Class 6 - Mathematics : Mensuration
Exercise : 10.1
Top Block 1
NCERT Solutions Class 6 Mathematics Mensuration Ex 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 x 50
= 100 cm = 1 m
Mddle block 1
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 x (length + breadth)
= 2 x (2.25 + 1.50)
= 2 x 3.75 = 7.50 m
Thus, perimeter of table top 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 x 53 cm = 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 x (length + breadth) = 2 x (0.7 + 0.5) = 2 x 1.2 = 2.4 km
= 2.4 x 1000 m = 2400 m
Thus, the length of wire = 4 x 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 ΔABCΔABC = AB + BC + CA
= 3 cm + 5 cm = 4 cm
= 12 cm
= 3 x 9 cm
= 27 cm
= 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, perimeter of triangle is 39 cm.
Question: 8.Find the perimeter of a regular hexagon with each side measuring 8 cm.
Answer :
Perimeter of Hexagon = 6 x length of one side
= 6 x 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 x side
⇒ 20 = 4 x side
⇒ side = 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 x side = 100 cm
⇒ side = 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 x side = 30 cm
⇒ side = 30⁄4 = 7.5 cm
Thus, the length of each side of square is 7.5 cm.
(b)Perimeter of equilateral triangle = 30 cm
⇒ 3 x side = 30 cm
⇒ side = 30⁄3 = 10 cm
Thus, the length of each side of equilateral triangle is 10 cm.
(c)Perimeter of hexagon = 30 cm
⇒ 6 x side = 30 cm
⇒ side = 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 the third side?
Answer :
Let the length of third side be xx 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=36–26x
⇒ x=10
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 meter.
Answer :
Side of square = 250 m
Perimeter of square = 4 x side
= 4 x 250 = 1000 m
Since, cost of fencing of per meter = Rs. 20
Therefore, cost of fencing of 1000 meters = 20 x 1000 = = Rs. 20,000
Question: 14.Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs. 12 per meter.
Answer :
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park = 2 x (length + breadth)
= 2 x (175 + 125)
= 2 x 300 = 600 m
Since, cost of fencing park per meter = = Rs. 12
Therefore, cost of fencing park of 600 m = 12 x 600 = Rs. 7,200
Question: 15Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length of 60 m and breadth 45 m. Who covers less distance?
Answer :
Distance covered by Sweety = Perimeter of square park
Perimeter of square = 4 x side
= 4 x 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 x (length + breadth)
= 2 x (60 + 45)
= 2 x 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 answer?
Answer :
(a) Perimeter of square = 4 x side
= 4 x 25 = 100 cm
(b) Perimeter of rectangle = 2 x (length + breadth)
= 2 x (40 + 10)
= 2 x 50 = 100 cm
(c) Perimeter of rectangle = 2 x (length + breadth)
= 2 x (30 + 20)
= 2 x 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 1212 m. He lays them in the form of a square
(b)Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement?
(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) 6 m
(b) 10 m
(c) Second arrangement has greater perimeter.
(d) Yes, if all the squares are arranged in row, the perimeter be 10 cm.