By using Ganita Prakash Book Class 6 Solutions and Chapter 6 Perimeter and Area Class 6 NCERT Solutions Question Answer, students can improve their problem-solving skills.
Class 6 Maths Chapter 6 Perimeter and Area Solutions
Perimeter and Area Class 6 Solutions Questions and Answers
6.1 Perimeter Figure it Out (Page No. 132)
Question 1.
Find the missing terms:
(a) Perimeter of a rectangle = 14 cm; breadth = 2 cm; length = ?
Solution:
Given: Perimeter = 14 cm and breadth = 2 cm
perimeter of a rectangle = 2(l + b)
⇒ 14 = 2(l + 2)
⇒ 7 = l + 2
⇒ l = 5 cm
(b) Perimeter of a square = 20 cm; side of a length = ?
Solution:
Given: Perimeter of square = 20 cm
side = \(\frac{Perimeter}{4}\)
= \(\frac{20}{4}\)
= 5 cm
(c) Perimeter of a rectangle = 12 m; length = 3 m; breadth = ?
Solution:
Given: Perimeter of a rectangle = 12 m and length = 3 m
perimeter of a rectangle = 2(l + b)
⇒ 12 = 2(l + 3)
⇒ 6 = 3 + b
⇒ b = 3 cm
Question 2.
A rectangle with sidelengths 5 cm and 3 cm is made using a wire. If the wire is straightened and then bent to form a square, what will be the length of the side of the square?
Solution:
Here perimeter of rectangle = 2(5 + 3)
= 2 × 8
= 16 cm
Now the wire is straightened and then bent to form a square.
∴ Perimeter of square = 16 cm
⇒ 4a = 16 cm
⇒ a = 4 cm, the required length of the side of the square.
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:
Perimeter of triangle = Sum of the lengths of its three sides = First side + Second side + Third side.
⇒ 55 cm = 20 cm + 14 cm + Third side
⇒ Third side = 55 cm – 20 cm – 14 cm = 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:
Given, length = 150 m and breadth = m
Perimeter of the park = 2 × (Length + Breadth)
= 2 × (150 +120)
= 540m
Given, cost of fencing the rectangular park = ₹ 40 per metre
∴ Cost of fencing the rectangular park = 540 × ₹ 40
= ₹ 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) Perimeter of square = length of given string
⇒ 4 × side = 36
⇒ side = 9
∴ Side of a square = 9 cm.
(b) Perimeter of triangle = length of given string
⇒ 3 × sides = 36 (As all sides are of equal length)
∴ Side of a triangle = 12 cm
(c) As the hexagon has sides of equal length it is a regular hexagon.
⇒ Perimeter of regular hexagon = length of the given string.
⇒ 6 × side = 36
∴ The side of a regular hexagon = 6 cm.
Question 6.
A farmer has a rectangular field with having length of 230 m and a breadth of 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 rectangular field = 2(l + b)
= 2(230 + 160)
= 780 m
Total length of rope = 3 × 780 = 2,340 m
6.1 Perimeter Figure it Out (Page No. 133 – 134)
Question 1.
Find out the total distance Akshi has covered in 5 rounds.
Solution:
Distance covered by Akshi in 5 rounds = 5 × perimeter of PQRS
= 5 × 220
= 1100 m
Question 2.
Find out the total distance Toshi has covered in 7 rounds. Who ran a longer distance?
Solution:
Distance covered by Toshi in 7 rounds = 7 × 180 = 1260 m
∴ 1260 m > 1100 m
Hence Toshi ran the 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), (b), (c), (d), (e), (f)
(c) Since Akshi ran 1000 m.
So, she has finished 4 full rounds running around her track. Her position as ‘C’ marked in above figure.
(f) Toshi ran 1000 m.
So, she has finished 5 full rounds running around her track. Her position as ‘Z’ marked in above figure.
6.2 Area Figure it Out (Page No. 138)
Question 1.
The area of a rectangular garden 25 m long is 300 m2. What is the width of the garden?
