Pandit Hariprasad Chaurasia is renowned for his mastery of the classical Indian bamboo flute, while A. R. Rahman is celebrated globally for pioneering modern orchestral arrangements and cross-cultural fusion compositions. I chose these artists because their creations embody the transformative potentRead more
Pandit Hariprasad Chaurasia is renowned for his mastery of the classical Indian bamboo flute, while A. R. Rahman is celebrated globally for pioneering modern orchestral arrangements and cross-cultural fusion compositions.
I chose these artists because their creations embody the transformative potential of music. Chaurasia’s bansuri recitals carry pure, meditative serenity that eases mental stress and connects listeners with nature. Rahman’s creative genius lies in his daring synthesis of Eastern devotional warmth and Western orchestral grandeur, showing that music can remain firmly rooted in cultural heritage while innovating for contemporary global audiences.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 A Friend Found in Music Question Answer (2026-27)
I prefer instrumental music because it transcends linguistic barriers, communicating emotion directly through pitch, harmony and rhythm. Without lyrics directing specific thoughts, instruments like the flute, sitar or violin invite listeners to interpret the melody personally. This wordless clarityRead more
I prefer instrumental music because it transcends linguistic barriers, communicating emotion directly through pitch, harmony and rhythm. Without lyrics directing specific thoughts, instruments like the flute, sitar or violin invite listeners to interpret the melody personally. This wordless clarity calms an overstimulated mind, relieves anxiety and provides a restorative background for study, thoughtful introspection and emotional tranquility.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 A Friend Found in Music Question Answer (2026-27)
Let the radius of the large circle with centre O be r, so OA = OB = r. Region AOB is a quarter circle of radius r: Area(quarter circle AOB) = (1/4) x π x r² = (π/4)r². In right-angled triangle AOB, the hypotenuse is AB = √(r² + r²) = r√2. The semicircle AEB is drawn on diameter AB = r√2, so its radiRead more
Let the radius of the large circle with centre O be r, so OA = OB = r.
Region AOB is a quarter circle of radius r:
Area(quarter circle AOB) = (1/4) x π x r² = (π/4)r².
In right-angled triangle AOB, the hypotenuse is AB = √(r² + r²) = r√2.
The semicircle AEB is drawn on diameter AB = r√2, so its radius is R = (r√2) / 2 = r / √2.
Area(semicircle AEB) = (1/2) x π x R² = (1/2) x π x (r² / 2) = (π/4)r².
Notice that:
Area(semicircle AEB) = Area(quarter circle AOB) = (π/4)r².
Let the unshaded circular segment between chord AB and arc AFB be S.
The upper shaded region (lune AEBFA) = Area(semicircle AEB) – S = (π/4)r² – S.
The lower shaded region (triangle AOB) = Area(quarter circle AOB) – S = (π/4)r² – S.
Since both expressions equal (π/4)r² – S, their areas are equal.
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let the rectangle have horizontal length X and vertical height Y, so total Area = XY. From Fig. 6.54, the common vertical segment shared by triangles A, B, and C has length h. Let the horizontal distance from the left edge to this segment be x₁, and to the right edge be x₂, so X = x₁ + x₂. TrianglesRead more
Let the rectangle have horizontal length X and vertical height Y, so total Area = XY.
From Fig. 6.54, the common vertical segment shared by triangles A, B, and C has length h.
Let the horizontal distance from the left edge to this segment be x₁, and to the right edge be x₂,
so X = x₁ + x₂.
Triangles A and C together form a triangle of base h and altitude X:
A + C = (1/2) x h x X.
Triangles B and C together form a triangle on base h with altitude Y:
B + C = (1/2) x h x Y (or using horizontal base x₂ with total height Y: B + C = (1/2) x x₂ x Y).
Triangle C has base h and horizontal width x₂:
C = (1/2) x h x x₂.
Multiplying (A + C) and (B + C):
(A + C)(B + C) = [(1/2)hX] x [(1/2)x₂Y]
= (1/2) x [(1/2)hx₂] x XY = (1/2) x C x (Area of rectangle).
Rearranging:
Area of rectangle = [2(A + C)(B + C)] / C.
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let the centres be A and B with distance AB = r. The circles intersect at C and D, forming equilateral triangles ABC and ABD of side r, each having area (√3/4)r². The central angle subtended at each centre is ∠CAD = 60° + 60° = 120°. Area of 120° sector = (120/360) x πr² = (1/3)πr². Area of rhombusRead more
Let the centres be A and B with distance AB = r. The circles intersect at C and D, forming equilateral triangles ABC and ABD of side r, each having area (√3/4)r².
The central angle subtended at each centre is ∠CAD = 60° + 60° = 120°.
Area of 120° sector = (120/360) x πr² = (1/3)πr².
Area of rhombus ACBD = 2 x (√3/4)r² = (√3/2)r².
Area of intersection (shaded region) = 2 x Area(Sector) – Area(Rhombus)
= 2 x (1/3)πr² – (√3/2)r² = (2π/3 – √3/2)r².
Total area enclosed by the two circles (their union):
Name your favourite musician(s). Give reasons for your choice.
Pandit Hariprasad Chaurasia is renowned for his mastery of the classical Indian bamboo flute, while A. R. Rahman is celebrated globally for pioneering modern orchestral arrangements and cross-cultural fusion compositions. I chose these artists because their creations embody the transformative potentRead more
Pandit Hariprasad Chaurasia is renowned for his mastery of the classical Indian bamboo flute, while A. R. Rahman is celebrated globally for pioneering modern orchestral arrangements and cross-cultural fusion compositions.
