CBSE Class 6

These NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.4

Question 1.
Write shaded portion as fraction. Arrange them in ascending and descending order using correct sign ‘<‘, ‘>’ between the fractions:

(c) Show \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}, \frac{6}{6}\) on the number line. Put appropriate signs between the fractions given.

Answer:
The fractions represented by the shaded portions are as follows:
(a) \(\frac{3}{8}, \frac{6}{8}, \frac{4}{8}, \frac{1}{8}\)
(b) \(\frac{8}{9}, \frac{4}{9}, \frac{6}{9}, \frac{6}{9}\)
When we compare two fractions
Note:
Like Fractions: If the denominator is same, the fraction with the greater numerator is greater.
Unlike Fractions: If the denominator is not same, we first get their equivalent fractions with a denominator which is the L.C.M of the denominators of both the fractions.
The above fractions in (a) and (b) are like fractions as they have the same denominator.
When comparing fractions with same denominator,
So, arranging the fractions in ascending order, we have
(a) \(\frac{1}{8}<\frac{3}{8}<\frac{4}{8}<\frac{6}{8}\)
(b) \(\frac{3}{9}<\frac{4}{9}<\frac{6}{9}<\frac{8}{9}\)
(c) Let us represent the fractions \(\frac{2}{6}, \frac{4}{6}, \frac{8}{6}, \frac{6}{6}\) on the number line.

Again these comparisons are between the like fractions, so numerator will dictate the sign:

Question 2.
Compare the fractions and put an appropriate sign.

Answer:
(a) Here, we have Tike fractions’ so we compare them by their numerators only.

(b) Here, we have ‘unlike fractions’ with same numerators, so we compare them with their denominators only.

(c) Like fractions, so we compare by their numerators only.

(d) ‘Unlike fractions’ with same nemerators so we compare them by their denominators, only.

Question 3.
Make five more such pairs and put appropriate signs.
Answer:

Question 4.
Look at the figures and write ‘<’ or ‘>’, ‘=’ between the given pairs of fractions.

Make five more such problems and solve them with you friends.
Answer:
(a) <
(b) >
(c) >
(d) =
(e) <

Question 5.
How quickly can you do this? Fill appropriate sign (‘<’ , ‘=’ , ‘>’)

Answer:

Question 6.
The following fractions represent just three different numbers. Separate them into three groups of equivalent fractions, by changing each one to its simplest form.
Answer:
Simplifying the fractions into simplest form.

Totally these are three graps of equivalent fractions 1
\(\frac { 1 }{ 6 }\) = (i) (v) (viii) (x) (xi)
\(\frac { 1 }{ 5 }\) = (ii) (vi) (vii)
\(\frac { 4 }{ 25 }\) = (iii) (iv) (ix) (xii)

Question 7.
Find answers to the following. Write and indicate how you solved them.
(a) Is \(\frac { 5 }{ 9 }\) equal to \(\frac { 4 }{ 5 }\) ?
(b) Is \(\frac { 9 }{ 16 }\) equal to \(\frac { 5 }{ 9 }\) ?
(c) Is \(\frac { 4 }{ 5 }\) equal to \(\frac { 16 }{ 20 }\) ?
(d) Is \(\frac { 1 }{ 15 }\) equal to \(\frac { 4 }{ 30 }\) ?
Answer:
We need to first simplify each fraction. After simplification if they are equal, else it is not equal
(a) Not equal
(b) Not equal
(c) Equal
(d) Not equal

Question 8.
Ila read 25 pages of a book containing 100 pages. Lalita read \(\frac { 2 }{ 5 }\) of the same book. who read less?
Answer:
Total number of pages in the book = 100 pages
Number of pages read by Ila = 25 pages
Fraction of pages read by Ila = \(\frac { 25 }{ 100 }\)
= \(\frac { 1 }{ 4 }\)
Lauta read \(\frac { 2 }{ 5 }\) of the same book
So, basically we need to compare the fraction \(\frac { 1 }{ 4 }\) and \(\frac { 2 }{ 5 }\)
Since, they are unlike fractions, we need to find LCM of 4 and 5 which is 20.
So \(\frac{1}{4}=\frac{5}{20}\)
\(\frac{2}{5}=\frac{8}{20}\)
Now, 8 > 5
So Ila read lesser than Lauta.

