The inscribed square base has diagonal 2r and area (2r)² / 2 = 2r². Pyramid volume is (1/3)(2r²)h = (2/3)r²h. Cylinder volume is πr²h. Their volume ratio is 2 : 3π.
Cbse Class 9 Maths Ganita Manjari Part 2 Solutions
Class 9 maths ganita manjari part 2 chapter 14 question answer
The square base of the pyramid is inscribed inside the circular base of radius r, making its diagonal equal to the diameter 2r. The area of this square base is (1/2) × d² = (1/2)(2r)² = 2r². The pyramid has height h, giving volume V_pyramid = (1/3) × base area × h = (2/3)r²h. The cylinder volume is V_cylinder = πr²h. Thus, the ratio of their volumes is [(2/3)r²h] / [πr²h] = 2 / (3π) or 2 : 3π.
For more NCERT Solutions of Class 9 Maths Ganita Manjari Part 2 Chapter 14 Math of Space: Surface Area and Volume Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-14/