RBSE Class 6 Maths Chapter 6 Perimeter and Area Solutions

Practicing RBSE Class 6 Maths Solutions and Class 6 Maths Chapter 6 Perimeter and Area Solutions Question Answer helps develop logical thinking and accuracy.

Perimeter and Area Class 6 Solutions

Ganita Prakash Class 6 Chapter 6 Solutions Perimeter and Area

Figure it Out (Page 132)

Question 1.
Find the missing terms :
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?.
(b) Perimeter of a square = 20 cm; side of a length = ?.
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?.
Solution:
(a) Perimeter of the rectangle = 2 × (length + breadth)
or 14 = 2 × (length + 2)
or 14 ÷ 2 = length + 2
or 7 = length + 2
or length = 7 – 2 = 5 cm

(b) Perimeter of the square = 4 × length of a side
or 20 = 4 × length of a side
or 20 ÷ 4 = length of a side
∴ length of a side = 5 cm

(c) Perimeter of the rectangle = 2 × (length + breadth)
or 12 = 2 × (3 + breadth)
or 12 ÷ 2 = 3 + breadth
or 6 = 3 + breadth
or breadth = 6 – 3
breadth = 3 m

Question 2.
A rectangle having sidelengths 5 cm and 3 cm is made using a piece of wire. If the wire is straightened and then bent to form a square, what will be the length of a side of the square?
Solution:
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × (5 + 3)
= 2 × 8 = 16 cm
Now, the wire is straightened and then bent to form a square.
∴ Perimeter of the square = 16 cm
or 4 × length of a side = 16 cm
or Length of a side = \(\frac{16}{4}\) = 4 cm
Hence, required length of a side of the square = 4 cm

Question 3.
Find the length of the third side of a triangle having a perimeter of 55 cm and having two sides of length 20 cm and 14 cm, respectively.
Solution:
Let the length of the third side be x cm.
Perimeter of the triangle = Sum of the lengths of its three sides
or 55 = 20 + 14 + x
or x = 55 – 20 – 14 = 21 cm

Question 4.
What would be the cost of fencing a rectangular park whose length is 150 m and breadth is 120 m, if the fence costs ₹40 per metre?
Solution:
Length of the rectangular park = 150 m
Breadth of the rectangular park = 120 m
then, Perimeter = 2 × (length + breadth)
or Perimeter = 2 × (150 + 120)
= 2 × 270
= 540 m
Cost of fencing the park = ₹ 40 per metre
∴ Total cost = 40 × 540 = ₹ 21600

Question 5.
A piece of string is 36 cm long. What will be the length of each side, if it is used to form:
(a) A square,
(b) A triangle with all sides of equal length, and
(c) A hexagon (a six sided closed figure) with sides of equal length?
Solution:
(a) Length of the piece of string = 36 cm
Let the length of each side of square be a, then
4a = perimeter = 36
∴ a = \(\frac{36}{4}\) = 9 cm

(b) Let the length of each side of the triangle be a, then
3a = perimeter = 36
∴ a = \(\frac{36}{3}\) = 12 cm

(c) Let the length of each side of the hexagon be a, then
6a = perimeter = 36
∴ a = \(\frac{36}{6}\) = 6 cm

Question 6.
A farmer has a rectangular field having length 230 m and breadth 160 m. He wants to fence it with 3 rounds of rope as shown. What is the total length of rope needed?

Solution:
Perimeter of the rectangular field = 2 × (length + breadth)
Here, length = 230 m, breadth = 160 m
∴ Perimeter = 2 × (230 + 160)
= 2 × 390
= 780 m
Distance covered by the farmer in one round = 780m
∴ The total length of the rope required by the farmer for making a three-splash fence
= 3 × 780 = 2340 m

Figure it Out (Page 133)

Each track is a rectangle. Akshi’s track has length 70 m and breadth 40 m. Running one complete round on this track would cover 220 m, i.e., 2 × (70 + 40) m = 220 m. This is the distance covered by Akshi in one round.

Question 1.
Find out the total distance Akshi has covered in 5 rounds.
Solution:
Total distance covered by Akshi in 5 rounds
= 5 × distance covered in one round
= 5 × 220 = 1100 m

Question 2.
Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
Solution:
Total distance covered by Toshi in 7 rounds
= 7 × distance covered in one round
= 7 × 180 = 1260 m
∴ 1260 m > 1100 m
Hence, Toshi ran a longer distance.

