

Question 7.
Solution:
The ratio of the radii of the circles = 5 : 3
Let radius of first circle = 5x
and radius of second circle = 3x
Circumference of first circle = 2πr = 2π x 5x = 10πx
and circumference of second circle = 2π x 3x = 6πx
Ratio = 10πx : 6πx = 10 : 6 = 5 : 3

Question 8.
Solution:
Radius of circular field (r) = 21 m
Circumference = 2πr
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 8

Question 9.
Solution:
Outer circumference = 616 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 9
Width of track = R – r = 98 – 84 = 14m

Question 10.
Solution:
Inner circumference of the circular track = 330 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 10
Rate of fencing = Rs. 20 per metre
Total cost = Rs. 20 x 396 = Rs. 7920

Question 11.
Solution:
Radius of inner circle (r) = 98 cm
Inner circumference = 2πr = 2 x \(\frac { 22 }{ 7 }\) x 98 cm = 616 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 11
Radius of the outer circle (R) = 1 m 26cm = 126 cm
Outer circumference = 2πR = 2 x \(\frac { 22 }{ 7 }\) x 126 cm = 792 cm

Question 12.
Solution:
Side of equilateral triangle = 8.8 cm
Its perimeter = 8.8 x 3 = 26.4 cm
By bending their wire into a circular shape,
the circumference = 26.4 cm
Let d be the diameter,
Then C = dπ
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 12

Question 13.
Solution:
Each side of rhombus = 33 cm
Perimeter = 4 x 33 = 132 cm
Perimeter of circle = 132 cm
Let r be the radius
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 13

Question 14.
Solution:
Length of rectangle (l) = 18.7 cm
Breadth (b) = 14.3 cm
Perimeter = 2 (l + b) = 2 (18.7 + 14.3) cm = 2 x 33 = 66 cm
Circumference of the so formed circle = 66 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 14

Question 15.
Solution:
Radius of the circle (r) = 35 cm
Its circumference (c) = 2πr
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 15

Question 16.
Solution:
Diameter of well (d) = 140 cm
Outer circumference of parapet = 616 cm
Let D be the diameter, then
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 16
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 17

Question 17.
Solution:
Diameter of wheel (d) = 98 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 18

Question 18.
Solution:
Diameter of cycle wheel (d) = 70 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 19

Question 19.
Solution:
Diameter of car wheel (d) = 77 cm
Circumference = πd = \(\frac { 22 }{ 7 }\) x 77 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 20

Question 20.
Solution:
No. of revolutions = 5000
Distance covered = 11 km = 11 x 1000 = 11000 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 21

Question 21.
Solution:
Length of hour hand (r) = 4.2 cm
and length of minutes hand (R) = 7cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20E 22
Distance covered by hour hand in 24 hours = 2πR
= 2 x \(\frac { 22 }{ 7 }\) x 4.2
= 26.4 cm
But distance covered by minute hand in one hour = 2πR = 2 x \(\frac { 22 }{ 7 }\) x 7 = 44 cm
and distance covered by minute hand in 24 hours = 44 x 24 cm = 1056 cm
Sum of distance covered by these hands = 26.4 + 1056 = 1082.4 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D.

Other Exercises

Question 1.
Solution:
(i) Base of the triangle = 42 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 1
Height = 25 cm
Area = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x 42 x 25 = 525 cm²
(ii) Base of the triangle = 16.8 m
and height = 75 cm = 0.75 m
Area = \(\frac { 1 }{ 2 }\) x Base x height
= \(\frac { 1 }{ 2 }\) x 16.8 x 0.75 m2 = 6.3 m²
(iii) Base of a triangle (b) = 8 m = 80 cm
and height (h) = 35 cm
Area = \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) x 80 x 35 = 1400 cm²

Question 2.
Solution:
Base of triangle = 16 cm
area of the triangle = 72 cm²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 2

Question 3.
Solution:
Area of triangular region = 224 m²
Base = 28 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 3