Solution:
Area of a rectangular garden = 300 m2
Length of garden = 25 m
∴ Breadth = \(\frac{Area}{Length}\)
= \(\frac{300}{25}\)
= 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:
Length of a plot = 500 m
Breadth of a plot = 200 m
Area = 500 × 200 = 100000 m2
Cost per m2 = \(\frac{Area}{Rate}\)
= \(\frac{100}{8}\)
= 12.5
Cost for 10000 m2 = ₹ 12,500
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:
Given, length of the coconut grove = 100 m
and breadth of the coconut grove = 50 m
∴ Area of the coconut grove = 100 m × 50 m = 5000 m²
∵ Each coconut tree requires 25 sq. m
∴ Number of tree = \(\frac{\text { Area of the coconut grove }}{\text { Area of each coconut tree }}\)
= \(\frac{5000}{25}\) = 200
Area of each coconut tree 5000 = 200
Hence, 200 trees can be planted in this grove.
Question 4.
By splitting the following figures into rectangles, find their areas (all measures are given in metres):
Solution:
Let the given figure be divided into rectangles A, B,C and D and their length and breadth be written on the figure.
For Rectangle A,
Length = 4 m and breadth = 2 m
Now, area of the Rectangle A = Length × Breadth
= 4 × 2 = 8m²
For Rectangle B,
Length = 3 m and breadth = 3 m
Area of the Rectangle B = Length × Breadth
= 3 × 3 = 9m²
For Rectangle C,
Length = 2 m and breadth = 1 m
Area of the Rectangle C = Length × Breadth
= 2 × 1 = 2 m²
For Rectangle D, length = 3 m and breadth = 3 m
∴ Area of the Rectangle D = Length × Breadth
= 3 × 3= 9m²
Now, total area of the given Figure
= Area of the Rectangle A + Area of the Rectangle B + Area of the Rectangle A – Area of the Rectangle D
= 8 + 9 + 2 + 9 = 28 m²
Hence, the required area is 28 m².
(b) Let the given figure is divided into rectangles A, B and C and their length and breadth are written on the figure.
For rectangle A,
length = 2 m and breadth = 1 m
Area of the rectangle A = Length × Breadth
= 2 × 1 = 2 m²
For rectangle B,
length = 5 m and breadth = 1 m
Area of the rectangle B = Length x Breadth
= 5 x 1 = 5m²
For rectangle C, length = 2 m and breadth = 1 m
∴ Area of the Rectangle C = Length × Breadth
= 2 × 1 = 2 m²
Now, total area of the given figure
= Area of the rectangle A + Area of the rectangle B + Area of the rectangle C
= 2 + 5 + 2 = 9m²
Hence, the area of the given figure is 9 m².
6.2 Area Figure it Out (Page No. 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:
There are two pieces (A and B) that 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 than shape C. Clearly from the figure, the area of shapes C and E 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:
Shape D is composed of shape C and E and they both have equal area. So shape D has twice the area of shape C. Similarly, shape F is composed of shape C and E. So shape D and F have same area.
Question 4.
Which shape has more area: Shape F or G? Give reasons for your answer.
Solution:
Shape F is composed of twice the area of shape C.
Now if we cut the shape F (equal triangle) into 2 small triangles and by placing the shape over shape G, we will find out that the shapes F and G 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:
When comparing the area of shape A to shape G, we get shape A has the same area of shape G. Both shapes are identical in size and cover the same amount of space. So, neither is bigger than the other, they are equal.
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:
The big square formed by all seven tangram pieces has an area that is 8 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:
From the given figure, we can see that,
Length of rectangle = two times side of shape B
and, Breadth of rectangle = \(\frac{1}{2}\) of side of shape A.
As Shape A = 4 C and Shape B = 4 C.
Area of rectangle = 16 C
The area remains the same as the same pieces of the figure have been rearranged to form a new figure.
Question 8.
Are the perimeters of the square and the rectangle forced from these 7 pieces different or the same? Explain your answer.
Solution:
From the given figures, In the case of a rectangle,
length = 8C and breadth = 2C
∴ Perimeter = 2(l + b)
= 2(8C + 2C)
= 2 × 10 C
= 20 C units
Now, in case of square,
Side of square = 4C
∴ Perimeter of square = 4 × 4C = 16 C units.
Hence, we can see that the perimeter of both the figures is different.
6.2 Area Figure it Out (Page No. 140)
Question 1.