Question 9.
Rafiq exercised for \(\frac { 3 }{ 6 }\) of an hour, while Rohit exercised for \(\frac { 3 }{ 4 }\) of an hour. Who exercised for a longer time?
Answer:
Rafiq = \(\frac { 3 }{ 6 }\)
Rohit = \(\frac { 3 }{ 4 }\)
We need to compare the fractions. Since these are unlike fraction, we need to calculate the LCM and 4 and 6. LCM of 4 and 6 is 12.
Rafiq = \(\frac{3}{6}=\frac{6}{12}\)
Rahit = \(\frac{3}{4}=\frac{9}{12}\)
\(\frac{9}{12}>\frac{6}{12}\)
∴ Rohit exercised for a longer time than Rafiq.

Question 10.
In a class A of 25 students, 20 passed with 60% or more marks; in another class B of 30 students, 24 passed with 60% or more marks. In which class was a greater fraction of students getting with 60% or more marks?
Answer:
Total number of students in class A = 25 Number of students passed in first class in class A = 20.
Fraction of students of class A who passed in first class = \(\frac{20}{25}=\frac{4}{5}\)
Total number of students in class B = 30 Number of students passed in first class = 24.
Fraction of students of class B who passed in first class = \(\frac{24}{30}=\frac{4}{5}\)
Both class A and B have the same fraction of students who passed in first class.

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration InText Questions

NCERT In-text Question Page No. 206
Question 1.
Measure and write the length of the four sides of the top of your study table.
AB = …………… cm BC = ………….. cm
CD = ………….. cm DA = …………… cm

Now, the sum of the lengths of the four sides
= AB + BC + CD + DA
= ……….. cm + ……….. cm + ………….. cm + ……………. cm
= …………… cm
What is the perimeter?
Answer:
AB = 140 cm, BC = 80 cm,
AB = 140 cm, DA = 80 cm
∴ AB + BC + CD + DA
= 140 cm + 80 cm + 140 cm + 80 cm
= 140 cm
The perimeter 440 cm.
Note :This is an activity so the measures of AB, BC, CD and DA will be different for different table tops.

Question 2.
Measure and write the lengths of the four sides of a page of your notebook. The sum of the lengths of the four sides
= AB + BC + CD + DA
= ………….. cm + ………………. cm + ……………. cm + ……………. cm
= ……………… cm
What is the perimeter of the page?
Answer:
Please try yourself.

Question 3.
Meera went to a park 150 m long and 80 m wide. She took one complete round
on its boundary. What is the distance covered by her?
Answer:
The park is in the shape of rectangle.
Since she takes one complete round which is equal to perimeter of the rectangular park.
Perimeter of rectangular park
= 2 (length + breadth)
= 2 ( 150+ 80)
= 2 × 230
= 460 m
Therefore, the distance covered by her is 460m.

Question 4.
Find the perimeter of the following figures:
(a) Perimeter = AB + BC + CD + DA
= …………… + …………… + ……………. + ……………….+
= ………………

(b) Perimeter = AB + BC + CD + DA
= ……………. + …………. + ………….. + …………… +
= ……………..

(c) Perimeter = AB + BC + CD + DE + EF + FG + GH + HI+ IJ + JK+ KL + LA
= ……………… + ……………… + …………… + …………… +
……………… + …………….. + …………….. + ……………. +
……………. + …………….. + ……………. + …………….. +

(d) Perimeter=AB + BC + CD + DE + EF + FA
= ……………. + ……………. + ……………… + …………….. +
……………… + ……………. +
= ………………

Answer:
(a) AB = 40 cm, BC =10 cm,
AD = 10 cm and DC = 40 cm
∴ Perimeter = AB + BC + AD + DC
= 40 cm + 10 cm + 10 cm + 40 cm
= 100 cm

(b) AB = 5 cm, BC = 5 cm,
AD = 5 cm and AD = 5 cm
∴ Perimeter = AB + BC + CD + AD
= 5 cm + 5 cm + 5 cm + 5 cm = 20 cm

(c) Here, perimeter = AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LA
= 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3cm + 1 cm + 3 cm + 3 cm + 1 cm + 3 cm + 3 cm = 28 cm

(d) Here, perimeter = AB + BC + CD + DE + EF + FA
= 100 m + 120 m + 90 m + 45 m + 60 m + 80 m = 495m

NCERT In-text Question Page No. 208
Question 1.
Find the perimeter of the following rectangle :

Answer:
Do yourself.

