Class 6 - Maths - Mensuration

Exercise 10.1

Question 1:

Find the perimeter of each of the following figures:

  Class_6_Mensuration_To_Find_Perimeter_Of_Different_Shapes1                                      

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 * 50

                                                    = 100 cm

                                                    = 1 m           {Since 1 m = 100 cm}

Thus, the total length of tape required is 100 cm or 1 m.

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 m + 0.25 cm = 2.25 m

Breadth of table top = 1 m 50 cm =  1m + 0.50 m = 1.50 m

Perimeter of table top = 2(length + breadth)

= 2(2.25 + 1.50)

= 2 * 3.75

= 7.50 m

Thus, the 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(length + breadth)

 = 2(32 + 21)

 = 2 * 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(length + breadth)

                                 = 2(0.7 + 0.5)

                                 = 2 * 1.2

                                 = 2.4 km

                                 = 2.4 * 1000 m

                                 = 2400 m

Thus, the length of wire = 4 * 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 triangle = Sum of all sides

                                            = 3 cm + 5 cm + 4 cm

                                            = 12 cm

(b) Perimeter of equilateral triangle = 3 * side

                                                                 = 3 * 9 cm

                                                                 = 27 cm

(c) Perimeter of isosceles triangle = Sum of all sides

                                                             = 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, the 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 * length of one side

                                        = 6 * 8

                                        = 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 * side

=> 20 = 4 * side

=> Side = 20/4

=> Side = 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 * side = 100 cm

=> Side = 100/5

=> Side = 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 * 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 * 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 * 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 x cm.

Lengths of other two sides are 12 cm and 14 cm.

Now, Perimeter of triangle = 36 cm

=> 12 + 14 + x = 36

 => 26 + x = 36

=> x = 36 - 26

=> x = 10 cm

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 Rs 20 per meter.

Answer:

Side of square = 250 m

Perimeter of square = 4 * side

                                     = 4 * 250

                                     = 1000 m

Since, cost of fencing of per meter = Rs 20

Therefore, the cost of fencing of 1000 meters = 20 * 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(length + breadth)

                                 = 2(175 + 125)

                                 = 2 * 300 = 600 m

Since, the cost of fencing park per meter = Rs 12

Therefore, the cost of fencing park of 600 m = 12 * 600 = Rs 7,200

Question 15:

Sweety 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 * side

                                     = 4 * 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(length + breadth)

                                                      = 2 * (60 + 45)

                                                      = 2 * 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?

                               Class_6_Mensuration_To_Find_Perimeter_Of_Different_Shapes        

Answer:

(a) Perimeter of square = 4 * side

                                           = 4 * 25 = 100 cm

(b) Perimeter of rectangle = 2(length + breadth)

                                                = 2(40 + 10)

                                                = 2 * 50

                                                = 100 cm

(c) Perimeter of rectangle = 2(length + breadth)

                                               = 2(30 + 20)

                                               = 2 * 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 ½ m. He lays them in the form of a square.

       Class_6_Mensuration_To_Find_Area_Of_Squares1          

(a) What is the perimeter of his arrangement?

(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) Side of square = 3 * (1/2) = 3/2 m

Perimeter of square = 4 * (3/2) = (4 * 3)/2 = 12/2 = 6 m

(b) Perimeter of cross = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1

                                        = 10 m

(c) The arrangement in the form of a cross has a greater perimeter.

(d) Arrangements with perimeters greater than 10 m cannot be determined.

                                                  Exercise 10.2

Question 1:

Find the areas of the following figures by counting squares:

          Class_6_Mensuration_To_Find_Area_By_Counting_Squares              

Class_6_Mensuration_To_Find_Area_By_Counting_Squares1

Answer:

