Given: - Difference in circumferences = 132 m. - Formula: 2π(R - r) = 132, where R - r = w (width). Substitute π = 22/7: 2 × (22/7) × w = 132. Simplify: (44/7) × w = 132. Solve for w: w = (132 × 7) / 44 = 21 m. This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 AreRead more
Given:
– Difference in circumferences = 132 m.
– Formula: 2π(R – r) = 132, where R – r = w (width).
Substitute π = 22/7:
2 × (22/7) × w = 132.
Simplify:
(44/7) × w = 132.
Solve for w:
w = (132 × 7) / 44 = 21 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 wheel = 0.25 m, - Total distance = 11 km = 11000 m. Circumference = 2πr = 2 × (22/7) × 0.25 = 11/7 m. Number of revolutions = Total distance / Circumference: = 11000 / (11/7) = 11000 × (7/11) = 7000. This question related to Chapter 11 Mathematics Class 10th NCERT. From the ChapteRead more
Given:
– Radius of wheel = 0.25 m,
– Total distance = 11 km = 11000 m.
Circumference = 2πr = 2 × (22/7) × 0.25 = 11/7 m.
Number of revolutions = Total distance / Circumference:
= 11000 / (11/7) = 11000 × (7/11) = 7000.
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: - BC = 6 cm, AB = 8 cm, AC = 10 cm (using Pythagoras theorem). Radius of incircle: r = (a + b - c) / 2, where a = 6, b = 8, c = 10. Substitute: r = (6 + 8 - 10) / 2 = 4 / 2 = 2 cm. This question is related to Chapter 10 of the Class 10th NCERT Mathematics textbook, which covers the topic of "Read more
Given:
– BC = 6 cm, AB = 8 cm, AC = 10 cm (using Pythagoras theorem).
Radius of incircle:
r = (a + b – c) / 2,
where a = 6, b = 8, c = 10.
Substitute:
r = (6 + 8 – 10) / 2 = 4 / 2 = 2 cm.
This question is related to Chapter 10 of the Class 10th NCERT Mathematics textbook, which covers the topic of “Circles.” Provide your answer based on your understanding of the chapter.
We are given: - The angle between two radii of a circle is 130°. Key property of tangents The tangents at the ends of the two radii are perpendicular to their respective radii. Therefore, the angle between the tangents is the **supplement** of the angle between the radii. Calculate the angle betweenRead more
We are given:
– The angle between two radii of a circle is 130°.
Key property of tangents
The tangents at the ends of the two radii are perpendicular to their respective radii. Therefore, the angle between the tangents is the **supplement** of the angle between the radii.
Calculate the angle between the tangents
The sum of the angle between the radii and the angle between the tangents is 180°. Thus:
Angle between tangents = 180° – Angle between radii.
Substitute the given angle between the radii:
Angle between tangents = 180° – 130° = 50°.
This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your understanding.
When two circles intersect at two distinct points, the maximum number of common tangents that can be drawn is 2. These tangents are the **external tangents**, as the circles are close enough to each other and intersect, so no internal tangents can exist. Explanation: - If two circles intersect at twRead more
When two circles intersect at two distinct points, the maximum number of common tangents that can be drawn is 2. These tangents are the **external tangents**, as the circles are close enough to each other and intersect, so no internal tangents can exist.
Explanation:
– If two circles intersect at two points, they share a region of overlap.
– Only two external tangents can be drawn, one on each side of the circles.
– No internal tangents are possible because the circles are not separate or disjoint.
This question is connected to Chapter 10 of the Class 10th NCERT Mathematics textbook, which focuses on “Circles.” Provide your response based on your comprehension of the concepts discussed in this chapter.
A circular park has a path of uniform width around it. The difference between the outer and inner circumferences of the circular path is 132m. Its width is
Given: - Difference in circumferences = 132 m. - Formula: 2π(R - r) = 132, where R - r = w (width). Substitute π = 22/7: 2 × (22/7) × w = 132. Simplify: (44/7) × w = 132. Solve for w: w = (132 × 7) / 44 = 21 m. This question related to Chapter 11 Mathematics Class 10th NCERT. From the Chapter 11 AreRead more
Given:
– Difference in circumferences = 132 m.