Question 3.
Think and mark the positions as directed—
(a) Mark ‘A’ at the point where Akshi will be after she ran 250 m.
(b) Mark ‘B’ at the point where Akshi will be after she ran 500 m.
(c) Now, Akshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘C’.
(d) Mark ‘X’ at the point where Toshi will be after she ran 250 m.
(e) Mark ‘Y’ at the point where Toshi will be after she ran 500 m.
(f) Now, Toshi ran 1000 m. How many full rounds has she finished running around her track? Mark her position as ‘Z’.
Solution:

(a) Distance covered by Akshi in one round = 220 m
Total distance covered by Akshi = 250 m
After one round Akshi runs 250 – 220 = 30 m
more distance and reaches at point A (See figure).

(b) Total distance covered by Akshi = 500 m
Number of rounds taken by Akshi = \(\frac{500}{220}\) = 2.27
(Approx.)
After taking tw o full rounds and running 500 m total distance, Akshi reaches at point B (See figure).

(c) Total distance covered by Akshi = 1000 m
Number of rounds taken by Akshi = \(\frac{1000}{220}\) = 4.54
(Approx.)
After taking 4 full rounds and running 1000 m total distance, Akshi reaches at point C (See figure).

(d) Distance covered by Toshi in one round = 180m
Total distance covered by Toshi = 250 m
After one round, Toshi covers 250 – 180 = 70 m
more distance and reaches at the point X (See figure).

(e) Total distance covered by Toshi = 500 m
Number of rounds taken by Toshi = \(\frac{500}{180}\) = 2.77
(Approx.)
∴ To run 500 m distance, Toshi takes 2 full rounds and reaches at the point Y (See figure).

(f) Total distance covered by Toshi = 1000 m
Number of rounds taken by Toshi = \(\frac{1000}{180}\) = 5.56
(Approx.)
∴ To run 1000 m distance, Toshi takes 5 full rounds and covers 1000 – 5 × 180 = 1000 – 900 = 100 m more distance and reaches at point Z (See figure).

Figure it Out (Page 138)

Question 1.
The area of a rectangular garden 25 m long is 300 sq m. What is the width of the garden?
Solution:
Given, length of the rectangular garden = 25 m
Area of the rectangular garden = 300 sq m
Area of a rectangle = length × width
∴ 300 = 25 × width
∴ Width = \(\frac{300}{25}\) = 12 m

Question 2.
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?
Solution:
Here, length of the rectangular plot of land = 500 m
and, width = 200 m
∴ Area of the rectangular plot of land = length × width
= 500 × 200 = 1,00,000 sq m
Cost of tiling the rectangular plot of land = ₹8
per hundred sq m
= ₹ \(\frac{8}{100}\) per sq m
Total cost of tiling the rectangular plot of land = \(\frac{8}{100}\) × 1,00, 000 = ₹ 8000

Question 3.
A rectangular coconut grove is 100 m long and 50 m wide. If each coconut tree requires 25 sq m, what is the maximum number of trees that can be planted in this grove?
Solution:
Length of the rectangular coconut grove = 100 m
and, breadth = 50 m
Area of the grove = 100 × 50 = 5000 sq m
Each coconut tree requires space = 25 sq m
∴ The maximum number of trees that can be planted in the grove \(\frac{5000}{25}\) = 200

Question 4.
By splitting the following figures into rectangles, find their areas (all measures are given in metres) :

Solution:
(a) Let the figure be divided into rectangles A. B. C and D.
Area of rectangle A = (3 × 3) sq m = 9 sq m
Area of rectangle B = (1 × 2) sq m = 2 sq m
Area of rectangle C = (3 × 3) sq m = 9 sq m
Area of rectangle D = (4 × 2) sq m = 8 sq m

∴ Total area of the figure
= (9 + 2 + 9 + 8) sq m
= 28 sq m

(b) Let the figure be divided into rectangles A, B and C as shown in the figure.

Area of the rectangle A = (2 × 1) sq m = 2 sq m
Area of the rectangle B = (5 × 1) sq m = 5 sq m
Area of the rectangle C = (2 × 1) sq m = 2 sq m
∴ Total area of the figure = (2 + 5 + 2) sq m = 9 sq m

Figure it Out (Page 139)

Cut out the tangram pieces given at the end of your textbook.

Question 1.
Explore and figure out how many pieces have the same area.
Solution:
Here, two pieces A and B have the same area. Similarly, shape C and E have the same area.

Question 2.
How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D and E?
Solution:
Shape D is two times bigger as compared to shape C. Clearly, the areas of shape C and E added together is equal to the area of shape D.

