Use the formula for the area of a rectangle: Area = Length × Width. Here, Area = 1000 square meters, and Length = 50 meters. Rearrange to find the width: Width = Area ÷ Length Width = 1000 ÷ 50 = 20 meters. Thus, the width of the garden is 20 meters. For more NCERT Solutions for Class 6 Math ChapterRead more
Use the formula for the area of a rectangle:
Area = Length × Width.
Here, Area = 1000 square meters, and Length = 50 meters. Rearrange to find the width:
Width = Area ÷ Length
Width = 1000 ÷ 50 = 20 meters.
Thus, the width of the garden is 20 meters.
First, calculate the area of the room: Area of room = 5 × 4 = 20 square meters. Next, calculate the area of the square carpet: Area of carpet = 3 × 3 = 9 square meters. The area that is not carpeted is: Uncovered area = Area of room − Area of carpet Uncovered area = 20 − 9 = 11 square meters. Thus,Read more
First, calculate the area of the room:
Area of room = 5 × 4 = 20 square meters.
Next, calculate the area of the square carpet:
Area of carpet = 3 × 3 = 9 square meters.
The area that is not carpeted is:
Uncovered area = Area of room − Area of carpet
Uncovered area = 20 − 9 = 11 square meters.
Thus, 11 square meters of the floor remains uncovered.
First, calculate the total area of the garden: Garden area = 15 × 12 = 180 square meters. Next, calculate the area of one flower bed: Area of one flower bed = 2 × 1 = 2 square meters. For four flower beds: Total flower bed area = 4 × 2 = 8 square meters. The area available for laying a lawn is: RemaRead more
First, calculate the total area of the garden:
Garden area = 15 × 12 = 180 square meters.
Next, calculate the area of one flower bed:
Area of one flower bed = 2 × 1 = 2 square meters.
For four flower beds:
Total flower bed area = 4 × 2 = 8 square meters.
The area available for laying a lawn is:
Remaining area = Garden area − Flower bed area
Remaining area = 180 − 8 = 172 square meters.
For Shape A: Dimensions = 1 unit × 18 units Area = 1 × 18 = 18 square units Perimeter = 2 × (1 + 18) = 38 units. For Shape B: Dimensions = 4 units × 5 units Area = 4 × 5 = 20 square units Perimeter = 2 × (4 + 5) = 18 units. Thus, Shape A with a longer perimeter and Shape B with a smaller perimeter sRead more
For Shape A:
Dimensions = 1 unit × 18 units
Area = 1 × 18 = 18 square units
Perimeter = 2 × (1 + 18) = 38 units.
For Shape B:
Dimensions = 4 units × 5 units
Area = 4 × 5 = 20 square units
Perimeter = 2 × (4 + 5) = 18 units.
Thus, Shape A with a longer perimeter and Shape B with a smaller perimeter satisfy the given conditions.
Let the original dimensions of the page be Length (L) and Width (W). Reduced dimensions: Reduced Length = L − 2 (1 cm from top and bottom = 2 cm) Reduced Width = W − 3 (1.5 cm from left and right = 3 cm) Perimeter of the border: Perimeter = 2 × (Reduced Length + Reduced Width) Substitute the given vRead more
Let the original dimensions of the page be Length (L) and Width (W).
Reduced dimensions:
Reduced Length = L − 2 (1 cm from top and bottom = 2 cm)
Reduced Width = W − 3 (1.5 cm from left and right = 3 cm)
Perimeter of the border:
Perimeter = 2 × (Reduced Length + Reduced Width)
Substitute the given values of Length and Width to find the final perimeter.
The area of the outer rectangle is calculated as: Area = Length × Width = 12 × 8 = 96 square units. Half the area is: Half Area = 96 ÷ 2 = 48 square units. For the inner rectangle to have this area, one possible set of dimensions is: Length = 8 units, Width = 6 units. Verification: Area of inner recRead more
The area of the outer rectangle is calculated as:
Area = Length × Width = 12 × 8 = 96 square units.
Half the area is:
Half Area = 96 ÷ 2 = 48 square units.
For the inner rectangle to have this area, one possible set of dimensions is:
Length = 8 units, Width = 6 units.
Verification:
Area of inner rectangle = 8 × 6 = 48 square units, which is exactly half of 96.
Thus, the inner rectangle can be 8 units × 6 units.
The square's perimeter is 4 × side. After folding, two rectangles are formed with lengths equal to half the square's side. The total perimeter of both rectangles is 6 × side, which is 1.5 times the square's perimeter. Thus, the correct option is (c): The perimeters of both rectangles added togetherRead more
The square’s perimeter is 4 × side. After folding, two rectangles are formed with lengths equal to half the square’s side. The total perimeter of both rectangles is 6 × side, which is 1.5 times the square’s perimeter.
Thus, the correct option is (c): The perimeters of both rectangles added together are always 1.5 times the perimeter of the square.