Question 4.
Solution:
Area of triangle = 90 cm²
and height (h) = 12 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 4

Question 5.
Solution:
Let height of a triangular field = x m
Then base (b) = 3x m
and area = \(\frac { 1 }{ 2 }\) bh = \(\frac { 1 }{ 2 }\) x 3x x x
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 5

Question 6.
Solution:
Area of the right angled triangle = 129.5 cm²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 6

Question 7.
Solution:
In right angled ∆ABC,
Base BC = 1.2 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 7
and hypotenuse AC = 3.7 m
But AC² = AB² + BC² (Pythagoras Theorem)
⇒ (3.7)² = AB² + (1.2)²
⇒ 13.69 = AB² + 1.44
⇒ AB² = 13.69 – 1.44
⇒ AB² = 12.25 = (3.5)²
⇒ AB = 3.5 m
Now, area of ∆ABC = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x 1.2 x 3.5 m² = 2.1 m²

Question 8.
Solution:
Legs of a right angled triangle = 3 : 4
Let one leg (base) = 3x
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 8
Then second leg (altitude) = 4x
Area = \(\frac { 1 }{ 2 }\) x base x altitude
= \(\frac { 1 }{ 2 }\) x 3x x 4x = 6x²
6x² = 1014
⇒ x² = \(\frac { 1014 }{ 6 }\) = 169 = (13)²
x = 13
one leg'(Base) = 3x = 3 x 13 = 39 cm
and second leg (altitude) = 4x = 4 x 13 = 52 cm

Question 9.
Solution:
One side BC of a right triangular scarf = 80 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 9
and longest side AC = 1 m = 100 cm
By Pythagoras Theorem,
AC² = AB² + BC²
⇒ (100)² = AB² + (80)²
⇒ 10000 = AB² + 6400
⇒ AB² = 10000 – 6400
⇒ AB² = 3600 = (60)²
⇒ AB = 60
Second side = 60 cm
Area of the scarf = \(\frac { 1 }{ 2 }\) x b x h
= \(\frac { 1 }{ 2 }\) x 80 x 60 cm2 = 2400 cm²
Rate of cost = Rs. 250 per m²
Total cost =\(\frac { 2400 }{ 100 x 100 }\) x 250 = Rs. 60

Question 10.
Solution:
(i) Side of the equilateral triangle (a) = 18 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 10

Question 11.
Solution:
Area of equilateral triangle = 16√3 cm²
Let each side = a
then \(\frac { \surd 3 }{ 4 }\) a² = 16√3
⇒ a² = \(\frac { 16\surd 3\times 4 }{ \surd 3 }\)
⇒ a² = 64 = (8)²
a = 8 cm
Each side = 8 cm

Question 12.
Solution:
Each side of an equilateral triangle = 24cm
Length of altitude = \(\frac { \surd 3 }{ 2 }\) a = \(\frac { \surd 3 }{ 2 }\) x 24
= 12√3 cm = 12 (1.73) = 20.76 cm

Question 13.
Solution:
(i) a = 13 m, b = 14 m, c = 15 m
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 11
= 2 x 2 x 3 x 7 x 7 x 7 = 4116 m²

Question 14.
Solution:
Let a = 33 cm, b = 44 cm, c = 55 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 12

Question 15.
Solution:
Perimeter of the triangle = 84 cm
Ratio in side = 13 : 14 : 15
Sum of ratios =13 + 14 + 15 = 42
Let then first side = \(\frac { 84 x 13 }{ 42 }\) = 26 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 13

Question 16.
Solution:
Let a = 42 cm, b = 34 cm, c = 20 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 14

Question 17.
Solution:
In isosceles ∆ABC
Base BC = 48 cm.
and AB = AC = 30cm.
Let AD ⊥ BC
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 15