Question 3.
Which shape has more area: Shape D or F? Give reasons for your answer.
Solution:
Here, the middle triangle and the square are made up of two small triangles, each of them having area twice the area of the surrounding triangle so, shape D and F have the same area.

Question 4.
Which shape has more area: Shape F or G? Give reasons for your answer.
Solution:
Here, the middle triangle and rhombus are made up of two small tangram triangles, each of them having twice the area of the small triangle. So F and G both have the same area.

Question 5.
What is the area of Shape A as compared to Shape G? Is it twice as big? Four times as big?
Hint: In the tangram pieces, by placing the shapes over each other, we can find out that Shapes A and B have the same area, Shapes C and E have the same area. You would have also figured out that Shape D can be exactly covered using Shapes C and E, which means Shape D has twice the area of Shape C or Shape E, etc.
Solution:
The area of shape A is twice the area of the shape G.

Question 6.
Can you now figure out the area of the big square formed with all seven pieces in terms of the area of Shape C?
Solution:
Let the area of shape C = x
Area of D = 2 × area of C = 2x
Area of E = Area of C = x
Area of F = 2 × area of C = 2x
Area of G = 2 × area of C = 2x
Area of A = 2 × area of F = 2 × 2x = 4x
Area of B = Area of A = 4x
∴ Area of the big square = Total area of shapes A, B, C, D, E, F and G
= 4x + 4x + x + 2x + x + 2x + 2x
= 16x
= 16 (area of shape C)
Hence, the area of the big square is 16 times the area of shape C.

Question 7.
Arrange these 7 pieces to form a rectangle. What will be the area of this rectangle in terms of the area of Shape C now? Give reasons for your answer.
Solution:
The tangram with all seven pieces is a 5- piece tangram square with two large triangles. All seven pieces together form a rectangle. Now the area of this rectangle is 16 times the area of shape C as explained in Q. 6.

Question 8.
Are the perimeters of the square and the rectangle formed from these 7 pieces different or the same? Give an explanation for your answer.
Solution:
The perimeters of the square and the rectangle formed from these 7 pieces are different as shown in figure. Because all shapes are arranged differently.

Figure it Out (Page 144)

Question 1.
Find the areas of the figures below by dividing them into rectangles and triangles.

Solution:
(a)

Covered area	Number	Estimated area (sq m)
Full small squares	20	20 × 1 = 20
Exactly half of a square	–	–
More than half of a square	4	4 × 1 = 4
Less than half of a square	4	0

∴ Total area of the figure = 20 + 4 = 24 sq m.

(b)

Covered area	Number	Estimated area (sq m)
Full small squares	25	25 × 1 = 25
Exactly half of a square	0	0
More than half of a square	4	4 × 1 = 4
Less than half of a square	4	0

∴ Total area of the figure = 25 + 4 = 29 sq m.

(c)

Covered area	Number	Estimated area (sq m)
Full small squares	36	36 × 1 = 36
Exactly half of a square	2	2 × \(\frac{1}{2}\) = 1
More than half of a square	8	8 × 1 = 8
Lesss than half of a square	6	6 × 0 = 0

∴ Total area of the figure = 36 + 1 + 8 = 45 sq m.
(d)

Covered area	Number	Estimated area (sq m)
Full small squares	13	13 × 1 = 13
Exactly half of a square	0	0
More than half of a square	3	3 × 1 = 3
Less than half of a square	2	2 × 0 = 0

∴ Total area of the figure = 13 + 3 = 16 sq m.

(e)

Covered area	Number	Estimated area (sq m)
Full small squares	5	5 × 1 = 5
Exactly half of a square	4	4 × \(\frac{1}{2}\) = 2
More than half of a square	4	4 × 1 = 4
Less than half of a square	3	3 × 0 = 0

∴ Total area of the figure = 5 + 2 + 4 = 11 sqm.

Figure it Out (Page 149)

Question 1.
Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
Solution:
Area of the first rectangle = 5 × 10 = 50 sq m
Area of the second rectangle = 2 × 7 = 14 sq m
Sum of the areas of both rectangles = 50 + 14 = 64 sq m
Let the one side of the required rectangle be x and other side be y.
Then, area of the rectangle – x × y = xy
So, xy = 64
When x = 1, y = \(\frac{64}{1}\) = 64
When x = 2, y = \(\frac{64}{2}\) = 32
When x = 4, y = \(\frac{64}{2}\) = 16
When x = 8, y = \(\frac{64}{8}\) = 8
Hence, the required dimensions of the rectangle are (8 m × 8 m), (4 m × 16 m), (2 m × 32 m) and (1 m × 64 m).

