CBSE Class 7

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

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

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

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

Question 5.
Solution:
First 50 whole numbers
0, 1, 2, 3, 4, …, 49
Here, number of terms = 50, which is even

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

Question 8.
Solution:
Preparing the cumulative frequency table

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.

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 :

Here, number of terms (N) = 50, which is even

Hence median = 154.5 cm

Hope given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21B are helpful to complete your math homework.

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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

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

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:

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.

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.

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

Question 7.
Solution:
First six odd natural numbers are 1, 3, 5, 7, 9, 11

Question 8.
Solution:
First seven even natural numbers are 2, 4, 6, 8, 10, 12, 14

Question 9.
Solution:
First five prime numbers are 2, 3, 5, 7, 11

Question 10.
Solution:
First six multiples of 5 are 5, 10, 15, 20, 25, 30

Question 11.
Solution:

Question 12.
Solution:

Mean = Rs. 159

Question 13.
Solution:

Mean = Rs. 318

Question 14.
Solution:

Question 15.
Solution:

Hope given RS Aggarwal Solutions Class 7 Chapter 21 Collection and Organisation of Data Ex 21A are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

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

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

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

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

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

Question 12.
Solution:
(b) 154 cm²
Let the radius of the circle be r cm.
Circumference = (2πr) cm
(2πr) = 44

Question 13.
Solution:
(c) 98 cm²
Given that the diagonal of a square is 14 cm
Area of a square

Question 14.
Solution:
(b) 10 cm
Given that the area of the square is 50 cm²
We know:
Area of a square

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

Question 17.
Solution:
(a) 16√3 cm²
Given that each side of an equilateral triangle is 8 cm
Area of the equilateral triangle

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

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper are helpful to complete your math homework.

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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

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

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

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

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

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

Question 10.
Solution:
(a)
Total cost of carpet = Rs. 6000
Rate per metre = Rs. 50

Question 11.
Solution:
(a)
Sides are 13 cm, 14 cm, 15 cm

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

Question 14.
Solution:
(c)
Side of an equilateral triangle = 8cm

Question 15.
Solution:
(b)
Let a be the side of an equilateral triangle

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

Question 18.
Solution:
(c)
Let r be the radius of the circle Then
c = 2πr
2πr – r = 37

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

Question 21.
Solution:
(c)
Perimeter = 46 cm

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

Question 25.
Solution:
(d)
Side of a square = a
Area = a²
Side of equilateral triangle = a

Question 26.
Solution:
(a)
Let a be the side of a square
Area = a²
Then area of circle = a²
Let r be the radius

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

Question 28.
Solution:
(c)
Area of rhombus = 36 cm²
Length of one diagonal = 6 cm
Length of second diagonal

Question 29.
Solution:
(d)
Area of a rhombus =144 cm²
Let one diagonal (d1) = a
then Second diagonal (d2) = 2a

Larger diagonal = 2a = 2 x 12 = 24 cm

Question 30.
Solution:
(c)
Area of a circle = 24.64 m²

Question 31.
Solution:
(c)
Let original radius = r
Then its area = πr²
Radius of increased circle = r + 1

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

Hope given RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G 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 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

• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20A
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20B
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20C
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20D
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20E
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20F
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration Ex 20G
• RS Aggarwal Solutions Class 7 Chapter 20 Mensuration CCE Test Paper

Question 1.
Solution:
(i) Radius of the circle (r) = 21 cm

Question 2.
Solution:
(i) Diameter of the circle = 28 cm

Question 3.
Solution:
Circumference of a circle = 264 cm

Question 4.
Solution:
Circumference of the circle (c) = 35.2 m

Question 5.
Solution:
Area of the circle = 616 cm²

Question 6.
Solution:
Area of a circle = 1386 m²

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

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

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

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

= 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

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

Question 15.
Solution:
Diameter of largest circle (outer circle) = 21 cm


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

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

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