Question 18.
Solution:
Perimeter of an isosceles triangle = 32 cm
Base = 12 cm
Sum of two equal sides = 32 – 12 = 20 cm
Length of each equal side = \(\frac { 20 }{ 2 }\) = 10cm
Let AD ⊥ BC
BD = DC = \(\frac { 12 }{ 2 }\) = 6 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 16

Question 19.
Solution:
In quadrilateral ABCD,
diagonal AC = 26 cm.
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 17
and perpendiculars DL = 12.8cm, BM = 11.2 cm
Area of quadrilateral ABCD
= \(\frac { 1 }{ 2 }\) (Sum of perpendicular) x diagonal
= \(\frac { 1 }{ 2 }\) (12.8 + 11.2) x 26 cm²
= \(\frac { 1 }{ 2 }\) x 24 x 26 = 312 cm²

Question 20.
Solution:
In quad. ABCD,
AB = 28 cm, BC = 26 cm, CD = 50 cm, DA = 40 cm
and diagonal AC = 30 cm
In ∆ABC,
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 18
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 19

Question 21.
Solution:
ABCD is a rectangle in which AB = 36 m
and BC = 24m
In ∆AED,
EF = 15 m
AD = BC = 24 m.
Now area of rectangle ABCD = l x b = 36 x 24 cm² = 864 cm²
Area of ∆AED = \(\frac { 1 }{ 2 }\) x AD x EF
= \(\frac { 1 }{ 2 }\) x 24 x 15 cm² = 180 cm²
Area of shaded portion = 864 – 180 = 684 m²

Question 22.
Solution:
In the fig. ABCD is a rectangle in which AB = 40 cm, BC = 25 cm.
P, Q, R and S and the mid points of sides, PQ, QR, RS and SP respectively
Then PQRS is a rhombus.
Now, join PR and QS.
PR = BC = 25cm and QS = AB = 40cm
Area of PQRS = \(\frac { 1 }{ 2 }\) x PR x QS
= \(\frac { 1 }{ 2 }\) x 25 x 40 = 500 cm²

Question 23.
Solution:
(i) Length of rectangle (l) = 18 cm
and breadth (b) = 10 cm
Area = l x b = 18 x 10 = 180 cm²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 20
Area of right ∆EBC = \(\frac { 1 }{ 2 }\) x 10 x 8 = 40 cm²
and area of right ∆EDF = \(\frac { 1 }{ 2 }\) x 10 x 6 = 30 cm²
Area of shaded region = 180 – (40 + 30) = 180 – 70 = 110 cm²
(ii) Side of square = 20 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 21
Area of square = a² = (20)² = 400 cm²
Area of right ∆LPM = \(\frac { 1 }{ 2 }\) x 10 x 10 cm² = 50 cm²
Area of right ∆RMQ = \(\frac { 1 }{ 2 }\) x 10 x 20 = 100 cm²
and area of right ∆RSL = \(\frac { 1 }{ 2 }\) x 20 x 10 = 100 cm²
Area of shaded region = 400 – (50 + 100 + 100) cm2 = 400 – 250 = 150 cm²

Question 24.
Solution:
In the quadrilateral ABCD
BD = 24 cm
AL ⊥ BD and CM ⊥ BD
AL = 5 cm and CM = 8 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20D 22

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D are helpful to complete your math homework.

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RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C

RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C

These Solutions are part of RS Aggarwal Solutions Class 7. Here we have given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C.

Other Exercises

Question 1.
Solution:
In parallelogram ABCD,
Base AB = 32cm
Height DL = 16.5cm.
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 1
Area = Base x height = 32 x 16.5 cm² = 528 cm²

Question 2.
Solution:
Base of parallelogram = 1 m 60m = 160 cm
and height = 75 cm
Area = Base x height = 160 x 75 = 12000 cm²
= \(\frac { 12000 }{ 10000 }\) m² = 1.2m²

