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.
In 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 lessIn 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 less