Share & grow the world's knowledge!
We want to connect the people who have knowledge to the people who need it, to bring together people with different perspectives so they can understand each other better, and to empower everyone to share their knowledge.
Explain how arranging objects differently can change the narrative of a still life artwork.
In still life art, the arrangement of objects plays a vital role in shaping the narrative and emotional tone. Carefully placed objects can convey harmony, while scattered or tilted elements suggest disorder or dynamism. Choosing contrasting textures or varying sizes adds depth and complexity. The poRead more
In still life art, the arrangement of objects plays a vital role in shaping the narrative and emotional tone. Carefully placed objects can convey harmony, while scattered or tilted elements suggest disorder or dynamism. Choosing contrasting textures or varying sizes adds depth and complexity. The positioning of objects, their proximity and their relationship to light and space all contribute to the mood and meaning. A simple change in arrangement can shift the viewer’s interpretation of the entire artwork.
See lessIf 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/
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/