NCERT Exemplar Solutions for Class 6 Maths Chapter 6 Mensuration are available here. These solutions are prepared by the subject experts at BYJU’S to guide students to tackle questions in many methods in the exam and manage their time accordingly. Therefore, we suggest that students refer to NCERT Exemplar Solutions for Class 6 Maths to obtain good marks in the annual exam.
Chapter 6, Mensuration, deals with measurement, especially the derivation and use of algebraic formulas to measure the areas, volumes and different parameters of geometric figures. In this Chapter, students will also learn about the perimeters of a triangle, square, rectangle and regular polygon. The perimeter of a closed figure is the distance covered in one round along the boundary of the figure.
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Exercise Page: 93
In questions 1 to 6, out of the four options, only one is correct. Write the correct answer.
1. Following figures are formed by joining six unit squares. Which figure has the smallest perimeter in Fig. 6.4?
Solution:-
Figure (i) has the smallest perimeter.
The perimeter of a closed figure is the distance covered in one round along the boundary of the figure.
We know that the perimeter of the figure = numbers of sides × length of each side
Consider the given figures,
(i) figure (i) has 10 units, and each side has length 1 cm
So, perimeter = 10 × 1
= 10 cm
(ii) figure (ii) has 12 units, and each side has length 1 cm
So, perimeter = 12 × 1
= 12 cm
(iii) figure (iii) has 14 sides, and each side has length 1 cm
So, perimeter = 14 × 1
= 14 cm
(iv) figure (iv) has 14 sides, and each side has length 1 cm
So, perimeter = 14 × 1
= 14 cm
By comparing all the perimeters of the given figures, figure (i) has the smallest perimeter.
2. A square-shaped park ABCD of side 100m has two equal rectangular flower beds, each of size 10m × 5m (Fig. 6.5). Length of the boundary of the remaining park is
(A) 360m (B) 400m (C) 340m (D) 460m
Solution:-
(B) 400m
We know that, Length of the boundary = Perimeter of the boundary
Given, two rectangular flower beds, each of size 10m × 5m
So, perimeter of remaining park = 90 + 5 + 10 + 95 + 90 + 5 + 10 + 95
= 400 m
3. The side of a square is 10cm. How many times will the new perimeter become if the side of the square is doubled?
(A) 2 times (B) 4 times (C) 6 times (D) 8 times
Solution:-
(A) 2 times
As we know, the perimeter of the square = side × 4
= 10 × 4
= 40 cm
Given, the side of the square = 10 cm
The side of the square is doubled = 10 + 10
= 20 cm
Therefore, the new perimeter becomes 2 times the side of the square is doubled.
4. Length and breadth of a rectangular sheet of paper are 20cm and 10cm, respectively. A rectangular piece is cut from the sheet, as shown in Fig. 6.6. Which of the following statements is correct for the remaining sheet?
(A) Perimeter remains the same but the area changes.
(B) Area remains the same, but the perimeter changes.
(C) Both area and perimeter are changing.
(D) Both area and perimeter remain the same.
Solution:-
(A) Perimeter remains the same but the area changes.
We know that the area of a big rectangle = length × breadth
= 10 × 20
= 200 cm2
Area of small rectangle = 5 × 2
= 10 cm2
Perimeter of rectangle = 2(length + breadth)
= 2(20 + 10)
= 2 × 30
= 60 cm
Then, perimeter of new figure = 20 + 8 + 5 + 2 + 15 + 10
= 60
Area of new figure = Area of the big rectangle – Area of the new figure
= 200 – 10
= 190 cm2
By comparing all the results, Perimeter remains the same but the area changes.
5. Two regular Hexagons of perimeter 30cm each are joined as shown in Fig. 6.7. The perimeter of the new figure is
(A) 65cm (B) 60cm (C) 55cm (D) 50cm
Solution:-
(D) 50cm
From the question, it is given that the perimeter of each hexagon = 30 cm
We know that, the hexagon has 6 sides.
So, the length of each side of the hexagon = 30/6
= 5 cm
Now, consider two hexagons are joined together.
Then, the perimeter of the new figure = number of sides × length of each side
= 10 × 5
= 50 cm
6. In Fig. 6.8, which of the following is a regular polygon? All have equal sides except (i)
(A) (i) (B) (ii) (C) (iii) (D) (iv)
Solution:-
(B) (ii)
A closed figure in which all sides and angles are equal is called a regular polygon.
7. Match the shapes (each side measures 2cm) in column I with the corresponding perimeters in column II:
Solution:-
Given, each side measures 2cm.
Figure (A) has 14 sides.
Perimeter of figure = number of sides × length of each side
= 14 × 2
= 28 cm
Figure (B) has 8 sides.