Question 2.
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Solution:
Width of the rectangular garden = \(\frac{\text { Area of the rectangular garden }}{\text { Length of the rectangular garden }}\)
= \(\frac{1000}{50}\) = 20 m

Question 3.
The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
Solution:
Length of the floor of the room = 5m
and breadth = 4 m
Side of the square carpet = 3 m

Area of the floor = length × breadth
= 5 × 4 = 20 sq m
Area of the carpet = side × side
= 3 × 3 = 9 sq m
∴ Area that is not carpeted = 20 – 9 = 11 sqm

Question 4.
Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
Solution:
Length of the garden = 15 m
Breadth of the garden = 12 m
Area of the garden = 15 × 12 = 180 sq m
Length of the flower bed = 2m
Width of the flower bed = 1 m
Area of the flow er bed = 2 × 1 = 2 sq m
Area of 4 flower beds = 4 × 2 = 8 sq m
Area available for laying down a lawn = (180 – 8) sq m
= 172 sq m

Question 5.
Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape
B. Draw two such shapes satisfying the given conditions.
Solution:
Area of the shape A = 18 sq m
Possible dimensions of the shape A = 18 × 1, 9 × 2, 6 × 3
Area of the shape B = 20 sq m
Possible dimensions of the shape B = 20 × 1, 10 × 2, 4 × 5
Possible perimeters of shape A = 2 × (18 + 1), 2 × (9 + 2), 2 × (6 + 3)
= 2 × 19, 2 × 11, 2 × 9
= 38, 22, 18
Possible perimeters of shape B = 2 × (20 + 1), 2 × (10 + 2), 2 × (4 + 5)
= 2 × 21, 2 × 12, 2 × 9
= 42, 24, 18
It is given that the perimeter of figure A is greater than the perimeter of figure B. Clearly, when the perimeter of figure A is 38, the perimeter of B can be 24 or 18 and w hen the perimeter of A is 22, the perimeter of B can be 18.
Hence,

Question 6.
On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
Solution:
Perimeter of the border
= 2 × (length + breadth)
= 2 × (1 + 1.5)
= 2 × 2.5
= 5 cm

Question 7.
Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
Solution:
Area of the rectangle = 12 × 8 = 96 sq units
Area of the rectangle drawn inside = \(\frac{1}{2}\) × 96 = 48
square units
∴ Dimensions of the new rectangle
= 12 × 4, 16 × 3, 8 × 6, 1 × 48, 2 × 24
Hence, the dimensions of the rectangle that can fit inside the first rectangle without touching will be 8 × 6.

Question 8.
A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here?
(a) The area of each rectangle is larger than the area of the square.
(b) The perimeter of the square is greater than the perimeters of both the rectangles added together.
(c) The perimeters of both the rectangles added together is always 1\(\frac{1}{2}\) times the perimeter of the square.
(d) The area of the square is always three times as large as the areas of both rectangles added together.
Solution:
Here we take a square paper of side 1 unit and fold it in half.

Area of the square = 1 × 1 = 1 sq unit
Perimeter of the square = 4 × 1 = 4 units
Perimeter of each rectangle formed by folding the paper in half
= 1 + \(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) = 3 units
Area of each rectangle = 1 × \(\frac{1}{2}\) = \(\frac{1}{2}\) sq unit

(a) Area of one rectangle (\(\frac{1}{2}\) unit) < area of square (1 unit). So, the statement (a) is false.

(b) Perimeter of the square = 4 units
Perimeters of both the rectangles added together = 3 + 3 = 6 sq units
So, the perimeters of both the rectangles added together is greater than the perimeter of the square. Hence, the statement (b) is false.

(c) Perimeters of both the rectangles added together = 6 units
Perimeter of the square = 4 units
1\(\frac{1}{2}\) times the perimeter of the square = 1\(\frac{1}{2}\) × 4

= \(\frac{3}{2}\) × 4 = 6 units.
Hence, the statement (c) is true.

(d) Area of the square = 1 sq unit
Area of both rectangles added together = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1 sq unit
Here, both the areas are equal. So, the statement

(d) is false.

