Chapter 10: Mensuration
EXERCISE 10.1
1. Find the perimeter of each of the following figures:
Solution:
(a) Perimeter = sum of all sides
= AB + BC + CD + AD
= (4 + 2 + 1 + 5) cm
= 12 cm
(b) perimeter = sum of all sides
= AB + BC + CD + AD
= (23 + 35 + 40 + 35) cm
= 133 cm
(c) Perimeter = sum of all sides
= AB + BC + CD + AD
= (15 + 15 + 15 + 15) cm
= 60 cm
(d) Perimeter = sum of all sides
= AB + BC + CD + DE + EA
= (4 + 4 + 4 + 4 + 4) cm
= 20 cm
(e) Perimeter = sum of all sides
= AB + BC + CD + DE + EF + FG + GA
= (2.5 + 2.5 + 0.5 + 4 + 1 + 4 + 0.5) cm
= 15 cm
(f) Perimeter = sum of all sides
= AB + BC + CD + DE + EF + FG + GH + HI + IJ + JK + KL + LM + MN + ON + OP + PQ + QR + RS + ST + TA
= (4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3) cm
= (13 + 13 + 13 + 13) cm
= 52 cm
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?
Solution:
The length of the required tape is equal to the perimeter of the rectangle.
We know that,
Perimeter of rectangle = 2(length + breadth)
= 2(40 + 10) cm
= 2(50) cm
= 100 cm
3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the table-top?
Solution:
Given,
Length of the table top (l) = 2 m 25 cm = 2.25m
Breadth of the table top (b) = 1 m 50 cm = 1.50 m
Perimeter of the table top = 2(l + b)
= 2(2.25 + 1.50)
= 2(3.75)
= 7.50 m
4. What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution:
The length of the photo frame is 32
The breadth of the photo frame is 21
The perimeter of the rectangular photograph = 2(l + b)
= 2(32 + 21)
= 2(53)
= 106 cm
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?
Solution:
Since the rectangular piece of land measuring 0.7 km by 0.5 km has to be fenced with 4 rows of wires on each side, the length of the wire required will be equal to four times the perimeter of the rectangular piece of land.
So, Length of the wire required = 4 x Perimeter of the rectangle = 4 x 2 (Length + Breadth)
Length = 0.7 km, Breadth = 0.5 km
Substituting the values we get,
⇒ 4 × 2 (0.7 + 0.5)
⇒ 4 × 2 × 1.2 = 9.6 km
Therefore, the length of the wire required to fence the park is 9.6 km.
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
Solution:
(a) Perimeter = 3 + 4 + 5 = 12cm
(b) Perimeter = 9 + 9 + 9 = 27cm
(c) Perimeter = 8 + 8 + 6 = 22cm
7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution:
The perimeter of triangle = Sum of all three sides.
= 10 cm + 14 cm + 15 cm.
= 39 cm.
Thus, the perimeter of triangle is 39 cm.
8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:
Perimeter of Hexagon = 6 × \times × length of one side.
= 6 × \times × 8 m.
= 48 m.
Thus, the perimeter of the hexagon is 48 m.
9. Find the side of the square whose perimeter is 20 m.
Solution:
Perimeter of square = 4 x side
20 = 4×Side
Side = 20/4
Side = 5m
Each side of square is 5m.
10. The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution:
The perimeter of a closed figure is the length of its boundary.
Given: Perimeter of regular pentagon = 100 cm.
⇒ 5 × Length of each side = 100 cm.
⇒ Length of each side = 1005 = 20 cm.
Thus, the length of each side of regular pentagon is 20 cm.
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?
Solution:
(a) The string is bent in the form of a square.
So, the perimeter of the square = length of the string.
Perimeter of square = 30 cm.
⇒ 4 ×side = 30 cm.
⇒ side = 304= 7.5 cm.
Thus, the length of each side of square is 7.5 cm.
(b) The string is bent in the form of an equilateral triangle.
So, the perimeter of an equilateral triangle = length of the string.
⇒ 3× Length of side = 30 cm.
⇒ Length of side = 303 = 10 cm.
Thus, the length of each side of equilateral triangle is 10 cm.
(c) Perimeter of the regular hexagon = 6 x Length of a side
⇒ Length of a side
=6Perimeter of the regular hexagon
=630cm=5c
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?
Solution:
Let the length of the third side be x cm.
Length of other two sides is 12 cm and 14 cm.
Now, Perimeter of triangle = Sum of all sides = 36 m.
⇒12+14+x=36
⇒26+x=36
⇒x=36−26
⇒x=10 cm
Thus, the length of third side is 10 cm.
13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per meter.
Solution:
The length of each side of square = 250 m.
Perimeter of square = 4× length of side.
= 4×250.
= 1000 m.
So, required fencing length is 1000 m.
The cost of fencing of per meter = ₹ 20.
The cost of fencing of 1000 meters = 20×1000 = ₹ 20,000.
