Here one parallel side of the trapezium (a) = 1 m And second side (b) = 1.2 m and height (h) = 0.8 m ∴ Area of top surface of the table = 1/2(a+b)xh = 1/2x(1+1.2)x0.8 = 1/2x(1+1.2)x0.8 = 1/2x2.2x0.8=0.88m² Hence, the surface area of the table is 88m² Class 8 Maths Chapter 11 Exercise 11.2 Solution iRead more
Here one parallel side of the trapezium (a) = 1 m
And second side (b) = 1.2 m and height (h) = 0.8 m
∴ Area of top surface of the table = 1/2(a+b)xh
= 1/2x(1+1.2)x0.8
= 1/2x(1+1.2)x0.8
= 1/2×2.2×0.8=0.88m²
Hence, the surface area of the table is 88m²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
(a) Radius = Diameter/2 = 2.8/2 = 1.4 cm Circumference of semi circle = πr=22/7x1.4=4.4 cm Total distance covered by the ant = Circumference of semi circle + Diameter = 4.4 + 2.8 = 7.2 cm (b) Diameter of semi circle = 2.8 cm ∴ Radius = Diameter/2 = 2.8/2 = 1.4 cm Circumference of semi circle = πr=22Read more
(a) Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr=22/7×1.4=4.4 cm
Total distance covered by the ant = Circumference of semi circle + Diameter
= 4.4 + 2.8 = 7.2 cm
(b) Diameter of semi circle = 2.8 cm
∴ Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr=22/7×1.4=4.4 cm
Total distance covered by the ant = 1.5 + 2.8 + 1.5 + 4.4 = 10.2 cm
(c) Diameter of semi circle = 2.8 cm
∴ Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr=22/7×1.4=4.4 cm
Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
Given: Base of flooring tile = 24 cm = 0.24 m Corresponding height of a flooring tile = 10 cm = 0.10 m Now Area of flooring tile = Base x Altitude = 0.24 x 0.10 = 0.024 m² ∴ Number of tiles required to cover the floor = Area of floor/Area of one tile = 1080/0.024 45000 tiles Hence 45000 tiles are reRead more
Given: Base of flooring tile = 24 cm = 0.24 m
Corresponding height of a flooring tile = 10 cm = 0.10 m
Now Area of flooring tile = Base x Altitude = 0.24 x 0.10 = 0.024 m²
∴ Number of tiles required to cover the floor = Area of floor/Area of one tile
= 1080/0.024
45000 tiles
Hence 45000 tiles are required to cover the floor.
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
Given: Total length = 20 m Diameter of semi circle = 7 m ∴ Radius of semi circle = 7/2=3.5m Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m Breadth of the rectangular field = 7 m ∴ Area of rectangular field = l X b = 13 x 7 = 91 m² Area of two semi circles = 2x1/2(πr²) = 2x1/2x22/7x3.Read more
Given: Total length = 20 m
Diameter of semi circle = 7 m
∴ Radius of semi circle = 7/2=3.5m
Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
∴ Area of rectangular field = l X b = 13 x 7 = 91 m²
Area of two semi circles = 2×1/2(πr²) = 2×1/2×22/7×3.5×3.5=38.5m²
Area of garden = 91 + 38.5 = 129.5 m²
Now Perimeter of two semi circles = 2xπr=2×22/7×3.5=22 m
And Perimeter of garden = 22 + 13 + 13 = 48 m
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
Side of a square plot = 25 m ∴ Area of square plot = (side)² = (25)² = 625m² Length of the house = 20 m and Breadth of the house = 15 m ∴ Area of the house = length x breadth = 20 x 15 = 300 m² Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m² ∵ Cost of developing the gardenRead more
Side of a square plot = 25 m
∴ Area of square plot = (side)² = (25)² = 625m²
Length of the house = 20 m and
Breadth of the house = 15 m
∴ Area of the house = length x breadth = 20 x 15 = 300 m²
Area of garden = Area of square plot – Area of house
= 625 – 300 = 325 m²
∵ Cost of developing the garden per sq. m = ₹ 55
∴ Cost of developing the garden 325 sq. m = ₹ 55 x 325
= ₹ 17,875
Hence total cost of developing a garden around is ₹ 17,875.
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 and 1.2 m and perpendicular distance between them is 0.8 m.