NCERT In-text Question Page No. 211
Question 1.
Find various objects from your surroundings which have regular shapes and find their perimeters.
Answer:
It is a project work. You can find objects having regular shapes such as a ‘name plae’ a road signal’, etc. Measure their sides and find perimeters.

NCERT In-text Question Page No. 215
Question 1.
Draw any circle on a graph sheet. Count the squares and use them to estimate the area of the circular region.
Answer:
It is an ‘activity’. Please do it yourself.

Question 2.
Trace shapes of leaves, flower petals and other such objects on the graph paper and find their areas.
Answer:
It is an ‘activity’. Please do it yourself.

NCERT In-text Question Page No. 217
Question 1.
Find the area of the floor of your classroom.
Answer:
This is an activity. Please do it yourself.

Question 2.
Find the area of any one door in your house.
Answer:
This is an activity. Please do it yourself.

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.3

Question 1.
Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Answer:
(a) Area of rectangle
= length × breadth
= 3 cm × 4 cm = 12 cm²

(b) Area of rectangle
= length × breadth
= 12 m × 21 m = 252 m²

(c) Area of rectangle
= length × breadth
= 2 km × 3 km = 6 km²

(d) Area of rectangle
= lenght × breadth
= 2m ( lm = 100cm)
= 200 cm × 70 cm = 14000sq.cm

Question 2.
Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Answer:
(a) Area of square = side × side
= 10 cm × 10 cm = 100 cm²

(b) Area of square = side × side
= 14 cm × 14 cm = 196 cm²

(c) Area of square = side × side
= 5 m × 5 m = 25 m²

Question 3.
The length and breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Answer:
(a) Area of rectangle
= length × breadth
= 9 m × 6 m = 54 m²

(b) Area of rectangle
= length × breadth
= 3 m × 17 m = 51 m²

(c) Area of rectangle
= length × breadth
= 4 m × 14 m = 56 m²
Thus, the rectangle (c) has largest area, and rectangle (b) has smallest area.

Question 4.
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Answer:
Length of rectangle
= 50 m and Area of rectangle = 300 m² Since, Area of rectangle
= length × breadth
Therefore, breadth
\(\frac{\text { Area of rectangle }}{\text { Length }}\) = \(\frac{300}{50}\) = 6m
Thus, the breadth of the garden is 6 m.

Question 5.
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹ 8 per hundred sq m?
Answer:
Length of land
= 500 m and Breadth of land = 200 m
Area of land = length × breadth
= 500 m × 200 m = 1,00,000 m²
Cost of tilling 100 sq. m of land = 8
∴ Cost of tilling 1,00,000 sq. m of land
\(\frac{8 \times 100000}{100}\) = 8000

Question 6.
A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Answer:
Length of table = 2 m
Breadth of table = 1 m 50 cm = 1.50 m
Area of table = length × breadth
= 2 m × 1.50 m = 3 m²

Question 7.
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Answer:
Length of room = 4 m
Breadth of room = 3 m 50 cm = 3.50 m
Area of carpet = length × breadth
= 4 × 3.50= 14 m²

Question 8.
A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Answer:
Length of floor
= 5 m and breadth of floor = 4 m
Area of floor = length × breadth
= 5m × 4m = 20m²
Now, Side of square carpet = 3 m
Area of square carpet = side × side
= 3 × 3 = 9 m²
Area of floor that is not carpeted = 20m² – 9m² = 11m²

Question 9.
Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Answer:
Side of square bed = 1 m
Area of square bed = side × side
= 1m × 1m = 1m²
∴ Area of 5 square beds = 1 × 5 = 5 m²
Now, Length of land = 5 m
Breadth of land = 4 m
∴ Area of land = length × breadth
= 5 m × 4 m = 20 m²
Area of remaining part
= Area of land – Area of 5 flower beds = 20 m² – 5 m² = 15 m²

Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Answer:

Area of HKLM = 3 × 3 = 9 cm²
Area of IJGH = 1 × 2 = 2 cm²
Area of FEDG = 3 × 3 = 9 cm²
Area of ABCD = 2 × 4 = 8 cm²
Total area of the figure
= 9 + 2 +9 + 8
= 28 cm²

Area of ABCD = 3 × 1 = 3 cm²
Area of BDEF = 3 × 1 = 3 cm²
Area of FGHI = 3 × 1 = 3 cm²
Total area of the figure = 3 + 3 + 3 = 9 cm²

Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)

Answer:
(a) Area of rectangle ABCD
= 2 × 10 = 20 cm²
Area of rectangle DEFG
= 10 × 2 = 20 cm²
Total area of the figure
= 20 + 20 = 40 cm²

