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.
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.
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.
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)
Given equal sides a = 15 cm, b = 15 cm and perimeter = 40 cm. Third side (base) c = 40 - (15 + 15) = 40 - 30 = 10 cm. Semi-perimeter s = 40 / 2 = 20 cm. Using Heron's formula: Area = √(s x (s - a) x (s - b) x (s - c)) Area = √(20 x (20 - 15) x (20 - 15) x (20 - 10)) Area = √(20 x 5 x 5 x 10) Area =Read more
Given equal sides a = 15 cm, b = 15 cm and perimeter = 40 cm.
Third side (base) c = 40 – (15 + 15) = 40 – 30 = 10 cm.
Semi-perimeter s = 40 / 2 = 20 cm.
Using Heron’s formula:
Area = √(s x (s – a) x (s – b) x (s – c))
Area = √(20 x (20 – 15) x (20 – 15) x (20 – 10))
Area = √(20 x 5 x 5 x 10)
Area = √(5000) = √(2500 x 2) = 50√2 cm².
In decimal form, 50 x 1.414 is approximately 70.71 cm².
Hence, the area of the triangle is 50√2 cm² (or 70.71 cm²).
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
An 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 lessAn isosceles triangle has perimeter 40 cm; the equal sides are 15 cm each. Find the area of the triangle.
Given equal sides a = 15 cm, b = 15 cm and perimeter = 40 cm. Third side (base) c = 40 - (15 + 15) = 40 - 30 = 10 cm. Semi-perimeter s = 40 / 2 = 20 cm. Using Heron's formula: Area = √(s x (s - a) x (s - b) x (s - c)) Area = √(20 x (20 - 15) x (20 - 15) x (20 - 10)) Area = √(20 x 5 x 5 x 10) Area =Read more
Given equal sides a = 15 cm, b = 15 cm and perimeter = 40 cm.
Third side (base) c = 40 – (15 + 15) = 40 – 30 = 10 cm.
Semi-perimeter s = 40 / 2 = 20 cm.
Using Heron’s formula:
Area = √(s x (s – a) x (s – b) x (s – c))
Area = √(20 x (20 – 15) x (20 – 15) x (20 – 10))
Area = √(20 x 5 x 5 x 10)
Area = √(5000) = √(2500 x 2) = 50√2 cm².
In decimal form, 50 x 1.414 is approximately 70.71 cm².
Hence, the area of the triangle is 50√2 cm² (or 70.71 cm²).
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