Let the legs of the right-angled triangle C be a and b, and the hypotenuse be c. Area of semicircle on leg a = (1/2) x π x (a/2)² = (1/8)πa². Area of semicircle on leg b = (1/2) x π x (b/2)² = (1/8)πb². Area of semicircle on hypotenuse c = (1/8)πc². By Pythagoras theorem, a² + b² = c², so: Area(SemiRead more
Let the legs of the right-angled triangle C be a and b, and the hypotenuse be c.
Area of semicircle on leg a = (1/2) x π x (a/2)² = (1/8)πa².
Area of semicircle on leg b = (1/2) x π x (b/2)² = (1/8)πb².
Area of semicircle on hypotenuse c = (1/8)πc².
By Pythagoras theorem, a² + b² = c², so:
Area(Semicircle a) + Area(Semicircle b) = Area(Semicircle c).
The semicircle on hypotenuse c consists of triangle C plus two circular segments lying outside the legs.
The two lunes A and B are formed by subtracting these same two circular segments from the two smaller semicircles.
Therefore:
Area(A) + Area(B) = Area(Semicircle a) + Area(Semicircle b) – Segments
= Area(Semicircle c) – Segments = Area(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 radius of the outer circle be R and the inner circle be r. The green region is the circular ring (annulus) between them: Area = piR² - pir² = pi(R² - r²). Since the chord BC touches the inner circle at A, OA is a radius to the point of contact, so OA is perpendicular to BC. The perpendicularRead more
Let the radius of the outer circle be R and the inner circle be r.
The green region is the circular ring (annulus) between them:
Area = piR² – pir² = pi(R² – r²).
Since the chord BC touches the inner circle at A, OA is a radius to the point of contact, so OA is perpendicular to BC.
The perpendicular from the centre to a chord bisects the chord, so:
AB = AC = l / 2.
In right-angled triangle OAB:
OA² + AB² = OB²
r² + (l / 2)² = R²
R² – r² = (l / 2)² = l² / 4.
Substituting this into the area expression:
Area = pi(l² / 4) = (1/4)pi l².
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
The square has side s = 2 units, so its area is 2² = 4 units². Each semicircle is drawn on a side of length 2 as diameter, giving radius r = 1 unit. Perimeter of flower: The boundary of the 4 petals is made up of 4 semicircular arcs of radius 1. Total perimeter = 4 x (pi x r) = 4 x pi x 1 = 4pi unitRead more
The square has side s = 2 units, so its area is 2² = 4 units².
Each semicircle is drawn on a side of length 2 as diameter, giving radius r = 1 unit.
Perimeter of flower:
The boundary of the 4 petals is made up of 4 semicircular arcs of radius 1.
Total perimeter = 4 x (pi x r) = 4 x pi x 1 = 4pi units (approximately 12.57 units).
Area of flower:
The four semicircles overlap to create the 4 petals.
Sum of areas of 4 semicircles = 4 x (1/2 x pi x 1²) = 2pi units².
Area of flower = Sum of 4 semicircles – Area of square = 2pi – 4 units² (approximately 2.28 units²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let the side of the square be s. The quarter circle has radius s and centre at the top-left vertex, giving: Area(quarter circle) = (1/4) x π x s² = (π/4)s². Each semicircle is constructed on a side of length s as diameter, so each has radius r = s/2: Area of one semicircle = (1/2) x π x (s/2)² = (1/Read more
Let the side of the square be s.
The quarter circle has radius s and centre at the top-left vertex, giving:
Area(quarter circle) = (1/4) x π x s² = (π/4)s².
Each semicircle is constructed on a side of length s as diameter, so each has radius r = s/2:
Area of one semicircle = (1/2) x π x (s/2)² = (1/8)πs².
The sum of the areas of the two semicircles = (1/8)πs² + (1/8)πs² = (π/4)s².
Notice that the sum of the areas of the two semicircles equals the area of the quarter circle.
Let C be the unshaded region within the quarter circle that is covered by the two semicircles.
Area of two semicircles = Area(A) + Area(C).
Area of quarter circle = Area(B) + Area(C).
Since both total areas are (π/4)s²:
Area(A) + Area(C) = Area(B) + Area(C).
Subtracting Area(C) from both sides gives Area(A) = Area(B).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
Let the top vertex be V and the opposite base be divided into three equal segments of length x by trisection points. Both the blue triangle and the red triangle stand on a base of length x. Because they share the common apex V, their perpendicular height h from V to the base line is exactly the sameRead more
Let the top vertex be V and the opposite base be divided into three equal segments of length x by trisection points.
Both the blue triangle and the red triangle stand on a base of length x.
Because they share the common apex V, their perpendicular height h from V to the base line is exactly the same.
Area of blue triangle = (1/2) x x x h = (1/2)xh.
Area of red triangle = (1/2) x x x h = (1/2)xh.
Hence, their areas are equal.
By the Wallace-Bolyai-Gerwien theorem, two polygons of equal area can always be dissected into finitely many polygonal pieces using straight cuts and rearranged into each other. Dropping perpendicular cuts from corresponding vertices converts both into congruent sets of right-angled triangles and trapezia that match upon rearrangement.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
In Fig. 6.52, semicircles have been drawn on all the sides of a right-angled triangle as shown. Show that Area (A) + Area (B) = Area (C).
Let the legs of the right-angled triangle C be a and b, and the hypotenuse be c. Area of semicircle on leg a = (1/2) x π x (a/2)² = (1/8)πa². Area of semicircle on leg b = (1/2) x π x (b/2)² = (1/8)πb². Area of semicircle on hypotenuse c = (1/8)πc². By Pythagoras theorem, a² + b² = c², so: Area(SemiRead more
Let the legs of the right-angled triangle C be a and b, and the hypotenuse be c.
