Chapter 10. Mensuration CBSE Class VI (6th) Mathematics | Exercise 10.3 - Solutions
Things to remember...
(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.
Question 2. Find the areas of the squares whose sides are :
(a) 10 cm
(b) 14 cm
(c) 5 m
Solution.
Question 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.
Rectagle with length and breadth in option (c) has the largest area (56 sq m) and rectagle with length and breadth in option (b) has the smallest area (51 sq m).
Question 4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution.
Thus, the width of the rectangular park is 6 m.
Question 5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m.?
Solution. Total area of tiles must be equal to the area of rectangular plot.
Thus, the cost of tiling the rectangular plot is Rs 8000
Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution.
Thus, the Area of the table-top is 3 sq m
Exercise 10.3 Mensuration | CBSE Class VI Mathematics | Solutions
Question 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.
Total area of carpet must be equal to the floor area of room.
Thus, to cover the floor of the room, a total of 14 sq m of carpet is required
Question 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.
Thus, the area of the floor that is not carpeted, is 11 sq m
Question 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.
Thus, the area of the remaining part of the land is 15 sq m
Question 10.By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
Solution (a) :
The shape (a) can be split into 4 rectangles a1, a2, a3 and a4 . The lengths and breadths of the respective rectangles are shown in the figure (a).
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Solution (b) :
The shape (b) can be split into 3 rectangles a1, a2 and a3 . The lengths and breadths of the respective rectangles are shown in the figure (b).
.jpg)
Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
Solution (a) :
The shape (a) can be split into 2 rectangles a1, and a2 . The lengths and breadths of the respective rectangles are shown in the figure (a).
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Solution (b) :
The shape (b) can be split into three parts i.e. square b1, rectangle b2 and square b3. The respective sizes of the sides of rectangle and squares are shown in the figure (b).
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Solution (c) :
The shape (c) can be split into 2 rectangles c1, and c2 . The lengths and breadths of the respective rectangles are shown in the figure (c).
.jpg)
Question 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. (a) 100 cm and 144 cm
Total area of tiles must be equal to the area of the rectangular region.
Solution. (b) 70 cm and 36 cm
Total area of tiles must be equal to the area of the rectangular region.
CBSE Class 6th Mathematics | Chapter 10. Mensuration | Exercise 10.3
(This is not a part of exercise.. but this comes after the exercise !! )
On a centimetre squared paper, make as many rectangles as you can, such that the area of the rectangle is 16 sq cm (consider only natural number lengths).
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter?
If you take a rectangle of area 24 sq cm, what will be your answers?
Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter? With the least perimeter? Give example and reason.
Solution :
For area of the rectangles to be 16 sq cm: The possible rectangles of different shapes (sizes), which we can make on a centimetre squared paper are as given below :
Rectangle a1 : length = 16 cm, breadth = 1 cm
Rectangle a2 : length = 8 cm, breadth = 2 cm
Rectangle a3 : length = 4 cm, breadth = 4 cm
What we find :
(a) Rectangle a1 has the greatest perimeter of 34 cm
(a) Rectangle a3 has the least perimeter of 16 cm
If we take a rectangle of area 24 sq cm : The possible rectangles of different shapes (sizes), which we can make on a centimetre squared paper are as given below :
Rectangle a1 : length = 24 cm, breadth = 1 cm
Rectangle a2 : length = 12 cm, breadth = 2 cm
Rectangle a3 : length = 8 cm, breadth = 3 cm
Rectangle a4 : length = 6 cm, breadth = 4 cm
What we find :
(a) Rectangle a1 has the greatest perimeter of 50 cm
(b) Rectangle a4 has the least perimeter of 20 cm
In the same way, taking different value of area for the rectangle, we can find the greatest and least values of perimeters as shown in the table below :
From the above table, we observe that , given any area
The greatest perimeter = 2 × Area + 2
For examples:
Taking area 16 sq cm, the greatest perimeter will be : 2×16 + 2 = 34 cm.
Similarly, for area 24 sq cm, the greatest perimeter will be : 2×24 + 2 = 50 cm.
There for, it is possible to predict the shape of the rectangle with the greatest perimeter
Where as, it is not as easy to predict the shape of the rectangle with the least perimeter as it is beyond the scope of this chapter.