Among the tangram pieces, Shapes C and E have the same area because they are identical in size and shape. Similarly, Shapes A and B share equal areas, as they are congruent triangles. Shapes F and G are also equal in area. However, Shape D is larger and cannot be paired with another shape of the samRead more
Among the tangram pieces, Shapes C and E have the same area because they are identical in size and shape. Similarly, Shapes A and B share equal areas, as they are congruent triangles. Shapes F and G are also equal in area. However, Shape D is larger and cannot be paired with another shape of the same area. This comparison is evident by overlaying the shapes on each other.
Shape D is twice as large as Shape C. By placing the pieces together, we see that Shape D can be entirely formed by combining Shapes C and E. This indicates that Shapes C and E are equal in area, and their combined areas equal that of Shape D. Thus, Shape D represents the total area of two smaller iRead more
Shape D is twice as large as Shape C. By placing the pieces together, we see that Shape D can be entirely formed by combining Shapes C and E. This indicates that Shapes C and E are equal in area, and their combined areas equal that of Shape D. Thus, Shape D represents the total area of two smaller identical pieces, making its size double that of a single Shape C or E.
Shape D is larger than Shape F in terms of area. Shape D is made up of the combined areas of Shapes C and E, making it twice the size of Shape C. On the other hand, Shape F has an area equal to Shape C. Therefore, the area of Shape D is two times that of Shape F, as it is composed of more identicalRead more
Shape D is larger than Shape F in terms of area. Shape D is made up of the combined areas of Shapes C and E, making it twice the size of Shape C. On the other hand, Shape F has an area equal to Shape C. Therefore, the area of Shape D is two times that of Shape F, as it is composed of more identical smaller sections.
The area of a rectangular garden that is 50 m long is 1000 sq m. Find the width of the garden.
Use the formula for the area of a rectangle: Area = Length × Width. Here, Area = 1000 square meters, and Length = 50 meters. Rearrange to find the width: Width = Area ÷ Length Width = 1000 ÷ 50 = 20 meters. Thus, the width of the garden is 20 meters. For more NCERT Solutions for Class 6 Math ChapterRead more
Use the formula for the area of a rectangle:
Area = Length × Width.
Here, Area = 1000 square meters, and Length = 50 meters. Rearrange to find the width:
Width = Area ÷ Length
Width = 1000 ÷ 50 = 20 meters.
Thus, the width of the garden is 20 meters.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
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The floor of a room is 5 m long and 4 m wide. A square carpet whose sides are 3 m in length is laid on the floor. Find the area that is not carpeted.
First, calculate the area of the room: Area of room = 5 × 4 = 20 square meters. Next, calculate the area of the square carpet: Area of carpet = 3 × 3 = 9 square meters. The area that is not carpeted is: Uncovered area = Area of room − Area of carpet Uncovered area = 20 − 9 = 11 square meters. Thus,Read more
First, calculate the area of the room:
Area of room = 5 × 4 = 20 square meters.
Next, calculate the area of the square carpet:
Area of carpet = 3 × 3 = 9 square meters.
The area that is not carpeted is:
Uncovered area = Area of room − Area of carpet
Uncovered area = 20 − 9 = 11 square meters.
Thus, 11 square meters of the floor remains uncovered.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
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Four flower beds having sides 2 m long and 1 m wide are dug at the four corners of a garden that is 15 m long and 12 m wide. How much area is now available for laying down a lawn?
First, calculate the total area of the garden: Garden area = 15 × 12 = 180 square meters. Next, calculate the area of one flower bed: Area of one flower bed = 2 × 1 = 2 square meters. For four flower beds: Total flower bed area = 4 × 2 = 8 square meters. The area available for laying a lawn is: RemaRead more
First, calculate the total area of the garden:
Garden area = 15 × 12 = 180 square meters.
Next, calculate the area of one flower bed:
Area of one flower bed = 2 × 1 = 2 square meters.
For four flower beds:
Total flower bed area = 4 × 2 = 8 square meters.
The area available for laying a lawn is:
Remaining area = Garden area − Flower bed area
Remaining area = 180 − 8 = 172 square meters.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
Shape A has an area of 18 square units and Shape B has an area of 20 square units. Shape A has a longer perimeter than Shape B. Draw two such shapes satisfying the given conditions.
For Shape A: Dimensions = 1 unit × 18 units Area = 1 × 18 = 18 square units Perimeter = 2 × (1 + 18) = 38 units. For Shape B: Dimensions = 4 units × 5 units Area = 4 × 5 = 20 square units Perimeter = 2 × (4 + 5) = 18 units. Thus, Shape A with a longer perimeter and Shape B with a smaller perimeter sRead more
For Shape A:
Dimensions = 1 unit × 18 units
Area = 1 × 18 = 18 square units
Perimeter = 2 × (1 + 18) = 38 units.