Perimeter and Area Class 6 Question Answer

Perimeter and Area Class 6 Extra Questions

Multiple Choice Questions—

Question 1.
If the length of a rectangle is 15 cm and its width is 9 cm, then its perimeter will be :
(a) 24 cm
(b) 60 cm
(c) 36 cm
(d) 48 cm
Answer:
(d) 48 cm

Question 2.
The perimeter of square is :
(a) 2 × (length + breadth)
(b) 2 + (length × breadth)
(c) 4 × (length of one side)
(d) 4 × (length of the four sides)
Answer:
(c) 4 × (length of one side)

Question 3.
The length of each side of a regular pentagon is 3 cm. its perimeter will be :
(a) 15 cm
(b) 8 cm
(c) 16 cm
(d) 24 cm
Answer:
(a) 15 cm

Question 4.
If the side of a square is 10 m, then its perimeter is :
(a) 100 m
(b) 40 m
(c) 50 m
(d) 104 m
Answer:
(b) 40 m

Question 5.
If the length and breadth of a room are 9 meter and 6 meter respectively, then its area will be :
(a) 30 sq m
(b) 36 sq m
(c) 54 sq m
(d) 81 sq m
Answer:
(c) 54 sq m

Question 6.
The perimeter of the figure given below will be :

(a) 50 m
(b) 32 m
(c) 52 m
(d) 54 m
Answer:
(a) 50 m

Question 7.
The area of a rectangular field is 2000 sq m. If the width of the field is 50 m, then the length w ill be :
(a) 400 meter
(b) 40 meter 50
(c) 100 meter
(d) \(\frac{50}{2000}\) meter
Answer:
(b) 40 meter 50

Question 8.
The area of the given figure will be :

(a) 3 units
(b) 6 units
(c) 10 units
(d) 8 units
Answer:
(d) 8 units

Question 9.
The length and width of a rectangular field are 10 m and 8 m respectively. Wire fencing has to be done all around it. So, the total length of the wire will be :
(a) 80 m
(b) 18 m
(c) 72 m
(d) 36 m
Answer:
(d) 36 m

Question 10.
If one side of a square figure is 25 m, then its area will be :
(a) 100 sq m
(b) 50 sq m
(c) 625 sq m
(d) 25 sq m
Answer:
(c) 625 sq m

Fill in the blanks—

1. The perimeter of a regular octagon will be ……………………… .
Answer:
8 × length of one side

2. The measure of the region enclosed by the closed figure is called its ……………………… .
Answer:
area

3. Area of the square = ……………………… .
Answer:
side × side

4. Figures in which all sides and angles are equal, called ……………………… figures.
Answer:
closed regular figure

5. The perimeter of a regular pentagon is ……………………… .
Answer:
5 × side measure

6. The distance covered in completing one revolution around a closed figure is called ……………………… of that figure.
Answer:
perimeter

7. The length of one side of an equilateral triangle is 6.5 cm then its perimeter will be ……………………… .
Answer:
19.5 cm

8. The area of a rectangle is 120 sq m. If the length of the rectangle is 12 cm, then the width = ……………………… .
Answer:
10 cm

Write True/False for the following statements—

1. Perimeter of a rectangle = 4 × (length + width) (True/False)
2. Area of a square = 4 × side (True/False)
3. The perimeter of triangle whose sides are 2 cm, 3 cm and 4 cm is 9 cm. (True/False)
4. Area of a square of side 1 cm is 1 sq cm. (True/False)
Answer:
1. False
2. False
3. True
4. True

Make the right match—

Question 1.

Answer:
(1) – (D), (2) – (C), (3) – (B), (4) – (A).

Question 2.

Answer:
(1) – (C), (2) – (A), (3) – (B), (4) – (D).

Very Short Answer Type Questions—

Question 1.
Find the perimeter of an equilateral triangle of side 4 cm.
Solution:
Perimeter of equilateral triangle = 3 × length of one side
= 3 × 4 = 12 cm

Question 2.
The perimeter of a regular hexagon is 18 cm. Find the length of one of its sides.
Solution:
One side of regular hexagon = \(\frac{\text { Perimeter }}{\text { Number of sides }}\)
\(\frac{18 \mathrm{~cm}}{6}\) = 3 cm

Question 3.
Find the area of a square plot of side 8 meter.
Solution:
Area of a square plot = side × side
= 8 m × 8 m
= 64 sq m.

Question 4.
Meera goes to park of 150 m length and 80 m width. She takes one complete round of this park. Find the distance covered by her.
Solution:
Distance covered by Meera in completing one round of the park
= Sum of the measures of the sides of the park
= 150 m + 80 m + 150 m + 80 m
= 460 m.