- Area of a rectangle = length × breadth
- Area of the square = side × side
(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 | = 4 cm |
| Breadth of the rectangle | = 3 cm |
| Area of the rectangle | = (length × breadth)) |
| = 4 cm × 3 cm | |
| = 12 sq cm | |
| (b) 12 m and 21 m | |
| Length of the rectangle | = 12 m |
| Breadth of the rectangle | = 21 m |
| Area of the rectangle | = length × breadth |
| = 12 m × 21 m | |
| = 252 sq m | |
| (c) 2 km and 3 km | |
| Length of the rectangle | = 3 km |
| Breadth of the rectangle | = 2 km |
| Area of the rectangle | = length × breadth |
| = 3 km × 2 km | |
| = 6 sq km | |
| (d) 2 m and 70 cm | |
| Length of the rectangle | = 2 m = 200 cm |
| Breadth of the rectangle | = 70 cm |
| Area of the rectangle | = length × breadth |
| = 200 cm × 70 cm | |
| = 14000 sq cm | |
| = 1.40 sq m | |
Question 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 cm × 10 cm | |
| = 100 sq cm | |
| (b) 14 cm | |
| Side of the square | = 14 cm |
| Area of the square | = Side × Side |
| = 14 cm × 14 cm | |
| = 196 sq cm | |
| (c) 5 m | |
| Side of the square | = 5 m |
| Area of the square | = Side × Side |
| = 5 m × 5 m | |
| = 25 sq m | |
Question 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 | = length × breadth |
| = 9 m × 6 m | |
| = 54 sq m | |
| (b) 17 m and 3 m | |
| Length of the rectangle | = 17 m |
| Breadth of the rectangle | = 3 m |
| Area of the rectangle | = length × breadth |
| = 17 m × 3 m | |
| = 51 sq m | |
| (c) 4 m and 14 m | |
| Length of the rectangle | = 4 m |
| Breadth of the rectangle | = 14 m |
| Area of the rectangle | = length × breadth |
| = 4 m × 14 m | |
| = 56 sq m | |
Rectagle with length and breadth in option (c) has the largest area (56 sq m) and rectagle with length and breadth in option (b) has the smallest area (51 sq m).
Question 4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution.
| Area of the rectangle garden | = 300 sq m |
| Length of the garden | = 50 m |
| Width of the rectangle | = ? m |
| Area of the rectangle | = length × width |
| So, width the rectangle | = Area of the rectangle/length |
| = 300/50 m | |
| = 6 m |
Thus, the width of the rectangular park is 6 m.
Question 5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m.?
Solution. Total area of tiles must be equal to the area of rectangular plot.
| Length of the rectangular plot | = 500 m |
| Breadth of the rectangular plot | = ? |
| Area of the plot or tiling | = length × breadth |
| = 500 m × 200 m | |
| The cost of tiling per 100 sq m, | = Rs 8 |
| The cost of tiling 10,0000 sq m , | = Rs (8/100)× 10,0000 |
Thus, the cost of tiling the rectangular plot is Rs 8000
Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution.
| Length of the table-top | = 2 m |
| Breadth of the table-top | = 1 m 50 cm = 1.5 m |
| Area of the table-top | = length × breadth |
| = 2 m × 1.5 m | |
| = 3 sq m |
Thus, the Area of the table-top is 3 sq m
Exercise 10.3 Mensuration | CBSE Class VI Mathematics | Solutions
Question 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.
Total area of carpet must be equal to the floor area of room.
| Length of the room | = 4 m |
| Breadth of the room | = 3 m 50 cm = 3.50 m |
| Area of the room floor | = length × breadth |
| = 4 m × 3.50 m | |
| = 14 sq m |
Thus, to cover the floor of the room, a total of 14 sq m of carpet is required
Question 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 the floor | = 5 m |
| Breadth of the floor | = 4 m |
| Area of the floor | = length × breadth |
| = 5 m × 4 m | |
| = 20 sq m | |
| Side of the square carpet | = 3 m |
| Area of the square carpet | = Side × Side |
| = 3 m × 3 m | |
| = 9 sq m | |
| The area of the floor without carpet | = Area of the floor − Area of the carpet |
| = 20 sq m − 9 sq m | |
| = 11 sq m |
Thus, the area of the floor that is not carpeted, is 11 sq m
Question 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.