Here one parallel side of the trapezium (a) = 1 m And second side (b) = 1.2 m and height (h) = 0.8 m ∴ Area of top surface of the table = 1/2(a+b)xh = 1/2x(1+1.2)x0.8 = 1/2x(1+1.2)x0.8 = 1/2x2.2x0.8=0.88m² Hence, the surface area of the table is 88m² Class 8 Maths Chapter 11 Exercise 11.2 Solution iRead more
Here one parallel side of the trapezium (a) = 1 m
And second side (b) = 1.2 m and height (h) = 0.8 m
∴ Area of top surface of the table = 1/2(a+b)xh
= 1/2x(1+1.2)x0.8
= 1/2x(1+1.2)x0.8
= 1/2×2.2×0.8=0.88m²
Hence, the surface area of the table is 88m²
Class 8 Maths Chapter 11 Exercise 11.2 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c=2πr where r is the radius of the circle.
(a) Radius = Diameter/2 = 2.8/2 = 1.4 cm Circumference of semi circle = πr=22/7x1.4=4.4 cm Total distance covered by the ant = Circumference of semi circle + Diameter = 4.4 + 2.8 = 7.2 cm (b) Diameter of semi circle = 2.8 cm ∴ Radius = Diameter/2 = 2.8/2 = 1.4 cm Circumference of semi circle = πr=22Read more
(a) Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr=22/7×1.4=4.4 cm
Total distance covered by the ant = Circumference of semi circle + Diameter
= 4.4 + 2.8 = 7.2 cm
(b) Diameter of semi circle = 2.8 cm
∴ Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr=22/7×1.4=4.4 cm
Total distance covered by the ant = 1.5 + 2.8 + 1.5 + 4.4 = 10.2 cm
(c) Diameter of semi circle = 2.8 cm
∴ Radius = Diameter/2 = 2.8/2 = 1.4 cm
Circumference of semi circle = πr=22/7×1.4=4.4 cm
Total distance covered by the ant = 2 + 2 + 4.4 = 8.4 cm
Hence for figure (b) food piece, the ant would take a longer round.
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? [If required you can split the tiles in whatever way you want to fill up the corners]
Given: Base of flooring tile = 24 cm = 0.24 m Corresponding height of a flooring tile = 10 cm = 0.10 m Now Area of flooring tile = Base x Altitude = 0.24 x 0.10 = 0.024 m² ∴ Number of tiles required to cover the floor = Area of floor/Area of one tile = 1080/0.024 45000 tiles Hence 45000 tiles are reRead more
Given: Base of flooring tile = 24 cm = 0.24 m
Corresponding height of a flooring tile = 10 cm = 0.10 m
Now Area of flooring tile = Base x Altitude = 0.24 x 0.10 = 0.024 m²
∴ Number of tiles required to cover the floor = Area of floor/Area of one tile
= 1080/0.024
45000 tiles
Hence 45000 tiles are required to cover the floor.
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
The shape of a garden is rectangular in the middle and semi-circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5 meters]
Given: Total length = 20 m Diameter of semi circle = 7 m ∴ Radius of semi circle = 7/2=3.5m Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m Breadth of the rectangular field = 7 m ∴ Area of rectangular field = l X b = 13 x 7 = 91 m² Area of two semi circles = 2x1/2(πr²) = 2x1/2x22/7x3.Read more
Given: Total length = 20 m
Diameter of semi circle = 7 m
∴ Radius of semi circle = 7/2=3.5m
Length of rectangular field = 20 – (3.5 + 3.5) = 20 – 7 = 13 m
Breadth of the rectangular field = 7 m
∴ Area of rectangular field = l X b = 13 x 7 = 91 m²
Area of two semi circles = 2×1/2(πr²) = 2×1/2×22/7×3.5×3.5=38.5m²
Area of garden = 91 + 38.5 = 129.5 m²
Now Perimeter of two semi circles = 2xπr=2×22/7×3.5=22 m
And Perimeter of garden = 22 + 13 + 13 = 48 m
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/
Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ₹ 55 per m².
Side of a square plot = 25 m ∴ Area of square plot = (side)² = (25)² = 625m² Length of the house = 20 m and Breadth of the house = 15 m ∴ Area of the house = length x breadth = 20 x 15 = 300 m² Area of garden = Area of square plot – Area of house = 625 – 300 = 325 m² ∵ Cost of developing the gardenRead more
Side of a square plot = 25 m
∴ Area of square plot = (side)² = (25)² = 625m²
Length of the house = 20 m and
Breadth of the house = 15 m
∴ Area of the house = length x breadth = 20 x 15 = 300 m²
Area of garden = Area of square plot – Area of house
= 625 – 300 = 325 m²
∵ Cost of developing the garden per sq. m = ₹ 55
∴ Cost of developing the garden 325 sq. m = ₹ 55 x 325
= ₹ 17,875
Hence total cost of developing a garden around is ₹ 17,875.
Class 8 Maths Chapter 11 Exercise 11.1 Solution in Video
for more answers vist to:
See lesshttps://www.tiwariacademy.com/ncert-solutions/class-8/maths/chapter-11/