Question 5.
Find the perimeter of a triangle whose sides are 10 cm, 14 cm and 15 cm respectively.
Solution:
Perimeter of the triangle
= Sum of the length of its sides
= 10 cm + 14 cm + 15 cm = 39 cm

Question 6.
Find the perimeter of a regular hexagon whose each side measures 8 m.
Solution:
A regular hexagon has 6 sides, so its perimeter
= 6 × length of each side
= 6 × 8 m = 48 m.

Question 7.
Find the side of a square whose perimeter is 20 m.
Solution:
A square has 4 sides, so we get the length of one side by dividing the perimeter by 4.
∴ One side of the square = 20 m ÷ 4 = 5 m.

Question 8.
The perimeter of a regular pentagon is 100 cm. Find the length of its each side.
Solution:
Perimeter = 100 cm
A regular pentagon has 5 equal sides. So, we get the length of one side by dividing the perimeter by 5.
∴ One side = 100 cm ÷ 5 = 20 cm.

Short Answer Type Questions—

Question 1.
The lid of a rectangular box of 40 cm length and 10 cm width is completely closed by a tape all around. Find the length of the tape required.
Solution:
Length of tape required = Perimeter of the lid of the rectangular box
= 2 × (length + breadth)
= 2 × (40 cm + 10 cm)
= 2 × 50 cm = 100 cm or 1 m
Hence, length of the tape required is 100 cm.

Question 2.
The dimensions of a table top surface are 2 m 25 cm and 1 m 50 cm. Find the perimeter of the table top surface.
Solution:
Perimeter of the table top surface
= 2 × (length + breadth)
= 2 × (2 m 25 cm + 1 m 50 cm)
= 2 × (3 m 75 cm)
= 2 × 3.75 m = 7.50 m
So, the perimeter of the table top surface is 7.5 m.

Question 3.
A photograph of 32 cm length and 21 cm width is to be framed using a wooden strip. Find the length of the wooden strip required.
Solution:
Perimeter of the photograph = 2 × (length + breadth)
= 2 × (32 cm + 21 cm)
= 2 × 53 cm = 106 cm
∴ Length of the wooden strip required is 106 cm.

Question 4.
Find the perimeter of each of the following figures :
(a) A triangle whose sides are 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle with the length of one side being 9 cm.
(c) An isosceles triangle with each equal side 8 cm and the third side 6 cm.
Answer:
(a) Perimeter = Sum of sides
= 3 cm + 4 cm + 5 cm
= 12 cm

(b) Perimeter = 3 × length of one side
= 3 × 9 cm
= 27 cm

(c) Perimeter = Sum of sides
= 8 cm + 8 cm + 6 cm
= 22 cm.

Question 5.
The length and breadth of a rectangular plot are 0.7 km and 0.5 km respectively. It is to be fenced around it in 4 rows with a wire. Find the length of the required wire.
Solution:
Here, the length of wire required will be 4 times the perimeter of the plot.
Perimeter of the plot = 2 × (length + width)
= 2 × (0.7 km + 0.5 km)
= 2 × 1.2 km = 2.4 km
∴ Total length of wire required = 4 × 2.4 km = 9.6 km.

Essay Type Questions—

Question 1.
Split the following figures into rectangles/squares and find the area of each. (All measures are given in cm.)

Solution:
(a) On dividing the given shape into 5 squares of 7 cm side each.

∴ Total area of the figure
= 5 × (Area of a square of side 7 cm)
= 5 × (7 × 7) sq cm
= 5 × 49 sq cm = 245 sq cm

(b) On dividing the given shape into two rectangles A and B.

Area of the rectangle (A) = 5 × 1 sq cm = 5 sq cm
Area of the rectangle (B) = 4 × 1 sq cm = 4 sq cm
∴ Total area of the figure = (5 + 4) sq cm = 9 sq cm.

Question 2.
A tile measures 5 cm × 12 cm. How many such tiles will be required to completely cover an area whose length and breadth are respectively :
(a) 144 cm and 100 cm.
(b) 70 cm and 36 cm.
Solution:
Area of the field = 144 cm × 100 cm = 14400 sq cm
Area of one tile = 5 cm × 12 cm = 60 sq cm
Number of tiles = \(\frac{\text { Area of the field }}{\text { Area of one tile }}\)
= (\(\frac{1}{2}\)) = 240

(b) Area of field = 70 cm × 36 cm = 2520 sq cm
Area of one tile = 5 cm × 12 cm = 60 sq cm
Number of tiles = \(\frac{\text { Area of the field }}{\text { Area of one tile }}\)
= (\(\frac{2520}{60}\)) = 240