Given: - Ratio of outer to inner perimeter = 23:22, - Width of path = 5 m. From ratio: R/r = 23/22 ⇒ R = (23/22)r. Width: R - r = 5 ⇒ (23/22)r - r = 5. Simplify: (1/22)r = 5 ⇒ r = 5 × 22 = 110 m. Diameter of inner circle = 2r = 2 × 110 = 220 m. This question related to Chapter 11 Mathematics Class 1Read more
Given:
– Ratio of outer to inner perimeter = 23:22,
– Width of path = 5 m.
From ratio: R/r = 23/22 ⇒ R = (23/22)r.
Width: R – r = 5 ⇒ (23/22)r – r = 5.
Simplify:
(1/22)r = 5 ⇒ r = 5 × 22 = 110 m.
Diameter of inner circle = 2r = 2 × 110 = 220 m.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
Given: - Ratio of outer to inner perimeter = 23:22, - Width of path = 5 m. From ratio: R/r = 23/22 ⇒ R = (23/22)r. Width: R - r = 5 ⇒ (23/22)r - r = 5. Simplify: (1/22)r = 5 ⇒ r = 5 × 22 = 110 m. Diameter of inner circle = 2r = 2 × 110 = 220 m. This question related to Chapter 11 Mathematics Class 1Read more
Given:
– Ratio of outer to inner perimeter = 23:22,
– Width of path = 5 m.
From ratio: R/r = 23/22 ⇒ R = (23/22)r.
Width: R – r = 5 ⇒ (23/22)r – r = 5.
Simplify:
(1/22)r = 5 ⇒ r = 5 × 22 = 110 m.
Diameter of inner circle = 2r = 2 × 110 = 220 m.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
Given: - Ratio of outer to inner perimeter = 23:22, - Width of path = 5 m. From ratio: R/r = 23/22 ⇒ R = (23/22)r. Width: R - r = 5 ⇒ (23/22)r - r = 5. Simplify: (1/22)r = 5 ⇒ r = 5 × 22 = 110 m. Diameter of inner circle = 2r = 2 × 110 = 220 m. This question related to Chapter 11 Mathematics Class 1Read more
Given:
– Ratio of outer to inner perimeter = 23:22,
– Width of path = 5 m.
From ratio: R/r = 23/22 ⇒ R = (23/22)r.
Width: R – r = 5 ⇒ (23/22)r – r = 5.
Simplify:
(1/22)r = 5 ⇒ r = 5 × 22 = 110 m.
Diameter of inner circle = 2r = 2 × 110 = 220 m.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
Given: - Radius of circle = 56 cm. - Circumference of circle = 2πr = 2 × (22/7) × 56 = 352 cm. If bent into a square: - Perimeter of square = 352 cm. - Side of square = 352 / 4 = 88 cm. - Area of square = 88² = 7744 cm². This question is related to Chapter 11 of the Class 10th NCERT Mathematics textRead more
Given:
– Radius of circle = 56 cm.
– Circumference of circle = 2πr = 2 × (22/7) × 56 = 352 cm.
If bent into a square:
– Perimeter of square = 352 cm.
– Side of square = 352 / 4 = 88 cm.
– Area of square = 88² = 7744 cm².
This question is related to Chapter 11 of the Class 10th NCERT Mathematics textbook, which covers “Areas Related to Circles.” Provide your answer based on your understanding of the concepts discussed in this chapter.
Area of semi-circle:
– Area = (1/2)πr² = (1/2)(22/7)(7²) = (1/2)(22)(7) = 77 cm².
This question pertains to Chapter 11 of the Class 10th NCERT Mathematics textbook, titled “Areas Related to Circles.” Answer the question based on your comprehension of the chapter.
The ratio of the outer and inner perimeters of a circular path is 23:22. If the path is 5 metres wide, the diameter of the inner circle
Given: - Ratio of outer to inner perimeter = 23:22, - Width of path = 5 m. From ratio: R/r = 23/22 ⇒ R = (23/22)r. Width: R - r = 5 ⇒ (23/22)r - r = 5. Simplify: (1/22)r = 5 ⇒ r = 5 × 22 = 110 m. Diameter of inner circle = 2r = 2 × 110 = 220 m. This question related to Chapter 11 Mathematics Class 1Read more
Given:
– Ratio of outer to inner perimeter = 23:22,
– Width of path = 5 m.
From ratio: R/r = 23/22 ⇒ R = (23/22)r.