I chose these artists because their creations embody the transformative potential of music. Chaurasia’s bansuri recitals carry pure, meditative serenity that eases mental stress and connects listeners with nature. Rahman’s creative genius lies in his daring synthesis of Eastern devotional warmth and Western orchestral grandeur, showing that music can remain firmly rooted in cultural heritage while innovating for contemporary global audiences.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 A Friend Found in Music Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/english/kaveri-chapter-6/
See lessWhat kind of music do you prefer to listen to-vocal or instrumental?
I prefer instrumental music because it transcends linguistic barriers, communicating emotion directly through pitch, harmony and rhythm. Without lyrics directing specific thoughts, instruments like the flute, sitar or violin invite listeners to interpret the melody personally. This wordless clarityRead more
I prefer instrumental music because it transcends linguistic barriers, communicating emotion directly through pitch, harmony and rhythm. Without lyrics directing specific thoughts, instruments like the flute, sitar or violin invite listeners to interpret the melody personally. This wordless clarity calms an overstimulated mind, relieves anxiety and provides a restorative background for study, thoughtful introspection and emotional tranquility.
For more NCERT Solutions of Class 9 English Kaveri Chapter 6 A Friend Found in Music Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/english/kaveri-chapter-6/
See lessIn the figure we see two shaded regions formed by a quarter circle, a semicircle, and a triangle. Show that the areas of the two shaded regions are equal.
Let the radius of the large circle with centre O be r, so OA = OB = r. Region AOB is a quarter circle of radius r: Area(quarter circle AOB) = (1/4) x π x r² = (π/4)r². In right-angled triangle AOB, the hypotenuse is AB = √(r² + r²) = r√2. The semicircle AEB is drawn on diameter AB = r√2, so its radiRead more
Let the radius of the large circle with centre O be r, so OA = OB = r.
Region AOB is a quarter circle of radius r:
Area(quarter circle AOB) = (1/4) x π x r² = (π/4)r².
In right-angled triangle AOB, the hypotenuse is AB = √(r² + r²) = r√2.
The semicircle AEB is drawn on diameter AB = r√2, so its radius is R = (r√2) / 2 = r / √2.
Area(semicircle AEB) = (1/2) x π x R² = (1/2) x π x (r² / 2) = (π/4)r².
Notice that:
Area(semicircle AEB) = Area(quarter circle AOB) = (π/4)r².
Let the unshaded circular segment between chord AB and arc AFB be S.
The upper shaded region (lune AEBFA) = Area(semicircle AEB) – S = (π/4)r² – S.
The lower shaded region (triangle AOB) = Area(quarter circle AOB) – S = (π/4)r² – S.
Since both expressions equal (π/4)r² – S, their areas are equal.
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 lessIn Fig. 6.54, we see three triangles within a rectangle. The areas of the triangles are A, B, C, as marked. Show that the area of the rectangle is [2(A + C)(B + C)] / C.
Let the rectangle have horizontal length X and vertical height Y, so total Area = XY. From Fig. 6.54, the common vertical segment shared by triangles A, B, and C has length h. Let the horizontal distance from the left edge to this segment be x₁, and to the right edge be x₂, so X = x₁ + x₂. TrianglesRead more
Let the rectangle have horizontal length X and vertical height Y, so total Area = XY.
From Fig. 6.54, the common vertical segment shared by triangles A, B, and C has length h.
Let the horizontal distance from the left edge to this segment be x₁, and to the right edge be x₂,
so X = x₁ + x₂.
Triangles A and C together form a triangle of base h and altitude X:
A + C = (1/2) x h x X.
Triangles B and C together form a triangle on base h with altitude Y:
B + C = (1/2) x h x Y (or using horizontal base x₂ with total height Y: B + C = (1/2) x x₂ x Y).
Triangle C has base h and horizontal width x₂:
C = (1/2) x h x x₂.
Multiplying (A + C) and (B + C):
(A + C)(B + C) = [(1/2)hX] x [(1/2)x₂Y]
= (1/2) x [(1/2)hx₂] x XY = (1/2) x C x (Area of rectangle).
Rearranging:
Area of rectangle = [2(A + C)(B + C)] / C.
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 lessFig. 6.53 shows two circles passing through each other’s centres. Find the area of the region enclosed by the two circles in terms of the common radius r.
Let the centres be A and B with distance AB = r. The circles intersect at C and D, forming equilateral triangles ABC and ABD of side r, each having area (√3/4)r². The central angle subtended at each centre is ∠CAD = 60° + 60° = 120°. Area of 120° sector = (120/360) x πr² = (1/3)πr². Area of rhombusRead more
Let the centres be A and B with distance AB = r. The circles intersect at C and D, forming equilateral triangles ABC and ABD of side r, each having area (√3/4)r².
The central angle subtended at each centre is ∠CAD = 60° + 60° = 120°.
Area of 120° sector = (120/360) x πr² = (1/3)πr².
Area of rhombus ACBD = 2 x (√3/4)r² = (√3/2)r².
Area of intersection (shaded region) = 2 x Area(Sector) – Area(Rhombus)
= 2 x (1/3)πr² – (√3/2)r² = (2π/3 – √3/2)r².
Total area enclosed by the two circles (their union):
Area(Union) = Area(Circle 1) + Area(Circle 2) – Area(Intersection)
= 2πr² – [(2π/3 – √3/2)r²]
= (4π/3 + √3/2)r².
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