– Formula: 2π(R – r) = 132, where R – r = w (width).
Substitute π = 22/7:
2 × (22/7) × w = 132.
Simplify:
(44/7) × w = 132.
Solve for w:
w = (132 × 7) / 44 = 21 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 radius of a wheel is 0.25m. The number of revolution it will make to travel a distance of 11km will be
Given: - Radius of wheel = 0.25 m, - Total distance = 11 km = 11000 m. Circumference = 2πr = 2 × (22/7) × 0.25 = 11/7 m. Number of revolutions = Total distance / Circumference: = 11000 / (11/7) = 11000 × (7/11) = 7000. This question related to Chapter 11 Mathematics Class 10th NCERT. From the ChapteRead more
Given:
– Radius of wheel = 0.25 m,
– Total distance = 11 km = 11000 m.
Circumference = 2πr = 2 × (22/7) × 0.25 = 11/7 m.
Number of revolutions = Total distance / Circumference:
= 11000 / (11/7) = 11000 × (7/11) = 7000.
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/
ABC is a right angled triangle, right angled at B such that BC = 6cm and AB = 8cm. A circle with centre O is inscribed in Triangle ABC. The radius of the circle is
Given: - BC = 6 cm, AB = 8 cm, AC = 10 cm (using Pythagoras theorem). Radius of incircle: r = (a + b - c) / 2, where a = 6, b = 8, c = 10. Substitute: r = (6 + 8 - 10) / 2 = 4 / 2 = 2 cm. This question is related to Chapter 10 of the Class 10th NCERT Mathematics textbook, which covers the topic of "Read more
Given:
– BC = 6 cm, AB = 8 cm, AC = 10 cm (using Pythagoras theorem).
Radius of incircle:
r = (a + b – c) / 2,
where a = 6, b = 8, c = 10.
Substitute:
r = (6 + 8 – 10) / 2 = 4 / 2 = 2 cm.
This question is related to Chapter 10 of the Class 10th NCERT Mathematics textbook, which covers the topic of “Circles.” Provide your answer based on your understanding of the chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
If angle between two radii of a circle is 130°, the angle between the tangents at the ends of radii is
We are given: - The angle between two radii of a circle is 130°. Key property of tangents The tangents at the ends of the two radii are perpendicular to their respective radii. Therefore, the angle between the tangents is the **supplement** of the angle between the radii. Calculate the angle betweenRead more
We are given:
– The angle between two radii of a circle is 130°.
Key property of tangents
The tangents at the ends of the two radii are perpendicular to their respective radii. Therefore, the angle between the tangents is the **supplement** of the angle between the radii.
Calculate the angle between the tangents
The sum of the angle between the radii and the angle between the tangents is 180°. Thus:
Angle between tangents = 180° – Angle between radii.
Substitute the given angle between the radii:
Angle between tangents = 180° – 130° = 50°.
This question related to Chapter 10 Mathematics Class 10th NCERT. From the Chapter 10 Circle. Give answer according to your understanding.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/
The maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is
When two circles intersect at two distinct points, the maximum number of common tangents that can be drawn is 2. These tangents are the **external tangents**, as the circles are close enough to each other and intersect, so no internal tangents can exist. Explanation: - If two circles intersect at twRead more
When two circles intersect at two distinct points, the maximum number of common tangents that can be drawn is 2. These tangents are the **external tangents**, as the circles are close enough to each other and intersect, so no internal tangents can exist.
Explanation:
– If two circles intersect at two points, they share a region of overlap.
– Only two external tangents can be drawn, one on each side of the circles.
– No internal tangents are possible because the circles are not separate or disjoint.
This question is connected to Chapter 10 of the Class 10th NCERT Mathematics textbook, which focuses on “Circles.” Provide your response based on your comprehension of the concepts discussed in this chapter.
For more please visit here:
See lesshttps://www.tiwariacademy.in/ncert-solutions/class-10/maths/