(b) There are 5 squares each of side 7 cm.
Area of one square
= 7 × 7 = 49 cm²
Area of 5 square
= 49 × 5 = 245 cm2

(c) Area of rectangle ABCD
= 5 × 1 = 5 cm²
Area of rectangle EFGH
= 4 × 1 = 4 cm²
Total area of the figure
= 5 + 4 cm²

Question 12.
How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Answer:
(a) Area of region
= 100 cm × 144 cm = 14400 cm²
Area of one tile
= 5 cm × 12 cm = 60 cm²
Number of tiles = \(\frac{\text { Area of region }}{\text { Area of one tile }}\) = \(\frac{14400}{60}\) = 240
Thus, 240 tiles are required.

(b) Area of region
= 70 cm × 36 cm = 2520 cm²
Area of one tile
= 5 cm × 12 cm = 60 cm²
Number of tiles
= \(\frac{\text { Area of region }}{\text { Area of one tile }}\) = \(\frac{2520}{60}\) = 42
Thus, 42 tiles are required.

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.2

Question 1.
Find the areas of the following figures by counting square:


Answer:
(a) Number of filled square = 9
∴ Area covered by squares = 9 × 1 = 9 sq. units

(b) Number of filled squares = 5
∴ Area covered by filled squares = 5 × 1 = 5 sq. units

(c) Number of full filled squares = 2
Number of half-filled squares = 4
∴ Area covered by full filled squares = 2 × 1 = 2 sq. units
And Area covered by half-filled squares = 4 × \(\frac { 1 }{ 2 }\) = 2 sq. units
∴ Total area = 2 + 2 = 4 sq. units

(d) Number of filled squares = 8
∴ Area covered by filled squares = 8 × 1 = 8 sq. units

(e) Number of filled squares = 10
∴ Area covered by filled squares = 10 × 1 = 10 sq. units

(f) Number of full filled squares = 2
Number of half-filled squares = 4
∴ Area covered by full filled squares = 2 × 1 = 2 sq. units
And Area covered by half-filled squares = 4 × \(\frac { 1 }{ 2 }\) = 2 sq. units
∴ Total area = 2 + 2 = 4 sq. units

(g) Number of full filled squares = 4
Number of half-filled squares = 4
∴ Area covered by full filled squares = 4 × 1 = 4 sq. units
And Area covered by half-fille squares = 4 × \(\frac { 1 }{ 2 }\) = 2 sq. units
∴ Total area = 4 + 2 = 6 sq. units

(h) Number of filled squares = 5
∴ Area covered by filled squares = 5 × 1 = 5 sq. units

(i) Number of filled squares = 9
∴ Area covered by filled squares = 9 × 1 = 9 sq. units

(j) Number of full filled squares = 2
Number of half-filled squares = 4
∴ Area covered by full filled squares = 2 × 1 = 2 sq. units
And Area covered by half-fille squares = 4 × \(\frac { 1 }{ 2 }\) = 2 sq. units
∴ Total area = 2 + 2 = 4 sq. units

(k) Number of full filled squares = 4
Number of half-filled squares = 2
∴ Area covered by full filled squares = 4 × 1 = 4 sq. units
And Area covered by half-filled squares = 2 × \(\frac { 1 }{ 2 }\) = 1 sq. units
∴ Total area = 4 + 1 = 5 sq. units

(l) Number of full filled squares = 3
Number of half-filled squares = 10
∴ Area covered by full filled squares = 3 × 1 = 3 sq. units
And Area covered by half-filled squares = 10 × \(\frac { 1 }{ 2 }\) = 5 sq. units
∴ Total area = 3 + 5 = 8 sq. units

(m) Number of full filled squares = 7
Number of half-filled squares = 14
∴ Area covered by full filled squares = 7 × 1 = 7 sq. units
And Area covered by half-filled squares = 14 × \(\frac { 1 }{ 2 }\) = 7 sq. units
∴ Total area = 7 + 7 = 14 sq. units

(n) Number of full filled squares = 10 Number of half-filled squares = 16
∴ Area covered by full filled squares = 10 × 1 = 10 sq. units
And Area covered by half-filled squares = 16 × \(\frac { 1 }{ 2 }\) = 8 sq. units
∴ Total area = 10 + 8 = 18 sq. units

These NCERT Solutions for Class 6 Maths Chapter 7 Fractions Ex 7.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 7 Fractions Exercise 7.1

Question 1.
Write the fraction representing the shaded portion.

