ML Aggarwal CBSE Solutions Class 6 Math Eleventh Chapter Mensuration Exercise 11.2

ML Aggarwal CBSE Solutions Class 6 Math 11th Chapter Mensuration Exercise 11.2

Exercise 11.2

(1) Find the area of the region enclosed by the following figures by counting squares:

(i) The given figure is covered by complete squares only.

Number of complete covered squares  = 9.

Area covered by complete square = (9 x 1) sq. unit = 9 sq. units

∴ Area of the region enclosed by the figure = 9 sq. units.

(ii) The given squares is covered by complete squares only.

Number of complete covered squares = = 5

Area covered by complete square = (5 x 1) sq. unit = 5 sq. unit.

∴ Area of the region enclosed by the figure = 5 sq. units.

(iii) The given figure is covered by complete squares only.

Number of complete square = 10.

Area covered by complete square = (10 x 1) = 10 sq. unit.

∴ Area of the region enclosed by the figure = 10 sq. units.

(iv) Number of complete covered squares = 4

Number of exactly half covered squares = 4

Area covered by complete squares = (4 x 1) sq units = 4 sq. units.

Area covered by exactly half covered squares = (4 x ½) = 2 square units.

∴ Area of the region enclosed by the given figure = (4 + 2) square units = 6 sq. units.

(v) Number of complete covered squares = 2

Number of exactly half covered squares = 4

Area covered by complete squares = (2 x 1) = 2 sq. units.

Area covered by exactly half covered squares = (4 x ½) = 2 square units

∴ Area of the region enclosed by the given figure = (2 + 2) sq. units = 4 sq. units.

(vi) Number of complete covered squares = 3

Number of exactly half covered squares = 6

Area covered by complete covered squares = (3 x 1) sq. units = 3 sq. units.

Area covered by exactly half covered squares = (6 x ½) = 3 sq. units

∴ Area of the region enclosed by the given figure = (3 + 3) sq. units = 6 sq. units.

(2) Find the area of the following closed figures by counting squares:

(i) Number of complete covered squares = 4

Number of more than half covered squares = 4

And, there is no exactly half squares covered.

So, estimate area of the given closed figure = (4 + 4) sq. units = 8 sq. units.

(ii) Number of complete covered squares = 7

Number of more than half covered squares = 6

And there is no exactly half covered squares covered.

So, estimate area of the given closed figure = (7 + 6) 13 sq units.

(iii) Number of complete covered squares = 11

Number of more than half covered squares = 7

And there is no exactly half covered squares covered.

So, estimate area of the given closed figure = (11 + 7) sq. units = 18 sq. units.

(3) Find the area:

(i) 9 m and 6 m

Solution: The area of rectangle = 9 m x 6 m

= 54 m2

(ii) 17 m and 3 m

Solution: The area of rectangle = 17 m x 3 m

= 51 m square.

(iii) 14 m and 4 m

Solution: The area of rectangle = 14 m x 4 m

= 56 meter square.

Here, No. iii) 14 m and 4 m  is the largest area and No. ii) 17 m and 3 m  is the smallest area.

(4) Find the areas of the rectangles whose two adjacent sides are:

(i) 14 cm and 23 cm

Solution: The area of rectangle = 14 cm x 23 cm

= 322 cm2

(ii) 3 km and 4 km

Solution: The area of rectangle = 3 km x 4 km = 12 km2

(iii) 2 m and 90 m

Solution:  The area of rectangle = 2 m x  90 m  = 180 m2

(5) Find area of squares whose sides are:

(i) 8 cm

Solution: The area of square = (side)2

= (8 cm)2

= 64 cm2

(ii) 14 m

Solution: The area of square = (14 cm)2

= 196 cm2

(iii) 2 m 50 cm

Solution: 2 m 50 cm

= 2 x 100 + 50

= 200 + 50

= 250 cm

The area of square = (250 cm)2

= 62500 cm2

(6) A room is 4 m long and 3 m 25 cm wide. How many square meters of carpet is needed to cover the floor of the room.