| Length of the piece of land | = 5 m |
| Breadth of the piece of land | = 4 m |
| Area of the land | = length × breadth |
| = 5 m × 4 m | |
| = 20 sq m | |
| Side of the flower bed | = 1 m |
| Area of the flower bed | = Side × Side |
| = 1 m × 1 m | |
| = 1 sq m | |
| Area of the 5 flower beds | = 5 × 1 sq m |
| = 5 sq m | |
| The area of the remaining land | = Area of the land − Area of flower beds |
| = 20 sq m − 5 sq m | |
| = 15 sq m |
Thus, the area of the remaining part of the land is 15 sq m
Question 10.By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
Solution (a) :
The shape (a) can be split into 4 rectangles a1, a2, a3 and a4 . The lengths and breadths of the respective rectangles are shown in the figure (a).
| Area of rectangle a1 | = 3 cm × 4 cm | = 12 sq cm |
| Area of rectangle a2 | = 4 cm × 1 cm | = 4 sq cm |
| Area of rectangle a3 | = 5 cm × 2 cm | = 10 sq cm |
| Area of rectangle a3 | = 2 cm × 1 cm | = 2 sq cm |
| Area of shape (a) | = a1+ a2+ a3+ a4 | = (12 + 4 +10 +2 ) sq cm |
| = 28 sq cm |
Solution (b) :
The shape (b) can be split into 3 rectangles a1, a2 and a3 . The lengths and breadths of the respective rectangles are shown in the figure (b).
| Area of rectangle a1 | = 1 cm × 2 cm | = 2 sq cm |
| Area of rectangle a2 | = 5 cm × 1 cm | = 5 sq cm |
| Area of rectangle a3 | = 1 cm × 2 cm | = 2 sq cm |
| Area of rectangle a3 | = 2 cm × 1 cm | = 2 sq cm |
| Area of shape (b) | = a1+ a2+ a3 | = (2 + 5 +2 ) sq cm |
| = 9 sq cm |
Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
Solution (a) :
The shape (a) can be split into 2 rectangles a1, and a2 . The lengths and breadths of the respective rectangles are shown in the figure (a).
| Area of rectangle a1 | = 10 cm × 2 cm | = 20 sq cm |
| Area of rectangle a2 | = 2 cm × 10cm | = 20 sq cm |
| Area of shape (a) | = a1+ a2 | = (20 +20 ) sq cm |
| = 40 sq cm |
Solution (b) :
The shape (b) can be split into three parts i.e. square b1, rectangle b2 and square b3. The respective sizes of the sides of rectangle and squares are shown in the figure (b).
| Area of rectangle b1 | = 7 cm × 7 cm | = 49 sq cm |
| Area of rectangle b2 | = 21 cm × 7 cm | = 147 sq cm |
| Area of rectangle b3 | = 7 cm × 7 cm | = 49 sq cm |
| Area of shape (b) | = b1+ b2 + b3 | = (49 + 147 + 49 ) sq cm |
| = 245 sq cm |
Solution (c) :
The shape (c) can be split into 2 rectangles c1, and c2 . The lengths and breadths of the respective rectangles are shown in the figure (c).
.jpg)
| Area of rectangle c1 | = 1 cm × 4 cm | = 4 sq cm |
| Area of rectangle c2 | = 5 cm × 1 cm | = 5 sq cm |
| Area of shape (c) | = c1+ c2 | = (4 +5 ) sq cm |
| = 9 sq cm |
Question 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. (a) 100 cm and 144 cm
Total area of tiles must be equal to the area of the rectangular region.
| Length of the rectangular region | = 144 cm |
| Breadth of the rectangular region | = 100 cm |
| Area of the rectangular region | = length × breadth |
| = 144 cm × 100 cm | |
| = 14400 sq m | |
| Area of one tile | = length × breadth |
| = 12 cm × 5 cm | |
| = 60 sq m | |
| Number of tiles required to fit in | = Area of the region / Area of one tile |
| = 14400 / 60 | |
| = 240 tiles |
Solution. (b) 70 cm and 36 cm
Total area of tiles must be equal to the area of the rectangular region.
| Length of the rectangular region | = 70 cm |
| Breadth of the rectangular region | = 36 cm |
| Area of the rectangular region | = length × breadth |
| = 70 cm × 36 cm | |
| = 2520 sq m | |
| Area of one tile | = length × breadth |
| = 12 cm × 5 cm | |
| = 60 sq m | |
| Number of tiles required to fit in | = Area of the region / Area of one tile |
| = 2520 / 60 | |
| = 42 tiles |
CBSE Class 6th Mathematics | Chapter 10. Mensuration | Exercise 10.3
A challenge!