Perimeter of figure = number of sides × length of each side
= 8 × 2
= 16 cm
Figure (C) has 10 sides.
Perimeter of figure = number of sides × length of each side
= 10 × 2
= 20 cm
Figure (D) has 12 sides.
Perimeter of figure = number of sides × length of each side
= 12 × 2
= 24 cm
8. Match the following
Solution:-
(A) Perimeter of a rectangle = 2 × (length + breadth)
= 2 × (6 + 4)
= 2 × 10
= 20
(B) Perimeter of a square = 4 × length of its side
= 4 × 5
= 20
(C) Perimeter of an equilateral triangle = 3 × length of a side
= 3 × 6
= 18
(D) Perimeter of isosceles triangle = sum of the length of all sides
= 4 + 4 + 2
= 10
In questions 9 to 13, fill in the blanks to make the statements true.
9. Perimeter of the shaded portion in Fig. 6.9 is
AB + _ + _ + _ + _ + _ + _ + HA
Solution:-
We know that the perimeter of a figure = sum of the length of all sides
= AB + BM + MD + DE + EN + NG + GH + HA
10. The amount of region enclosed by a plane closed figure is called its _________.
Solution:-
The amount of region enclosed by a plane closed figure is called its area.
11. Area of a rectangle with length 5 cm and breadth 3cm is _________.
Solution:-
The area of a rectangle with length 5 cm and breadth 3cm is 15 cm.
Area of rectangle = length × breadth
= 5 cm × 3 cm
= 15 cm
12. A rectangle and a square have the same perimeter (Fig. 6.10).
(a) The area of the rectangle is _________.
Solution:-
The area of the rectangle is 12 sq. units.
From the figure, the length of the rectangle is 6, and the breadth is 2.
Then, area of rectangle = length × breadth
= 6 × 2
= 12 sq. units
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (6 + 2)
= 2 × 8
= 16 units
(b) The area of the square is _________.
Solution:-
The area of the square is 16 sq. Units.
From the question, it is given that a rectangle and a square have the same perimeter.
So, the length of each side of a square = 16/4
= 4 units.
Area of square = side × side
= 4 × 4
= 16 sq. Units
13. (a) 1m = _________ cm.
Solution:-
1m = 100 cm.
(b) 1sqcm = _________ cm × 1cm.
Solution:-
1sqcm = 1 cm × 1cm.
(c) 1sqm = 1m × _________ m = 100cm × _________ cm.
Solution:-
1sqm = 1m × 1 m
= 100cm × 100 cm.
(d) 1sqm = _________ sqcm.
Solution:-
1sqm = 10000 sqcm.
We know that 1 m = 100 cm
Then, 1 sqm = 100 × 100
= 10000sqcm
In questions 14 to 20, state which of the statements are true and which are false.
14. If the length of a rectangle is halved and the breadth is doubled, then the area of the rectangle obtained remains the same.
Solution:-
True.
We know that the area of the rectangle = length × breadth
As per the condition given in the question, the length of a rectangle is halved = l/2
Breadth is doubled = 2b
Then, new area = l/2 × 2b
= l × b.
Therefore, if the length of a rectangle is halved and the breadth is doubled, then the area of the rectangle obtained remains the same.
15. Area of a square is doubled if the side of the square is doubled.
Solution:-
False
We know that the area of a square = side × side
As per the condition given in the question, the side of the square is doubled = 2 × side
Then, new area = 2side × 2side
= 4 side2
Therefore, if the side of the square is doubled, then the area of the square obtained is 4 times the old area.
16. Perimeter of a regular octagon of side 6cm is 36cm.
Solution:-
False.
Perimeter of regular octagon = number of sides × length of reach sides
= 8 × 6
= 48 cm
17. A farmer who wants to fence his field must find the perimeter of the field.
Solution:-
True.
18. An engineer who plans to build a compound wall on all sides of a house must find the area of the compound.
Solution:-
False.
An engineer who plans to build a compound wall on all sides of a house must find the perimeter of the compound.
19. To find the cost of painting a wall, we need to find the perimeter of the wall.
Solution:-
False.
To find the cost of painting a wall, we need to find the area of the wall.
20. To find the cost of a frame of a picture, we need to find the perimeter of the picture.
Solution:-
True.
21. Four regular hexagons are drawn so as to form the design as shown in Fig. 6.11. If the perimeter of the design is 28cm, find the length of each side of the hexagon.
Solution:-
From the question, it is given that the perimeter of the design is 28cm.
Given design has 14 sides, so the length of each side = 28/14
= 2 cm
In the given figure, 4 regular hexagons are joined.