(a) Number of filled square = 9

      So, area covered by squares = 9 * 1 = 9 sq. units

(b) Number of filled squares = 5

      So, area covered by filled squares = 5 * 1 = 5 sq. units

(c) Number of full filled squares = 2

      Number of half-filled squares = 4

      So, area covered by full filled squares = 2 * 1 = 2 sq. units

      And area covered by half-filled squares = 4 * (1/2) = 2 sq. units

      Now, Total area = 2 + 2 = 4 sq. units

(d) Number of filled squares = 8

      So, area covered by filled squares = 8 * 1 = 8 sq. units

(e) Number of filled squares = 10

      So, area covered by filled squares = 10 * 1 = 10 sq. units

(f) Number of full filled squares = 2

     Number of half-filled squares = 4

     Area covered by full filled squares = 2 * 1 = 2 sq. units

     And Area covered by half-filled squares = 4 * (1/2) = 2 sq. units

     So, total area = 2 + 2 = 4 sq. units

(g) Number of full filled squares = 4

     Number of half-filled squares = 4

     Area covered by full filled squares = 4 * 1 = 4 sq. units

     And area covered by half-filled squares = 4 * (1/2) = 2 sq. units

     So, total area = 4 + 2 = 6 sq. units

(h) Number of filled squares = 5

      So, area covered by filled squares = 5 * 1 = 5 sq. units

(i) Number of filled squares = 9

     So, area covered by filled squares = 9 * 1 = 9 sq. units

(j) Number of full filled squares = 2

     Number of half-filled squares = 4

     Area covered by full filled squares = 2 * 1 = 2 sq. units

     And area covered by half-filled squares = 4 * (1/2) = 2 sq. units

     So, total area = 2 + 2 = 4 sq. units

(k) Number of full filled squares = 4

      Number of half-filled squares = 2

      Area covered by full filled squares = 4 * 1 = 4 sq. units

      And area covered by half-filled squares = 2 * (1/2) = 1 sq. units

      So, total area = 4 + 1 = 5 sq. units

(l) Number of full filled squares = 3

     Number of half-filled squares = 10

     Area covered by full filled squares = 3 * 1 = 3 sq. units

     And area covered by half-filled squares = 10 * (1/2) = 5 sq. units

     So, total area = 3 + 5 = 8 sq. units

(m) Number of full filled squares = 7

       Number of half-filled squares = 14

       Area covered by full filled squares = 7 * 1 = 7 sq. units

       And area covered by half-filled squares = 14 * (1/2) = 7 sq. units

       So, total area = 7 + 7 = 14 sq. units

(n) Number of full filled squares = 10

      Number of half-filled squares = 16

      Area covered by full filled squares = 10 * 1 = 10 sq. units

      And area covered by half-filled squares = 16 * (1/2) = 8 sq. units

     So, total area = 10 + 8 = 18 sq. units

 

 

                                                                Exercise 10.3

Question 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

Answer:

(a) Area of rectangle = length * breadth

                                      = 3 cm * 4 cm = 12 cm2

(b) Area of rectangle = length * breadth

                                      = 12 m * 21 m = 252 m2

(c) Area of rectangle = length * breadth

                                      = 2 km * 3 km = 6 km2

(d) Area of rectangle = length * breadth

                                      = 2 m * 70 cm = 2 m * 0.7 m = 1.4 m2

Question 2:

Find the areas of the squares whose sides are:

(a) 10 cm   (b) 14 cm   (c) 5 cm

Answer:

(a) Area of square = side * side = 10 cm * 10 cm = 100 cm2

(b) Area of square = side * side = 14 cm * 14 cm = 196 cm2

(c) Area of square = side * side = 5 m * 5 m = 25 m2

 

 

 

Question 3:

The length and the 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?

Answer:

(a) Area of rectangle = length * breadth = 9 m * 6 m = 54 m2

(b) Area of rectangle = length * breadth= 3 m * 17 m = 51 m2

(c) Area of rectangle = length * breadth= 4 m * 14 m = 56 m2

Thus, the rectangle (c) has largest area, and rectangle (b) has smallest area.

Question 4:

The area of a rectangle garden 50 m long is 300 m2, find the width of the garden.

Answer:

Length of rectangle = 50 m and Area of rectangle = 300 m2

Since, Area of rectangle = length * breadth

Therefore, Breadth = Area of rectangle/Length

                                   = 300/50

                                   = 6 m

Thus, the breadth of the garden is 6 m.

Question 5:

What is the cost of tilling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq. m?