Solution: The length of the room is 4 m

& The breadth of the rooms is 3 m 25 cm = 3 m + 25/100 m

= 3 m + 0.25 m

= 3.25 m

The square meter of carpet is needed to cover the floor = 4 m x 3.25 m = 13 m2

(7) What is the cost of tiling a rectangular field 500 m long and 200 m wide at the rate of Rs. 7.5 per hundred square metres?

Solution: Te area of rectangular field = 500 m x 200 m = 100000 m2

Total cost needed = 100000 x 7.5

= Rs. 750000

(8) A floor is 5 m long and 4 m wide. A square carpet of side 3 m is laid on the floor. Find the area of the floor that is not carpeted.

Solution: The area of the floor = 5 m x 4 m = 20 m2

The area of square carpet = (3 m )2 = 9 m2

Therefore, The area of the floor that is not carpeted = 20 m2 – 9 m2

= 11 m2

(9) In the adjoining figure, find the area of the path which is 2 m all around.

Solution: Total area of the given figures in books = 100 m x 60 m  = 6000 meter square.

Here it is given that, the path is 2 m all around.

So, here we have to find inner area, by subtracting 2 m path from total length and total breadth.

Therefore, Inner length is = (100 m – 2 – 2) = 96 m

Inner breadth is = (60 m – 2 – 2) = 56 m

Now, Inner area = 96 m x 56 m = 5376 m2

Therefore, area of the path = Total Area – Inner Area

= 6000 meter square – 5376 meter square

= 624 meter square.

(10) Four square flower beds of side 1 m 50 cm are dug on a rectangular piece of land 8 m long and 6 m 50 cm wide. What is the area of the remaining part of the land.

Solution: Here side of one flower beds = 1 m 50 cm = 150 cm

The area of four flower beds = 4 x (150)2

= 4 x 22500

= 90000 cm2

Here, the length of land = 8 m = 800 cm

Breadth of the land = 6 m 50 cm = 650 cm

The area of the land = 800 cm x 650 cm

= 520000 cm2

∴ The area of the remaining part of the land = 520000 cm2 – 90000 cm2

= 430000 cm2

= 43 m2 (divide the area value by 10000)

(11) How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed to cover a rectangular region whose length and breadth are respectively:

(i) 70 cm and 36 cm (ii) 144 cm and 1 m

Solution: (i) area of the tiles = 12 cm x 5 cm = 60 cm2

Area of rectangular region = 70 cm x 36 cm

= 2520 cm2

Total tiles needed = 2520 / 60 = 42 pieces.

(ii) area of the tiles = 12 cm x 5 cm = 60 cm2

Area of rectangular region = 144 cm x 100 cm (As 1 m = 100 cm)

= 14400 cm2

Total tiles needed = 14400 / 60 = 260 pieces.

(12) The area of rectangular plot is 340 sq. m and its length is 17 m. Find the breadth of the plot and the perimeter.

Solution: The area of the rectangular plot = 340 sq m.

 Length = 17 m

∴ Breadth = 340 / 12 = 20 m.

The perimeter of the field = 2 x (17 + 20)

= 2 x 37

= 74 m

 (13) If the area of a rectangular plot is 144 sq. m and its length is 16 m. Find the breadth of the plot and the cost of fencing  it at the rate of Rs. 6 per metre.

Solution: Here the area of the plot = 144 sq m.

Length is = 16 m.

∴ Breadth is = 144 / 16

= 9 m.

The perimeter of the rectangular plot = 2 x (16 + 9)

= 2 x 25

= 50 m

Cost needed for fencing = 50 m x 6

= Rs. 300

(14) split the following  shapes into rectangles and find their areas (The measure are given in centimetres)

Solution: (1st) Area = (2 x 12) + (2 x 8) = 24 + 16 = 40 cm square.

(2nd) Area = 5 x (7 x 7) = 5 x 49 = 245 cm square.

(3rd) Area = (5 x 1) + (4 x 1)  = 5 + 4 = 9 cm square.

Hope you Guyz Understand this chapter. Share this page to your friends.

Regards,

Net Explanations


Leave a Reply

Your email address will not be published. Required fields are marked *