Question 3.
Solution:
Base of parallelogram = 14dm = 140cm
and height = 6.5 dm = 65cm
Area (in cm²) = Base x height = 140 x 65 = 9100 cm²
Area (in m²) = \(\frac { 140 }{ 100 }\) x \(\frac { 65 }{ 100 }\) = \(\frac { 9100 }{ 10000 }\)
= 0.91 m²

Question 4.
Solution:
Area of parallelogram = 54 cm²
Base = 15 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 2

Question 5.
Solution:
Area of parallelogram ABCD = 153 cm²
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 3

Question 6.
Solution:
In parallelogram ABCD
AB || DC and AD || BC and AB = DC, AD = BC
AB = DC = 18cm, BC = 12cm
Area of parallelogram ABCD = Base x altitude = DC x AL = 18 x 6.4cm2 = 115.2 cm²
and area of parallelogram ABCD = BC x AM
⇒ 115.2 = 12 x AM
⇒ AM = 9.6 cm

Question 7.
Solution:
In parallelogram ABCD
AB = DC = 15 cm
BC = AD = 8 cm.
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 4
Distance between longer sides AB and DC is 4cm
i.e. perpendicular DL = 4cm.
DM ⊥ BC.
Area of parallelogram = Base x altitude = AB x DL = 15 x 4 = 60 cm²
Again let DM = x cm
area ABCD = BC x DM = 8 x x = 8x cm²
8x cm² = 60 cm²
⇒ x = 7.5 cm
Distance between shorter lines = 7.5 cm

Question 8.
Solution:
Let Base of the parallelogram = x
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 5
⇒ x² = 108 x 3 = 324 = (18)²
⇒ x = 18
Base = 18 cm
and altitude = \(\frac { 1 }{ 3 }\) x 18 = 6 cm

Question 9.
Solution:
Area of parallelogram = 512 cm²
Let height of the parallelogram = x
Then base = 2x
Area = Base x height
⇒ 512 = 2x x x
⇒ 2x² = 512
⇒ x² = 256 = (16)²
⇒ x = 16
Base = 2x = 2 x 16 = 32 cm
and height = x = 16 cm

Question 10.
Solution:
(i) Each side of rhombus = 12 cm
height = 7.5 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 6
Area = Base x height = 12 x 7.5 = 90 cm²
(ii) Each side = 2 cm = 20 cm
Height = 12.6 cm
Area = Base x height = 20 x 12.6 = 252 cm²

Question 11.
Solution:
(i) Diagonals of rhombus ABCD are 16 cm and 28 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 7

Question 12.
Solution:
In rhombus ABCD, diagonals AC and BD intersect each other at right angles at O.
AO = OC and BO = OD
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 8
AO = \(\frac { 1 }{ 2 }\) x AC = \(\frac { 1 }{ 2 }\) x 24 cm = 12cm
Let OB = x
Each side of rhombus = 20cm
In right ∆AOB
AO² + OB² = AB² (Pythagoras Theorem)
⇒ (12)² + OB² = (20)²
⇒ 144 + OB² = 400
⇒ OB² = 400 – 144 = 256 = (16)²
⇒ OB = 16
But BD = 2 BO = 2 x 16 = 32cm
Now, area of rhombus = \(\frac { Product of diagonals }{ 2 }\)
= \(\frac { 24 x 32 }{ 2 }\) cm2 = 384 cm²

Question 13.
Solution:
Area of rhombus = 148.8 cm²
one diagonal = 19.2 cm
Let second diagonal = x
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 9

Question 14.
Solution:
Area of rhombus = 119 cm²
Perimeter = 56 cm
Its side = \(\frac { 56 }{ 4 }\) = 14 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 10

Question 15.
Solution:
Area of rhombus = 441 cm²
Height = 17.5 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 11

Question 16.
Solution:
Base of a triangle = 24.8 cm
Corresponding height = 16.5 cm
RS Aggarwal Class 7 Solutions Chapter 20 Mensuration Ex 20C 12

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C are helpful to complete your math homework.

If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you.