To find the area of triangle ABE, divide it into two smaller triangles, AEF and BEF. Each of these smaller triangles is half the area of their respective rectangles (AFED and BFEC). By adding these two areas, you get the total area of triangle ABE. Since these smaller triangles together represent haRead more
To find the area of triangle ABE, divide it into two smaller triangles, AEF and BEF. Each of these smaller triangles is half the area of their respective rectangles (AFED and BFEC). By adding these two areas, you get the total area of triangle ABE. Since these smaller triangles together represent half of the rectangle ABCD, the total area of triangle ABE equals half the area of rectangle ABCD. This calculation is verified using grid paper.
The conclusion is that the diagonal of a rectangle divides it into two triangles of equal area, with each triangle's area being half the total area of the rectangle. When additional triangles are formed by further divisions within the rectangle, their areas can also be calculated as fractions of smaRead more
The conclusion is that the diagonal of a rectangle divides it into two triangles of equal area, with each triangle’s area being half the total area of the rectangle. When additional triangles are formed by further divisions within the rectangle, their areas can also be calculated as fractions of smaller rectangles. This relationship between triangles and rectangles remains consistent regardless of the rectangle’s size or the dimensions of the divided sections.
To calculate the area of each figure, divide the irregular shapes into smaller rectangles and triangles. Use the formula for the area of a rectangle (length multiplied by width) for the rectangular sections. For the triangular sections, apply the formula (base multiplied by height divided by 2). SumRead more
To calculate the area of each figure, divide the irregular shapes into smaller rectangles and triangles. Use the formula for the area of a rectangle (length multiplied by width) for the rectangular sections. For the triangular sections, apply the formula (base multiplied by height divided by 2). Sum the areas of all the sections to determine the total area of the figure. Using grid paper can simplify this process by counting squares and dividing accordingly.
Calculate the areas of both rectangles: Area 1 = 5 × 10 = 50 square meters Area 2 = 2 × 7 = 14 square meters Add them together to get the total area: Total Area = 50 + 14 = 64 square meters. A square with dimensions 8 meters × 8 meters satisfies this, as: Area = 8 × 8 = 64 square meters. Thus, the dRead more
Calculate the areas of both rectangles:
Area 1 = 5 × 10 = 50 square meters
Area 2 = 2 × 7 = 14 square meters
Add them together to get the total area:
Total Area = 50 + 14 = 64 square meters.
A square with dimensions 8 meters × 8 meters satisfies this, as:
Area = 8 × 8 = 64 square meters.
Thus, the dimensions of the required rectangle are 8 m × 8 m.
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.
Use your understanding from previous grades to calculate the area of any closed figure using grid paper and— 2. Find the area of red triangle ABE.
To find the area of triangle ABE, divide it into two smaller triangles, AEF and BEF. Each of these smaller triangles is half the area of their respective rectangles (AFED and BFEC). By adding these two areas, you get the total area of triangle ABE. Since these smaller triangles together represent haRead more
To find the area of triangle ABE, divide it into two smaller triangles, AEF and BEF. Each of these smaller triangles is half the area of their respective rectangles (AFED and BFEC). By adding these two areas, you get the total area of triangle ABE. Since these smaller triangles together represent half of the rectangle ABCD, the total area of triangle ABE equals half the area of rectangle ABCD. This calculation is verified using grid paper.
For more NCERT Solutions for Class 6 Math Chapter 6 Perimeter and Area Extra Questions and Answer:
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What is the conclusion based on the relationship between areas of triangles and rectangles?
The conclusion is that the diagonal of a rectangle divides it into two triangles of equal area, with each triangle's area being half the total area of the rectangle. When additional triangles are formed by further divisions within the rectangle, their areas can also be calculated as fractions of smaRead more
The conclusion is that the diagonal of a rectangle divides it into two triangles of equal area, with each triangle’s area being half the total area of the rectangle. When additional triangles are formed by further divisions within the rectangle, their areas can also be calculated as fractions of smaller rectangles. This relationship between triangles and rectangles remains consistent regardless of the rectangle’s size or the dimensions of the divided 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/
Find the areas of the figures below by dividing them into rectangles and triangles.
To calculate the area of each figure, divide the irregular shapes into smaller rectangles and triangles. Use the formula for the area of a rectangle (length multiplied by width) for the rectangular sections. For the triangular sections, apply the formula (base multiplied by height divided by 2). SumRead more
To calculate the area of each figure, divide the irregular shapes into smaller rectangles and triangles. Use the formula for the area of a rectangle (length multiplied by width) for the rectangular sections. For the triangular sections, apply the formula (base multiplied by height divided by 2). Sum the areas of all the sections to determine the total area of the figure. Using grid paper can simplify this process by counting squares and dividing accordingly.
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/
Give the dimensions of a rectangle whose area is the sum of the areas of these two rectangles having measurements: 5 m × 10 m and 2 m × 7 m.
Calculate the areas of both rectangles: Area 1 = 5 × 10 = 50 square meters Area 2 = 2 × 7 = 14 square meters Add them together to get the total area: Total Area = 50 + 14 = 64 square meters. A square with dimensions 8 meters × 8 meters satisfies this, as: Area = 8 × 8 = 64 square meters. Thus, the dRead more
Calculate the areas of both rectangles:
Area 1 = 5 × 10 = 50 square meters
Area 2 = 2 × 7 = 14 square meters
Add them together to get the total area:
Total Area = 50 + 14 = 64 square meters.
A square with dimensions 8 meters × 8 meters satisfies this, as:
Area = 8 × 8 = 64 square meters.
Thus, the dimensions of the required rectangle are 8 m × 8 m.
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/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/
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:
See lesshttps://www.tiwariacademy.com/ncert-solutions-class-6-maths-ganita-prakash-chapter-6/