Find the area of the following figures. (Page 140)
Solution:
(i) In the figure, full small square = 3
Exactly half square = 2
∴ Total area = 3 × 1 + 2 × \(\frac{1}{2}\)
= 3 + 1
= 4 sq. units
(ii) In the figure, full small square = 6
Exactly half square = 6
∴ Total area = 6 × 1 + 6 × \(\frac{1}{2}\)
. = 6 + 31= 9 sq. units
(iii) In the figure, full small square = 7
Exactly half square = 6
∴ Total area = 7 × 1 + 6 × \(\frac{1}{2}\)
= 7 + 3 = 10 sq. units
(iv) In the figure, full small square = 8
Exactly half square = 6
Total area = 8 × 1 + 6 × \(\frac{1}{2}\)
= 8 + 3 = 11 sq. units
6.3 Area of A Triangle Figure it Out (Page 144)
Question 1.
Find the areas of the figures below by dividing them into rectangles and triangles.
Solution:
(a) Area of rectangle BCEF = (4 × 5) = 20 sq units
Area of triangle = ABF = \(\frac{1}{2}\) (4 × 1) = 2 sq units
Area of triangle CDE = \(\frac{1}{2}\) (4 × 1) = 2 sq units
Total required area = 20 + 2 + 2 = 24 sq units
(b) Area of rectangle BCFG = (4 × 5) = 20 sq units
Area of triangle ABG = \(\frac{1}{2}\) × (Area of rectangle ABGH)
= \(\frac{1}{2}\) × (4 × 3) = 6 sq units
Area of triangle FCD = \(\frac{1}{2}\) × Area of rectangle FEDC
= \(\frac{1}{2}\) × (4 × 2) = 4 sq units
Total area = (20 + 6 + 4) = 30 sq units
(c) Area of triangle BIJ = \(\frac{1}{2}\) × Area of rectangle ABJI
= \(\frac{1}{2}\) × (3 × 2) = 3 sq units
Area of triangle BJD = \(\frac{1}{2}\) × (Area of rectangle BCDJ)
= \(\frac{1}{2}\) × (3 × 2) = 3 sq units
Area of rectangle IJKH = (3 × 8) = 24 sq units
Area of triangle HKF = \(\frac{1}{2}\) × (3 × 2) = 3 sq units
Area of triangle DJF = \(\frac{1}{2}\) × (Area of rectangle JDEF)
= \(\frac{1}{2}\) × (3 × 10) = 15 sq units
Total area = (3 + 3 + 24 + 3 + 15) = 48 sq units
(d) Area of rectangle GDEF = (4 × 3) = 12sq units
Area of triangle AGH = \(\frac{1}{2}\) × (2 × 1) = ) Sq units
Area of triangle CHD = \(\frac{1}{2}\) × (Area of rectangle 2 BCDH)
= \(\frac{1}{2}\) × (3 × 2) = 3 sq units
Total area = (12 + 1 + 3) = 16 sq units
(e) Area ABHD = \(\frac{1}{2}\) x (Area of rectangle ACDH)
= \(\frac{1}{2}\) x (4 x 2)
= 4 sq units Area of AHIF = \(\frac{1}{2}\) x (Area of rectangle HIFG)
= \(\frac{1}{2}\) x (3 x 4) = 6 sq units
Area of ADIF = \(\frac{1}{2}\) x (Area of rectangle DEFI)
= \(\frac{1}{2}\) x (1 x 4) = 2 sq units
Total area = (4 + 6 + 2) = 12 sq units
6.3 Area of A Triangle Figure it Out (Page 145)
Question 1.
Making it ‘More’ or ‘Less’.
Observe these two figures. Is there any similarity or difference between the two?
Using 9 unit squares (having an area of 9 sq units), we have made figures with two different perimeters—the first figure has a perimeter of 12 units and the second has a perimeter of 20 units. Arrange or draw different figures with 9 sq units to get other perimeters. Each square should align with atleast one other square on atleast one side completely and together all squares should form a single connected figure with no holes.
Using 9 unit squares, solve the following.
(i) What is the smallest perimeter possible?
(ii) What is the largest perimeter possible?
(iii) Make a figure with a perimeter of 18 units.
(iv) Can you make other shaped figures for each of the above three perimeters, or is there only one shape with that perimeter? What is your reasoning?
Solution:
(i) The smallest perimeter possible using 9 unit squares is 12 units when arranged in a 3 × 3 square.
(ii) The largest perimeter possible using 9 unit squares is 20 units when arranged in a straight line of 1 by 9 squares.
(iii) A possible figure with a perimeter of 18 units can be arranged in a 2 × 4 rectangle with one additional square attached on any side.
(iv) Yes, different shaped figure can have the same perimeter. The perimeter depends on the arrangement of the squares and not just the number of squares.