(This is not a part of exercise.. but this comes after the exercise !! )
On a centimetre squared paper, make as many rectangles as you can, such that the area of the rectangle is 16 sq cm (consider only natural number lengths).
(a) Which rectangle has the greatest perimeter?
(b) Which rectangle has the least perimeter?
If you take a rectangle of area 24 sq cm, what will be your answers?
Given any area, is it possible to predict the shape of the rectangle with the greatest perimeter? With the least perimeter? Give example and reason.
Solution :
For area of the rectangles to be 16 sq cm: The possible rectangles of different shapes (sizes), which we can make on a centimetre squared paper are as given below :
Rectangle a1 : length = 16 cm, breadth = 1 cm
Rectangle a2 : length = 8 cm, breadth = 2 cm
Rectangle a3 : length = 4 cm, breadth = 4 cm
| Rectangle | Area | Perimeter |
| Rectangle a1 | 16cm × 1 cm = 16 sq cm | 2(16 + 1) = 34 cm |
| Rectangle a2 | 8cm × 2 cm = 16 sq cm | 2(8 + 2) = 20 cm |
| Rectangle a3 | 4 cm × 4 cm = 16 sq cm | 2(4 + 4) = 16 cm |
What we find :
(a) Rectangle a1 has the greatest perimeter of 34 cm
(a) Rectangle a3 has the least perimeter of 16 cm
If we take a rectangle of area 24 sq cm : The possible rectangles of different shapes (sizes), which we can make on a centimetre squared paper are as given below :
Rectangle a1 : length = 24 cm, breadth = 1 cm
Rectangle a2 : length = 12 cm, breadth = 2 cm
Rectangle a3 : length = 8 cm, breadth = 3 cm
Rectangle a4 : length = 6 cm, breadth = 4 cm
| Rectangle | Area | Perimeter |
| Rectangle a1 | 24cm × 1 cm = 24 sq cm | 2(24 + 1) = 50 cm |
| Rectangle a2 | 12cm × 2 cm = 24 sq cm | 2(12 + 2) = 28 cm |
| Rectangle a3 | 8 cm × 3 cm = 24 sq cm | 2(8 + 3) = 22 cm |
| Rectangle a4 | 6 cm × 4 cm = 24 sq cm | 2(6 + 4) = 20 cm |
What we find :
(a) Rectangle a1 has the greatest perimeter of 50 cm
(b) Rectangle a4 has the least perimeter of 20 cm
In the same way, taking different value of area for the rectangle, we can find the greatest and least values of perimeters as shown in the table below :
| Area of rectangle | Shortest Perimeter | Longest Perimeter |
| 1 | 4 | 4 |
| 2 | 6 | 6 |
| 3 | 8 | 8 |
| 4 | 8 | 10 |
| 5 | 10 | 12 |
| 6 | 10 | 14 |
| 7 | 12 | 16 |
| 8 | 12 | 18 |
| 9 | 12 | 20 |
| 10 | 14 | 22 |
| 11 | 14 | 24 |
| 12 | 14 | 26 |
| 13 | 16 | 28 |
| 14 | 16 | 30 |
| 15 | 16 | 32 |
| 16 | 16 | 34 |
| 17 | 18 | 36 |
| 18 | 18 | 38 |
| 19 | 18 | 40 |
| 20 | 18 | 42 |
| 21 | 20 | 44 |
| 22 | 20 | 46 |
| 23 | 20 | 48 |
| 24 | 20 | 50 |
From the above table, we observe that , given any area
The greatest perimeter = 2 × Area + 2
For examples:
Taking area 16 sq cm, the greatest perimeter will be : 2×16 + 2 = 34 cm.
Similarly, for area 24 sq cm, the greatest perimeter will be : 2×24 + 2 = 50 cm.
There for, it is possible to predict the shape of the rectangle with the greatest perimeter
Where as, it is not as easy to predict the shape of the rectangle with the least perimeter as it is beyond the scope of this chapter.