Therefore, the cost of fencing of the square park of side 250 m is ₹ 20,000.
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.
Solution:
rectangular park = 175 m.
Breadth of the rectangular park = 125 m.
Perimeter of park = 2 × (length + breadth).
= 2 × (175 + 125).
= 2 × 300 = 600 m.
So, the required fencing for a rectangular park is 600 m.
Since, the cost of fencing park per meter = ₹ 12.
Therefore, the cost of fencing the rectangular park of perimeter 600 m = 12×600 = ₹ 7,200.
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?
Solution:
Side of square park = 75 m
The perimeter of the square park = 4 × side
⇒ 4 × 75 = 300
Hence, Sweety covers 300 meters around the square park.
Now, the length of the rectangular park = 60 m
The breadth of the rectangular park = 45 m
The perimeter of rectangular park = 2 (Length + Breadth)
⇒ 2 (60 + 45)
⇒ 2 × 105 = 210
Hence, Bulbul covers 210 meters around the rectangular park.
Thus, we see that Bulbul covers a distance of 210 m which is lesser than the distance of 300 m covered by Sweety. Therefore, Bulbul covers a lesser distance.
16. What is the perimeter of each of the following figures? What do you infer from the answers?
Solution:
- Perimeter = sum of all side
= AB + BC + CD + DA
= (25 + 25 + 25 + 25) cm
= 100 cm
- Perimeter = sum of all side
= AB + BC + CD + DA
= (30 + 20 +30 + 20) cm
= 100 cm
- Perimeter = sum of all side
= AB + BC + CD + DA
= (40 + 10 + 40 + 10) cm
= 100 cm
- Perimeter = sum of all side
= AB + BC + CA
= (30 + 30 + 40) cm
= 100 cm
All the figures have same perimeter.
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 10.7(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 10.7 (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.)
Solution:
(a) Perimeter of the square = 4× side.
Length of the side of the square =3 square slabs × length of each side
=3×0.5m=1.5m [∵each side=12 m]
Perimeter of Avneet’s arrangement =4×1.5 m=6 m
(b) Consider the figure
Perimeter of the figure =5× length of each side =5×0.5 m=2.5 m [∵each side=12 m]
Perimeter of the cross =4× figure =4×2.5 m=10 m
(c) Shari’s arrangement (cross) has greater perimeter.
(d) No, but we can achieve the same perimeter as Shari’s arrangement by arranging all the squares in a row.
EXERCISE 10.2
- Find the areas of the following figures by counting square:
Solution:
We have 9 full squares
Area = 9(1)
= 9 square unit.
We have 5 full square
Area = 5(1)
= 5 square unit
We have 2 full squares and 4 half squares
Area = 2(1) + 4(1/2)
= 2 + 2
= 4 squares units
We have 8 full squares
Area = 8(1)
= 8 square units
We have 10 full squares
Area = 10(1)
= 10 square units
We have 2 full squares and 4 half square
Area = 2(1) + 4(1/2)
= 2+ 2
= 4 square units
We have 4 full square and 4 half square
Area = 4(1) + 4(1/2)
= 4 + 2
= 6 square units
We have 5 full square
Area = 5(1)
= 5 square units
We have 9 full square
Area = 9(1)
= 9 square units
We have 2 full squares
And 4 half squares
Area = 2(1) + 4 (1/2)
= 2+ 2
=4 square units
We have 4 full squares and 2 half squares
Area = 4(1) + 2(1/2)
= 4 + 1
= 5 square units
We have 2 full squares, 4 half square, 4 almost full square and 2 less than half squares
Area = 2(1) + 4(1/2) +4(1) + 2(0)
= 2 + 2 + 4 + 0
= 8 squares units
We have 7 full squares, 7 almost full squares and 8 less than half squares
Area = 7(1) + 0(1/2) + 7(1) + 8(0)
= 7 + 0 + 7 + 0
= 14 square units.
We have 9 full squares, 9 almost full squares and 7 less than half area
Area = 9(1) + 0(1/2) + 9(1) + 7(0)
= 9 + 0 + 9 + 0
= 18 square units
EXERCISE 10.3
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
Solution:
(a) 3 cm and 4 cm
Length of the rectangle = 3 cm
Breadth of the rectangle = 4 cm
Area of the rectangle = l * b
= 3 * 4 cm
= 12 cm
(b) 12 m and 21 m
Length of the rectangle = 12 m
Breadth of the rectangle = 21 m
Area of rectangle = l * b
= 12 * 21 cm
= 252 m
(c) 2 km and 3 km
Length of the rectangle = 2 km
Breadth of the rectangle = 3 km
Area of the rectangle = l * b
= 2 * 3
= 6 km
(d) 2 m and 70 cm
Length of rectangle = 2 m
Breadth of the rectangle = 70 cm = 70/100 = 0.7
Area of the rectangle = l * b
= 2 * 0.7
= 1.4 m
2. Find the areas of the squares whose sides are:
(a) 10 cm (b) 14 cm (c) 5 m
Solution:
(a) 10 cm
Side of the square = 10 cm
Area of the square = side * side
= 10 * 10
= 100 cm
(b) 14 cm
Side of the square = 14 cm
Area of the square = side * side
= 14 * 14
= 196
(c) 5 m
Side of the square = 5 m
Area of the square = side * side
= 5 * 5
= 25 m
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?
