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Do you think Shruti and her parents would understand one another? If yes, why? If no, why not?
Shruti and her parents reach a fulfilling resolution through genuine understanding and shared values. Shruti proves that modern fusion does not corrupt classical purity by preserving authentic raga notes during performance. Guided by Leela's historical context, Nabin recognizes his daughter's uniqueRead more
Shruti and her parents reach a fulfilling resolution through genuine understanding and shared values. Shruti proves that modern fusion does not corrupt classical purity by preserving authentic raga notes during performance. Guided by Leela’s historical context, Nabin recognizes his daughter’s unique artistic path, abandoning his rigid assumptions and offering his full parental trust and practical musical support.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 Twin Melodies Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/english/kaveri-chapter-6/
See lessAn equilateral triangle is inscribed in a circle of radius r. Show that the ratio of the area of the triangle to the area of the circle is equal to (3√3) / (4π) ≈ 0.413.
Let an equilateral triangle of side a be inscribed in a circle of radius r. The circumradius of an equilateral triangle is related to its side by: r = a / √3, so a = r√3. Area of the equilateral triangle: Area(triangle) = (√3 / 4) x a² = (√3 / 4) x (r√3)² = (3√3 / 4) x r². Area of the circumscribingRead more
Let an equilateral triangle of side a be inscribed in a circle of radius r.
The circumradius of an equilateral triangle is related to its side by:
r = a / √3, so a = r√3.
Area of the equilateral triangle:
Area(triangle) = (√3 / 4) x a² = (√3 / 4) x (r√3)² = (3√3 / 4) x r².
Area of the circumscribing circle:
Area(circle) = π x r².
Ratio of areas:
Ratio = Area(triangle) / Area(circle) = [(3√3 / 4)r²] / [πr²] = (3√3) / (4π).
Evaluating numerically:
(3 x 1.732) / (4 x 3.1416) = 5.196 / 12.566 ≈ 0.413.
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessA square is inscribed in a circle of radius r. Show that the ratio of the area of the square to the area of the circle is equal to 2 / π ≈ 0.637.
When a square is inscribed in a circle of radius r, the diagonal of the square passes through the centre and equals the diameter of the circle: Diagonal d = 2r. Area of the square in terms of its diagonal: Area(square) = (1/2) x d² = (1/2) x (2r)² = (1/2) x 4r² = 2r². Area of the circle: Area(circleRead more
When a square is inscribed in a circle of radius r, the diagonal of the square passes through the centre and equals the diameter of the circle:
Diagonal d = 2r.
Area of the square in terms of its diagonal:
Area(square) = (1/2) x d² = (1/2) x (2r)² = (1/2) x 4r² = 2r².
Area of the circle:
Area(circle) = π x r².
Ratio of the area of the square to the circle:
Ratio = Area(square) / Area(circle) = (2r²) / (πr²) = 2 / π.
Evaluating numerically:
2 / 3.1416 ≈ 0.6366 ≈ 0.637.
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessA hexagon is inscribed in a circle of radius r. Show that the ratio of the area of the hexagon to the area of the circle is equal to (3√3) / (2π) ≈ 0.827. Can you see why the answer is exactly twice the answer to Question 8?
A regular inscribed hexagon has side equal to the circle's radius r and consists of 6 equilateral triangles of side r: Area(hexagon) = 6 x [(√3 / 4) x r²] = (3√3 / 2) x r². Area of the circle = πr². Ratio = Area(hexagon) / Area(circle) = [(3√3 / 2)r²] / [πr²] = (3√3) / (2π) ≈ 0.827. Why it is twiceRead more
A regular inscribed hexagon has side equal to the circle’s radius r and consists of 6 equilateral triangles of side r:
Area(hexagon) = 6 x [(√3 / 4) x r²] = (3√3 / 2) x r².
Area of the circle = πr².
Ratio = Area(hexagon) / Area(circle) = [(3√3 / 2)r²] / [πr²] = (3√3) / (2π) ≈ 0.827.
Why it is twice the answer to Question 8:
Connecting alternating vertices of the hexagon forms an inscribed equilateral triangle whose area is exactly half of the hexagon’s area:
(3√3 / 2π) = 2 x [(3√3) / (4π)].
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See lessIdentities in algebra can sometimes be shown as area relationships. The figure shown corresponds to the identity (a + b)² = a² + 2ab + b². Do you see how? Draw figures corresponding to the identities (a + b)(a – b) = a² – b² and (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca.
In Fig. 6.41, side length (a + b) forms a square split into regions a², ab, ab and b², summing to a² + 2ab + b². For (a + b)(a - b) = a² - b²: Start with a square of side a (area a²). Cut out a corner square of side b (area b²). The remaining L-shaped region splits into two rectangles of dimensionsRead more
In Fig. 6.41, side length (a + b) forms a square split into regions a², ab, ab and b², summing to a² + 2ab + b².
For (a + b)(a – b) = a² – b²: Start with a square of side a (area a²). Cut out a corner square of side b (area b²). The remaining L-shaped region splits into two rectangles of dimensions (a – b) by a and (a – b) by b, which combine into one rectangle of sides (a + b) and (a – b).
For (a + b + c)²: Partition a square of side (a + b + c) into 9 sub-rectangles: three squares a², b², c² and six rectangular regions giving 2ab, 2bc, 2ca.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/
See less