6.3 Area of A Triangle Figure it Out (Page 146)
Question 1.
Below is the house plan of Charan. It is in a rectangular plot. Look at the plan. What do you notice?
Some of the measurements are given.
(i) Find the missing measurements.
(ii) Find out the areas of his house.
Solution:
(i) For utility
width = 15 – 12 = 3ft and length = 15 ft
Area = 15 × 3 = 45 sq. ft
For small bedroom 15ft × _ ft = 180
⇒ Missing term = \(\frac{180}{15}\) = 12 ft
For garden
Length = 15 + 5 = 20 ft
and width = 3 ft
∴ Area = 20 × 3 = 60 sq. ft
For parking
Length = 15 ft
and width = 3 ft
Area = 15 × 3 = 45 sq. ft
For hall
Area = (20 × 12) + (5 × 5)
= 240 sq. ft 4- 25 sq. ft
= 265 sq. ft
(ii) The area of house = Length × Width
= 35 × 30
= 1050 sq. ft
6.3 Area of A Triangle Figure it Out (Page 147)
Question 1.
Now, find out the missing dimensions and area of Sharan’s home. Below is the plan.
Some of the measurements are given.
(i) Find the missing measurements.
(ii) Find out the area of his house.
What are the dimensions of all the different rooms in Sharan’s house? Compare the areas and perimeters of Sharan’s house and Charan’s house.
Solution:
(i) For utility
Area = 70 sq. ft
∴ Dimensions = 10 × 7
For small bedroom
Area = 12 × 10 = 120 sq. ft
For hall
Dimensions = 23 × 15
∴ Area = 345 sq. ft For toilet
Dimensions = 5 × 10
∴ Area = 50 sq. ft
For entrance Dimensions = 7 × 15
∴ Area = 105 sq. ft
(ii) Length of house = 42 ft and width = 25 ft
∴ Area = Length × Width
= 42 × 25
= 1050 sq. ft
Hence, the area of Sharan’s and Charan’s house is same but perimeter of Sharan’s house is more than the Charan’s house.
6.3 Area of A Triangle 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 1st rectangle = 5 × 10 = 50 m2
Area of Ilnd rectangle = 2 × 7 = 14 m2
Area of new rectangle = 50 + 14 = 64 m2
Question 2.
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Solution:
Area of a rectangular garden = 50 × width
⇒ 1000 = 50 × width
⇒ Width = \(\frac{1000}{50}\) = 20 cm
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:
Given, length of the floor = 5 m
and width of the floor = 4 m .
∴ Area of the floor = Length × Width
= 5 × 4 = 20 sq. m
Also, given that length of the square carpet = 3 m
The area of the square carpet = 3 × 3 = 9 sq m
Hence, the area of the floor laid with carpet is 9 sq m.
Therefore, the area of the floor that is not carpeted = Area of the floor – Area of the floor laid with carpeted.
= 20 – 9
= 11 sq.m
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:
Here, Length of garden = 15 m
Width of garden = 12 m
So, the area of the garden = 15 × 12 sq m = 180 sq m
Now, the length of the flower bed = 2 m
Width of flower bed = 1 m
Area of the flower bed = 2 × 1 sq m = 2 sq m
Since, the area of four flower beds = 2 × 4 sq m = 8 sq m
Now the area is 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:
Shape A has an area of 18 sq. units.
∴ Possible sides are 18 × 1, 2 × 9, 6 × 3
Also, shape B has an area of 20 sq. units.
∴ Possible sides are 20 × 1, 4 × 5, 10 × 2
Given shape A has a longer perimeter than shape B, hence two such shapes satisfying the given conditions are:
Here Perimeter of shape A = 9 + 2 + 9 + 2 = 22 units
Here Perimeter of shape B = 5 + 4 + 5 + 4 = 18 units.
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:
Dimensions of borders: 1 cm from top and bottom; 1.5 cm from left and right sides.
So, the border dimensions is (L – 2), (W – 3)
Perimeter of the border = 2 [(L – 2) + (W – 3)]
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:
Given dimensions of rectangle are 12 units × 8 units
So, area of rectangle = 96 sq units
Area of inner rectangle (half the area of outer rectangle)
= \(\frac{96}{2}\) = 48 sq units
Possible dimensions of inner rectangle is = 8 units × 6 units
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:
(c) is true i.e., the perimeters of both the rectangles added together is always 1 \(\frac{1}{2}\) times the perimeter of the square.