Therefore, the length of each side of the hexagon = 2 cm
22. Perimeter of an isosceles triangle is 50cm. If one of the two equal sides is 18cm, find the third side.
Solution:-
We know that an isosceles triangle contains two equal sides.
From the question, it is given that one of the two equal sides is 18cm.
And Perimeter of an isosceles triangle is 50cm
We know that the perimeter of the isosceles triangle = sum of all sides
Let us assume the third side is x.
Then,
50 = 18 + 18 + x
50 = 36 + x
X = 50 – 36
X = 14 cm
Therefore, the length of the third side of an isosceles triangle is 14 cm.
23. Length of a rectangle is three times its breadth. The perimeter of the rectangle is 40cm. Find its length and width.
Solution:-
From the question, it is given that,
Perimeter of rectangle = 40 cm
The length of a rectangle is three times its breadth = 3b
We know that, perimeter of rectangle = 2 × (length + breadth)
40 = 2 × (3b + b)
40 = 2 × 4b
40 = 8b
40/8 = b
b = 5 cm
Therefore, the width of the rectangle = 5 cm
Length of rectangle = 3b
= 3 × 5
= 15 cm
24. There is a rectangular lawn 10m long and 4m wide in front of Meena’s house (Fig. 6.12). It is fenced along the two smaller sides and one long side leaving a gap of 1m for the entrance. Find the length of the fencing.
Solution:-
From the question, it is given that,
Length of rectangular lawn = 10m
Width of rectangular lawn = 4m
Then,
Perimeter of fencing = length of fencing
Perimeter of fencing = two smaller sides + one longer side – gap in the longer side
= 4 + 4 + 10 – 1
= 18 – 1
= 17 m
Therefore, the length of the fencing is 17 m.
Solution:-
From the question, it is given that a region measured by taking a rectangle is one unit.
Given figure contain 13 rectangles.
Therefore, the area of the region = 13 × 1
= 13 sq. Units
26. Tahir measured the distance around a square field as 200 rods (lathi). Later he found that the length of this rod was 140cm. Find the side of this field in metres.
Solution:-
From the question, it is given that,
Tahir measured the distance around a square field as 200 rods
Length of the rod = 140 cm
So, the total distance of the square field = 200 × 140
= 28000 cm
Then, the length of one side of the square field = 28000/4
= 7000 cm
We know that 1 m = 100 cm
Therefore, side of field = 7000/100
= 70 m
27. The length of a rectangular field is twice its breadth. Jamal jogged around it four times and covered a distance of 6km. What is the length of the field?
Solution:-
From the question, it is given that,
The length of a rectangular field is twice its breadth = 2b
Perimeter of rectangular field = 6km/4
= 1.5 km
We know that, perimeter of rectangle = 2 × (length + breadth)
1.5 = 2 × (2b + b)
1.5 = 2 × 3b
1.5 = 6b
b = 1.5/6
b = 0.25 km
length of rectangle field = 2 × 0.25 = 0.5km
= 0.5 × 1000
= 500 m
28. Three squares are joined together, as shown in Fig. 6.14. Their sides are 4cm, 10cm and 3cm. Find the perimeter of the figure.
Solution:-
We know that the perimeter of the figure = sum of all sides
= AB + BC + CD + DE + EF + FG + GH + HI + JI +JA
= 4 + 4 + (10 – 4) + 10 + (10 – 3) + 3 + 3 + 3 + 10 + 4
= 4 + 4 + 6 + 10 + 7 + 3 + 3 + 3 + 10 + 4
= 54 cm
29. In Fig. 6.15 all triangles are equilateral and AB = 8 units. Other triangles have been formed by taking the midpoints of the sides. What is the perimeter of the figure?
Solution:-
45 units
Perimeter of given figure = sum of the length of each side
= BN + NM + ML + LK + KJ + JI + IA + AH + HV + VG + GF + FE + ED + DC + UC + UT + TS + SR + RQ + PQ + BP
= 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4 + 4 + 2 + 1 + 1 + 1 + 2 + 4
= 45 units
30. Length of a rectangular field is 250m, and the width is 150m. Anuradha runs around this field 3 times. How far did she run? How many times should she run around the field to cover a distance of 4km?
Solution:-
From the question, it is given that,
Length of a rectangular field is 250m
Width of rectangular field = 150 m
Perimeter of rectangle = 2 × (length + breadth)
= 2 × (250 + 150)
= 2 × 400
= 800 m
Given that, Anuradha runs around this field 3 times = 3 × 800
= 2400 m
We know that 1km = 1000 m
So, 2400/1000
= 2.4 km i.e. 2 km 400 m
Number of times Anuradha should run around the field to cover a distance of 4km = 4000/800
= 5 times
Therefore, 5 times Anuradha should run around the field to cover a distance of 4km.
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