Answer:

Length of land = 500 m and Breadth of land = 200 m

Area of land = length * breadth = 500 m * 200 m = 1,00,000 m2

Cost of tilling 100 sq. m of land = Rs 8

Cost of tilling 1 sq. m of land = Rs 8/100

So, cost of tilling 1,00,000 sq. m of land = (8 * 1000 00) /100

                                                                       = 8 * 1000

                                                                       = Rs 8000

Question 6:

A table-top measures 2 m by 1 m 50 cm. What is its area in square meters?

Answer:

Length of table = 2 m

Breadth of table = 1 m 50 cm = 1 m + 0.50 cm = 1.50 m

Area of table = length * breadth

                        = 2 m * 1.50 m = 3 m2  

Question 7:

A room us 4 m long and 3 m 50 cm wide. How many square meters of carpet is needed to cover the floor of the room?      

Answer:

Length of room = 4 m

Breadth of room = 3 m 50 cm = 3 m + 0.50 m = 3.50 m

Area of carpet = length * breadth

                          = 4 * 3.50 = 14m2

 

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.

Answer:

Length of floor = 5 m and breadth of floor = 4 m

Area of floor = length * breadth

                        = 5 m * 4 m = 20 m2

Now, Side of square carpet = 3 m

Area of square carpet = side * side = 3 * 3 = 9 m2

Area of floor that is not carpeted = 20 m2 – 9 m2 = 11 m2

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?

Answer:

Side of square bed = 1 m

Area of square bed = side * side = 1 m * 1 m = 1 m2

Area of 5 square beds = 1 * 5 = 5 m2

Now, Length of land = 5 m

          Breadth of land = 4 m

Area of land = length * breadth

                       = 5 m * 4 m = 20 m2

Area of remaining part = Area of land – Area of 5 flower beds

                                          = 20 m2 – 5 m2 = 15 m2

 

Question 10:

By splitting the following figures into rectangles, find their areas. (The measures are given in centimeters)

 

 Class_6_Mensuration_To_Find_Area_Of_Different_Shapes_3

 

 

                             

Answer:

(a) Area of HKLM = 3 * 3 = 9 cm2                                                        

      Area of IJGH = 1 * 2 = 2 cm2

      Area of FEDG = 3 * 3 = 9 cm2

     Area of ABCD = 2 * 4 = 8 cm2

    Total area of the figure = 9 + 2 + 9 + 8 = 28 cm2

(b) Area of ABCD = 3 x 1 = 3 cm2

     Area of BDEF = 3 x 1 = 3 cm2

     Area of FGHI = 3 x 1 = 3 cm2

    Total area of the figure = 3 + 3 + 3 = 9 cm2

 

 Class_6_Mensuration_To_Find_Area_Of_Different_Shapes_2 

Question 11:

Split the following shapes into rectangles and find their areas. (The measures are given in centimeters)

     Class_6_Mensuration_To_Find_Area_Of_Different_Shapes_1                      

Answer:

(a) Area of rectangle ABCD = 2 * 10 = 20 cm2

      Area of rectangle DEFG = 10 * 2 = 20 cm2

      Total area of the figure = 20 + 20 = 40 cm2

 Class_6_Mensuration_Rectangle_In_Form_Of_L

(b) There are 5 squares each of side 7 cm.                         

      Area of one square = 7 x 7 = 49 cm2

      Area of 5 squares = 49 x 5 = 245 cm2

 

(c) Area of rectangle ABCD = 5 x 1 = 5 cm2

      Area of rectangle EFGH = 4 x 1 = 4 cm2                         

     Total area of the figure = 5 + 4 = 9 cm2

 Class_6_Mensuration_To_Find_Area_Of_Different_Shapes

 

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

Answer:

(a) Area of region = 100 cm * 144 cm = 14400 cm2

      Area of one tile = 5 cm * 12 cm = 60 cm2

      Number of tiles = Area of region/Area of one tile

                                   = 14400/60

                                   = 240

Thus, 240 tiles are required.

(b) Area of region = 70 cm * 36 cm = 2520 cm2

      Area of one tile = 5 cm * 12 cm = 60 cm2

      Number of tiles = Area of region/Area of one tile

                                   = 2520/60

                                   = 42

Thus, 42 tiles are required.

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