Solution:
(a) 9 m and 6 m
Length of the rectangle = 9 m
Breadth of the rectangle = 6 m
Area of the rectangle = l*b
= 9 * 6
= 54 m
(b) 17 m and 3 m
Length of the rectangle = 17 m
Breadth of the rectangle = 3 m
Area of the rectangle = l * b
= 17 * 3 m
= 51 m
(c) 4 m and 14 m
Length of the rectangle = 4 m
Breadth of the rectangle = 14 m
Area of the rectangle = l * b
= 4 * 14 m
= 56 m
largest area is (c) and (b) is smallest area
4. The area of a rectangular garden 50 m long is 300 sq. m. Find the width of the garden.
Solution:
Length of rectangle = 50 m
Breadth of rectangle = X
Area of rectangle = l * b
300 sq. m = 50 * X
X = 300/50 m
X = 6 m
Therefore, width of garden is 6 m
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.?
Solution:
length is given as 500 m and breadth is given as 200 m hence, the area can be calculated using the given formula.
Area of rectangular plot = length × breadth
Area of rectangular plot = 500 × 200
= 1,00,000 m2
Now, the cost of tilling the rectangular plot per hundred sq. m. = ₹8
So, the cost of tiling 1,00,000 sq. m. of land = (8/100) × 1,00,000 = ₹ 8000
Therefore, the cost of tiling 1,00,000 sq. m. of land is ₹ 8000
6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:
Given: Length of table =2 m.
1 m=100 cm
1 cm=1100 m
1 cm=0.01 m
Breadth of table =1 m 50 cm=1 m+50×0.01 m=1.50 m
Area of table =length × breadth.
=2 m×1.50 m=3 m2.
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?
Solution:
Length of the room = 4 m
Width of the room is 3 m 50 cm = 3.50 m
To carpet the room, we need to find the area of the floor.
So, Area of the room = length × breadth = 4 × 3.50 = 14 sq. m
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.
Solution:
Length of floor = 5 m and breadth of floor = 4 m
Area of floor = length x breadth
= 5 m x 4 m = 20 m. sq.
Now, Side of square carpet = 3 m
Area of square carpet = side x side = 3 x 3 = 9 m. sq.
Area of floor that is not carpeted = 20 m – 9 m = 11 m. sq.
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?
Solution:
Side of square bed = 1 m
Area of square bed = side x side = 1 m x 1 m = 1 m. sq.
\therefore∴ Area of 5 square beds = 1 x 5 = 5 m. sq.
Now, Length of land = 5 m and breadth of land = 4 m
\therefore∴ Area of land = length x breadth = 5 m x 4 m = 20 m. sq.
Area of remaining part = Area of land – Area of 5 flower beds
= 20 m – 5 m = 15 m. sq.
10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
Solution:
Area of yellow region = 3 × 3
= 9 cm2
Area of orange region = 1× 2
= 2 cm2
Area of grey region = 3 × 3
= 9 cm2
Area of brown region = 2 × 4
= 8 cm2
Total area = 9 + 2 + 9 + 8
= 28 cm2
∴ Total area is 28 cm2
Area of brown region = 3 × 1
= 3 cm2
Area of orange region = 3 × 1
= 3 cm2
Area of grey region = 3 × 1
= 3 cm2
Total area = 3 + 3 + 3
= 9 cm2
∴ Total area is 9 cm2
11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
Solution:
Total area of the figure = 12 × 2 + 8 × 2
= 40 cm2
There are 5 squares. Each side is 7 cm
Area of 5 squares = 5 × 72
= 245 cm2
Area of grey rectangle = 2 × 1
= 2 cm2
Area of brown rectangle = 2 × 1
= 2 cm2
Area of orange rectangle = 5 × 1
= 5 cm2
Total area = 2 + 2 + 5
= 9 cm2
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.
Solution:
We know that area of rectangle is given by
Area of rectangle = length × width
(a)Area of rectangle = 100 × 144
= 14400 cm
Area of one tile = 5 × 12
= 60 cm2
Number of tiles = (Area of rectangle) / (Area of one tile)
= 14400 / 60
= 240
Hence, 240 tiles are required
(b) Area of rectangle = 70 × 36
= 2520 cm2
Area of one tile = 5 × 12
= 60 cm2
Number of tiles = (Area of rectangle) / (Area of one tile)
= 2520 / 60
= 42
Hence, 42 tiles are required.
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