Answer:

Question 2.
Colour the part according to the given fraction.
Answer:

Question 3.

Identify the error if any.
Answer:
The given figures do not represent the correct fractions as the part are unequal.

Question 4.
What fraction of a day is 8 hours?
Answer:
We know that in a day there are 24 hours.
So, 8 hours in a 24-hours day is represented as \(\frac{8}{24}=\frac{1}{3}\)

Question 5.
What fraction of an hour is 40 minutes?
Answer:
We know that in an hour there are 60
minutes. So, 40 minutes in a 60 min is
represented as \(\frac{40}{60}=\frac{2}{3}\)

Question 6.
Arya, Abhimanyu, and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
Answer:
(a) Since there are three people to eat, Arya should divide his sandwiches into 3 equal parts and give 1 part of each sandwich so that each person has an equal share.
(b) Each boy will receive \(\frac{1}{3}\) of a sandwich.

Question 7.
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Answer:
Kanchan has finished 20 dresses out of 30.
So Fraction would be \(\frac{20}{30}=\frac{2}{3}\)

Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Answer:
Natural numbers from 2 to 12 are 2, 3,4, 5, 6, 7, 8, 9, 10, 11 and 12 .
Prime numbers – 2, 3, 5, 7, 11
There are 11 natural numbers from 2 to 12 out of which 5 are prime numbers.
∴ Fraction of numbers that are prime is \(\frac{5}{11} .\)

Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Answer:
Natural numbers from 102 to 113 are 102, 103, 104, 105, 106, 107, 108, 109, 110, 111,
112 and 113
Prime numbers—103, 107, 109, 113 There are 12 natural numbers from 102 to
113 out of which 4 are prime numbers.
So, fraction of numbers that are prime is \(\frac{4}{12}=\frac{1}{3}\)

Question 10.
What fractions of these circles have X’s in them?

Answer:
There are 8 circles out of which 4 have X’s in them.
So, fraction of circles having X’s is \(\frac{4}{8}=\frac{1}{2}\)

Question 11.
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Answer:
Number of CDs Kristin purchased = 3
Number of CDs she received as gifts = 5 So, total number of CDs = 8 Kristin purchased 3 CDs out of a total number of 8 CDs.
Fraction of CDs she purchased = \(\frac{3}{8}\)
Kristin received 5 CDs out of a total number of 8 CDs.
Fraction of CDs she received as gifts = \(\frac{5}{8}\)

These NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Exercise 10.1

Question 1.
Find the perimeter of each of the following figures:

Answer:
(a) Perimeter = Sum of all the sides = 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
(c) Perimeter = Sum of all the sides =15 cm + 15 cm + 15 cm + 15 cm = 60 cm
(d) Perimeter = Sum of all the sides = 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
(e) Perimeter = Sum of all the sides 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
(f) Perimeter = Sum of all the sides = 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm + 4
cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm

Question 2.
The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Answer:
Total length of tape required
= Perimeter of rectangle
= 2 (length + breadth) = 2 (40 + 10)
= 2 × 50 = 100 cm = 1 m
Thus, the total length of tape required is 100 cm or 1 m.

Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Answer:
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 × (length + breadth) = 2 × (2.25 + 1.50) = 2 × 3.75 = 7.5 m
Thus, the perimeter of tabletop is 7.5 m.

Question 4.
What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Answer:
Length of wooden strip = Perimeter of photograph
Perimeter of photograph
= 2 x (length + breadth)
= 2 (32 + 21) = 2 × 53cm =106 cm
Thus, the length of the wooden strip required is equal to 106 cm.

Question 5.
A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Answer:
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 × (length + breadth)
= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km
= 2.4 × 1000 m = 2400 m
Thus, the length of wire = 4 × 2400 = 9600 m = 9.6 km

Question 6.
Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.
Answer:
(a) Perimeter of ΔABC = AB + BC + CA = 3 cm + 5 cm + 4 cm = 12 cm

(b) Perimeter of equilateral ABC
= 3 × side = 3 × 9 cm = 27 cm

(c) Perimeter of AABC = AB + BC + CA = 8 cm + 6 cm + 8 cm = 22 cm

Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Answer:
Perimeter of triangle
= Sum of all three sides
= 10 cm + 14 cm + 15 cm = 39 cm
Thus, the perimeter of triangle is 39 cm

Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer:
Perimeter of Hexagon
= 6 × length of one side
= 6 × 8 m = 48 m
Thus, the perimeter of hexagon is 48 m.