Width: R – r = 5 ⇒ (23/22)r – r = 5.
Simplify:
(1/22)r = 5 ⇒ r = 5 × 22 = 110 m.
Diameter of inner circle = 2r = 2 × 110 = 220 m.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The ratio of the outer and inner perimeters of a circular path is 23:22. If the path is 5 metres wide, the diameter of the inner circle
Given: - Ratio of outer to inner perimeter = 23:22, - Width of path = 5 m. From ratio: R/r = 23/22 ⇒ R = (23/22)r. Width: R - r = 5 ⇒ (23/22)r - r = 5. Simplify: (1/22)r = 5 ⇒ r = 5 × 22 = 110 m. Diameter of inner circle = 2r = 2 × 110 = 220 m. This question related to Chapter 11 Mathematics Class 1Read more
Given:
– Ratio of outer to inner perimeter = 23:22,
– Width of path = 5 m.
From ratio: R/r = 23/22 ⇒ R = (23/22)r.
Width: R – r = 5 ⇒ (23/22)r – r = 5.
Simplify:
(1/22)r = 5 ⇒ r = 5 × 22 = 110 m.
Diameter of inner circle = 2r = 2 × 110 = 220 m.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The ratio of the outer and inner perimeters of a circular path is 23:22. If the path is 5 metres wide, the diameter of the inner circle
Given: - Ratio of outer to inner perimeter = 23:22, - Width of path = 5 m. From ratio: R/r = 23/22 ⇒ R = (23/22)r. Width: R - r = 5 ⇒ (23/22)r - r = 5. Simplify: (1/22)r = 5 ⇒ r = 5 × 22 = 110 m. Diameter of inner circle = 2r = 2 × 110 = 220 m. This question related to Chapter 11 Mathematics Class 1Read more
Given:
– Ratio of outer to inner perimeter = 23:22,
– Width of path = 5 m.
From ratio: R/r = 23/22 ⇒ R = (23/22)r.
Width: R – r = 5 ⇒ (23/22)r – r = 5.
Simplify:
(1/22)r = 5 ⇒ r = 5 × 22 = 110 m.
Diameter of inner circle = 2r = 2 × 110 = 220 m.
This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 Area related to Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
A wire can be bent in the form of a circle of radius 56 cm. If it is bent in the form of a square, then its area will be
Given: - Radius of circle = 56 cm. - Circumference of circle = 2πr = 2 × (22/7) × 56 = 352 cm. If bent into a square: - Perimeter of square = 352 cm. - Side of square = 352 / 4 = 88 cm. - Area of square = 88² = 7744 cm². This question is related to Chapter 11 of the Class 10th NCERT Mathematics textRead more
Given:
– Radius of circle = 56 cm.
– Circumference of circle = 2πr = 2 × (22/7) × 56 = 352 cm.
If bent into a square:
– Perimeter of square = 352 cm.
– Side of square = 352 / 4 = 88 cm.
– Area of square = 88² = 7744 cm².
This question is related to Chapter 11 of the Class 10th NCERT Mathematics textbook, which covers “Areas Related to Circles.” Provide your answer based on your understanding of the concepts discussed in this chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If a wire is bent into the shape of a square, then the area of the square is 81 cm². When wire is bent into a semi-circular shape, then the area of the semi-circle will be
Given: - Area of square = 81 cm² ⇒ Side = √81 = 9 cm. - Perimeter of square = 4 × 9 = 36 cm (length of wire). For semi-circle: - Perimeter = πr + 2r = 36 ⇒ r(π + 2) = 36. - Substitute π = 22/7: r(22/7 + 2) = 36 ⇒ r(36/7) = 36 ⇒ r = 7 cm. Area of semi-circle: - Area = (1/2)πr² = (1/2)(22/7)(7²) = (1/Read more
Given:
– Area of square = 81 cm² ⇒ Side = √81 = 9 cm.
– Perimeter of square = 4 × 9 = 36 cm (length of wire).
For semi-circle:
– Perimeter = πr + 2r = 36 ⇒ r(π + 2) = 36.
– Substitute π = 22/7: r(22/7 + 2) = 36 ⇒ r(36/7) = 36 ⇒ r = 7 cm.
Area of semi-circle:
– Area = (1/2)πr² = (1/2)(22/7)(7²) = (1/2)(22)(7) = 77 cm².
This question pertains to Chapter 11 of the Class 10th NCERT Mathematics textbook, titled “Areas Related to Circles.” Answer the question based on your comprehension of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/