Area of semicircle on leg a = (1/2) x π x (a/2)² = (1/8)πa².
Area of semicircle on leg b = (1/2) x π x (b/2)² = (1/8)πb².
Area of semicircle on hypotenuse c = (1/8)πc².
By Pythagoras theorem, a² + b² = c², so:
Area(Semicircle a) + Area(Semicircle b) = Area(Semicircle c).
The semicircle on hypotenuse c consists of triangle C plus two circular segments lying outside the legs.
The two lunes A and B are formed by subtracting these same two circular segments from the two smaller semicircles.
Therefore:
Area(A) + Area(B) = Area(Semicircle a) + Area(Semicircle b) – Segments
= Area(Semicircle c) – Segments = Area(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 lessIn Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is (1/4)pi l².
Let the radius of the outer circle be R and the inner circle be r. The green region is the circular ring (annulus) between them: Area = piR² - pir² = pi(R² - r²). Since the chord BC touches the inner circle at A, OA is a radius to the point of contact, so OA is perpendicular to BC. The perpendicularRead more
Let the radius of the outer circle be R and the inner circle be r.
The green region is the circular ring (annulus) between them:
Area = piR² – pir² = pi(R² – r²).
Since the chord BC touches the inner circle at A, OA is a radius to the point of contact, so OA is perpendicular to BC.
The perpendicular from the centre to a chord bisects the chord, so:
AB = AC = l / 2.
In right-angled triangle OAB:
OA² + AB² = OB²
r² + (l / 2)² = R²
R² – r² = (l / 2)² = l² / 4.
Substituting this into the area expression:
Area = pi(l² / 4) = (1/4)pi l².
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. 6.50, four semicircles have been drawn within the given square whose side is 2 units. The centres of these semicircles are the midpoints of the sides. They create a 4-petalled flower (shown in blue). Find the perimeter and the area of this flower.
The square has side s = 2 units, so its area is 2² = 4 units². Each semicircle is drawn on a side of length 2 as diameter, giving radius r = 1 unit. Perimeter of flower: The boundary of the 4 petals is made up of 4 semicircular arcs of radius 1. Total perimeter = 4 x (pi x r) = 4 x pi x 1 = 4pi unitRead more
The square has side s = 2 units, so its area is 2² = 4 units².
Each semicircle is drawn on a side of length 2 as diameter, giving radius r = 1 unit.
Perimeter of flower:
The boundary of the 4 petals is made up of 4 semicircular arcs of radius 1.
Total perimeter = 4 x (pi x r) = 4 x pi x 1 = 4pi units (approximately 12.57 units).
Area of flower:
The four semicircles overlap to create the 4 petals.
Sum of areas of 4 semicircles = 4 x (1/2 x pi x 1²) = 2pi units².
Area of flower = Sum of 4 semicircles – Area of square = 2pi – 4 units² (approximately 2.28 units²).
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 lessThe figure shows a quarter circle in a square. Its centre is at one vertex, and it passes through two adjacent vertices. There are two semicircles on two adjacent sides as diameters. They create the shaded regions A and B. Show that A and B have equal area.
Let the side of the square be s. The quarter circle has radius s and centre at the top-left vertex, giving: Area(quarter circle) = (1/4) x π x s² = (π/4)s². Each semicircle is constructed on a side of length s as diameter, so each has radius r = s/2: Area of one semicircle = (1/2) x π x (s/2)² = (1/Read more
Let the side of the square be s.
The quarter circle has radius s and centre at the top-left vertex, giving:
Area(quarter circle) = (1/4) x π x s² = (π/4)s².
Each semicircle is constructed on a side of length s as diameter, so each has radius r = s/2:
Area of one semicircle = (1/2) x π x (s/2)² = (1/8)πs².
The sum of the areas of the two semicircles = (1/8)πs² + (1/8)πs² = (π/4)s².
Notice that the sum of the areas of the two semicircles equals the area of the quarter circle.
Let C be the unshaded region within the quarter circle that is covered by the two semicircles.
Area of two semicircles = Area(A) + Area(C).
Area of quarter circle = Area(B) + Area(C).
Since both total areas are (π/4)s²:
Area(A) + Area(C) = Area(B) + Area(C).
Subtracting Area(C) from both sides gives Area(A) = Area(B).
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 lessShow that the areas of the shaded blue triangle and the shaded red triangle are equal. Find a way of cutting up the blue triangle into some number of pieces and rearranging the pieces to cover the red triangle.
Let the top vertex be V and the opposite base be divided into three equal segments of length x by trisection points. Both the blue triangle and the red triangle stand on a base of length x. Because they share the common apex V, their perpendicular height h from V to the base line is exactly the sameRead more
Let the top vertex be V and the opposite base be divided into three equal segments of length x by trisection points.
Both the blue triangle and the red triangle stand on a base of length x.
Because they share the common apex V, their perpendicular height h from V to the base line is exactly the same.
Area of blue triangle = (1/2) x x x h = (1/2)xh.
Area of red triangle = (1/2) x x x h = (1/2)xh.
Hence, their areas are equal.
By the Wallace-Bolyai-Gerwien theorem, two polygons of equal area can always be dissected into finitely many polygonal pieces using straight cuts and rearranged into each other. Dropping perpendicular cuts from corresponding vertices converts both into congruent sets of right-angled triangles and trapezia that match upon rearrangement.
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