For Shape B:
Dimensions = 4 units × 5 units
Area = 4 × 5 = 20 square units
Perimeter = 2 × (4 + 5) = 18 units.
Thus, Shape A with a longer perimeter and Shape B with a smaller perimeter satisfy the given conditions.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
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On a page in your book, draw a rectangular border that is 1 cm from the top and bottom and 1.5 cm from the left and right sides. What is the perimeter of the border?
Let the original dimensions of the page be Length (L) and Width (W). Reduced dimensions: Reduced Length = L − 2 (1 cm from top and bottom = 2 cm) Reduced Width = W − 3 (1.5 cm from left and right = 3 cm) Perimeter of the border: Perimeter = 2 × (Reduced Length + Reduced Width) Substitute the given vRead more
Let the original dimensions of the page be Length (L) and Width (W).
Reduced dimensions:
Reduced Length = L − 2 (1 cm from top and bottom = 2 cm)
Reduced Width = W − 3 (1.5 cm from left and right = 3 cm)
Perimeter of the border:
Perimeter = 2 × (Reduced Length + Reduced Width)
Substitute the given values of Length and Width to find the final perimeter.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
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Draw a rectangle of size 12 units × 8 units. Draw another rectangle inside it, without touching the outer rectangle that occupies exactly half the area.
The area of the outer rectangle is calculated as: Area = Length × Width = 12 × 8 = 96 square units. Half the area is: Half Area = 96 ÷ 2 = 48 square units. For the inner rectangle to have this area, one possible set of dimensions is: Length = 8 units, Width = 6 units. Verification: Area of inner recRead more
The area of the outer rectangle is calculated as:
Area = Length × Width = 12 × 8 = 96 square units.
Half the area is:
Half Area = 96 ÷ 2 = 48 square units.
For the inner rectangle to have this area, one possible set of dimensions is:
Length = 8 units, Width = 6 units.
Verification:
Area of inner rectangle = 8 × 6 = 48 square units, which is exactly half of 96.
Thus, the inner rectangle can be 8 units × 6 units.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
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A square piece of paper is folded in half. The square is then cut into two rectangles along the fold. Regardless of the size of the square, one of the following statements is always true. Which statement is true here? a. The area of each rectangle is larger than the area of the square. b. The perimeter of the square is greater than the perimeters of both the rectangles added together. c. The perimeters of both the rectangles added together is always 1 1/2 times the perimeter of the square. d. The area of the square is always three times as large as the areas of both rectangles added together.
The square's perimeter is 4 × side. After folding, two rectangles are formed with lengths equal to half the square's side. The total perimeter of both rectangles is 6 × side, which is 1.5 times the square's perimeter. Thus, the correct option is (c): The perimeters of both rectangles added togetherRead more
The square’s perimeter is 4 × side. After folding, two rectangles are formed with lengths equal to half the square’s side. The total perimeter of both rectangles is 6 × side, which is 1.5 times the square’s perimeter.
Thus, the correct option is (c): The perimeters of both rectangles added together are always 1.5 times the perimeter of the square.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
Explore and figure out how many pieces have the same area.
Among the tangram pieces, Shapes C and E have the same area because they are identical in size and shape. Similarly, Shapes A and B share equal areas, as they are congruent triangles. Shapes F and G are also equal in area. However, Shape D is larger and cannot be paired with another shape of the samRead more
Among the tangram pieces, Shapes C and E have the same area because they are identical in size and shape. Similarly, Shapes A and B share equal areas, as they are congruent triangles. Shapes F and G are also equal in area. However, Shape D is larger and cannot be paired with another shape of the same area. This comparison is evident by overlaying the shapes on each other.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
How many times bigger is Shape D as compared to Shape C? What is the relationship between Shapes C, D, and E?
Shape D is twice as large as Shape C. By placing the pieces together, we see that Shape D can be entirely formed by combining Shapes C and E. This indicates that Shapes C and E are equal in area, and their combined areas equal that of Shape D. Thus, Shape D represents the total area of two smaller iRead more
Shape D is twice as large as Shape C. By placing the pieces together, we see that Shape D can be entirely formed by combining Shapes C and E. This indicates that Shapes C and E are equal in area, and their combined areas equal that of Shape D. Thus, Shape D represents the total area of two smaller identical pieces, making its size double that of a single Shape C or E.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
Which shape has more area: Shape D or F? Give reasons for your answer.
Shape D is larger than Shape F in terms of area. Shape D is made up of the combined areas of Shapes C and E, making it twice the size of Shape C. On the other hand, Shape F has an area equal to Shape C. Therefore, the area of Shape D is two times that of Shape F, as it is composed of more identicalRead more
Shape D is larger than Shape F in terms of area. Shape D is made up of the combined areas of Shapes C and E, making it twice the size of Shape C. On the other hand, Shape F has an area equal to Shape C. Therefore, the area of Shape D is two times that of Shape F, as it is composed of more identical smaller sections.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/