Question 9.
Find the side of the square whose perimeter is 20 m.
Answer:
Perimeter of square = 4 × side
⇒ 20 = 4 × side
⇒ Side = \(\frac{20}{4}\) = 5 cm
Thus, the side of square is 5 cm.

Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Answer:
Perimeter of regular pentagon = 100 cm
⇒ 5 × side = 100 cm
⇒ Side = \(\frac{100}{5}\) = 20 cm
Thus, the side of regular pentagon is 20 cm.

Question 11.
A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square?
(b) an equilateral triangle?
(c) a regular hexagon?
Answer:
Length of string = Perimeter of each figure
(a) Perimeter of square = 30 cm
⇒ 4 × side = 30 cm
⇒ Side = \(\frac{30}{4}\) = 7.5 cm 4
Thus, the length of each side of square is 7.5 cm.

(b) Perimeter of equilateral triangle = 30 cm
⇒ 3 × side = 30 cm
⇒ Side = \(\frac{30}{3}\) = 10 cm
Thus, the length of each side of equilateral triangle is 10 cm.

(c) Perimeter of hexagon = 30 cm
⇒ 6 × side = 30 cm
⇒ Side = \(\frac{30}{6}\) = 5 cm
Thus, the side of each side of hexagon is 5 cm.

Question 12.
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Answer:
Let the length of third side be X cm.
Length of other two side are 12 cm and 14 cm.
Now, Perimeter of triangle = 36 cm
⇒ 12 +14 + X = 36
⇒ 26 + X = 36
⇒ X = 36 – 26 = 10 cm
Thus, the length of third side is 10 cm

Question 13.
Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.
Answer:
Side of square = 250 m
Perimeter of square
= 4 × side = 4 × 250 = 1000 m
Since, cost of fencing of per meter = ₹20
Therefore, the cost of fencing of 1000 meters = 20 × 1000 = ₹ 20,000

Question 14.
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of × 12 per metre.
Answer:
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park
= 2 × (length + breadth)
= 2 × (175 + 125) = 2 × 300 = 600 m
Since, the cost of fencing park per meter = ₹ 12
Therefore, the cost of fencing park of 600 m = 12 × 600 = ₹ 7,200

Question 15.
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less distance?
Answer:
Distance covered by Sweety
= Perimeter of square park
Perimeter of square = 4 × side
= 4 × 75 = 300 m
Thus, distance covered by Sweety is 300 m. Now, distance covered by Bulbul =
Perimeter of rectangular park
Perimeter of rectangular park
= 2 × (length + breadth)
= 2 × (60 + 45) = 2 × 105 = 210 m
Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance.

Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?

Answer:
(a) Perimeter of square
= 4 × side = 4 × 25 = 100 cm

(b) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (40 + 10) = 2 × 50 = 100 cm

(c) Perimeter of rectangle
= 2 × (length + breadth)
= 2 × (30 + 20) = 2 × 50 = 100 cm

(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm = 100 cm
Thus, all the figures have same perimeter.

Question 17.
Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [Fig (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [(Fig (ii)]?
(c) Which has greater perimeter?
(d) Avneet wonders if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges i.e. they cannot be broken.)

Answer:
(a) The arrangement so formed in Fig (i) is a square of side equal to 3 × side of the square slab
∴ Perimeter of his arrangement in Fig (i)
= 4 × ( 3 × side of the square slab)
= 4 × ( 3 × 0.5) m = 4 × 1.5 m = 6 m

(b) The cross arrangement so formed in Fig (ii) by 9 square slabs, has perimeter, which consists of 20 length segments each equal to side of I a square slab i.e. 0.5 m
∴ Perimeter of her arrangement in Fig (ii)
= 20 × ( side of the square slab)
= 20 × 0.5 m = 10 m

(c) ∵ 10m > 6m
∴ Cross arrangement has greater perimeter

(d) ∵ Total number of tiles = 9
∴ The following arrangement will also have greater perimeter.

Since, this arrangement is in the form of rectangle with length and breadth as \(\frac { 9 }{ 2 }\) m and \(\frac { 1 }{ 2 }\) respectively.
∴ Perimeter = 2 \(\left(\frac{9}{2}+\frac{1}{2}\right)\)
= 2 × = 10m \(\left[\frac{10}{2}\right]\) = 10m