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Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
Let original dimensions be l, w, h with volume V = lwh and total surface area S = 2(lw + wh + hl). If transformed to new dimensions l', w', h', we require V' = V/2 and S' = 2S. For instance, start with a 1 cm × 1 cm × 8 cm cuboid (V = 8 cm³, S = 34 cm²). Changing dimensions to 0.25 cm × 2 cm × 8 cmRead more
Let original dimensions be l, w, h with volume V = lwh and total surface area S = 2(lw + wh + hl). If transformed to new dimensions l’, w’, h’, we require V’ = V/2 and S’ = 2S. For instance, start with a 1 cm × 1 cm × 8 cm cuboid (V = 8 cm³, S = 34 cm²). Changing dimensions to 0.25 cm × 2 cm × 8 cm yields new volume V’ = 4 cm³ (halved) and new surface area S’ = 2(0.5 + 16 + 2) = 68 cm² (doubled).
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/
See lessThe Earth’s surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth’s surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth’s radius is ~6371 km. (i) Write an expression that gives the thickness of this water layer. (ii) Simplify the expression in (i) using a calculator.
(i) Since the water layer is extremely thin compared to Earth's radius, its volume equals surface area multiplied by thickness: V ≈ 4πR²t, giving thickness expression t = V/(4πR²). The exact expression is t = ∛(R³ + 3V/(4π)) − R. (ii) Earth's surface area is 4π(6371)² ≈ 5.10 × 10⁸ km². Evaluating tRead more
(i) Since the water layer is extremely thin compared to Earth’s radius, its volume equals surface area multiplied by thickness: V ≈ 4πR²t, giving thickness expression t = V/(4πR²). The exact expression is t = ∛(R³ + 3V/(4π)) − R. (ii) Earth’s surface area is 4π(6371)² ≈ 5.10 × 10⁸ km². Evaluating t = 1.38 × 10⁹/(5.10 × 10⁸) ≈ 2.705 km, meaning the uniform water layer is approximately 2.71 kilometres thick.
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/
See lessIf the entire human population decided to climb into one giant cube, how long would the side have to be?
Assuming an average human body volume of about 0.07 m³ (corresponding to an average weight around 70 kg, close to the density of water). For an estimated global population of 8 billion people, the total volume required is 8,000,000,000 × 0.07 = 560,000,000 m³. If packed into a cube of volume V = a³,Read more
Assuming an average human body volume of about 0.07 m³ (corresponding to an average weight around 70 kg, close to the density of water). For an estimated global population of 8 billion people, the total volume required is 8,000,000,000 × 0.07 = 560,000,000 m³. If packed into a cube of volume V = a³, side length is a = ∛560,000,000 ≈ 824 metres, well under 1 km.
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/
See lessLooking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders: (i) If all the pages of this textbook were laid out side-by-side on the floor would they cover the entire classroom floor? (ii) What is the maximum number of textbooks that can fit in an empty storeroom of dimensions 15 ft × 20 ft × 30 ft?
(i) The textbook has roughly 150 leaves measuring 0.26 m × 0.19 m ≈ 0.05 m² each. Total area covered is 150 × 0.05 = 7.5 m². A typical classroom floor is about 7 m × 7 m = 49 m², so the pages cannot cover it. (ii) Storeroom volume is 15 × 20 × 30 = 9000 cu ft ≈ 254.85 m³. A single book occupies 0.26Read more
(i) The textbook has roughly 150 leaves measuring 0.26 m × 0.19 m ≈ 0.05 m² each. Total area covered is 150 × 0.05 = 7.5 m². A typical classroom floor is about 7 m × 7 m = 49 m², so the pages cannot cover it. (ii) Storeroom volume is 15 × 20 × 30 = 9000 cu ft ≈ 254.85 m³. A single book occupies 0.26 × 0.19 × 0.012 m ≈ 0.00059 m³. Dividing gives approximately 400,000 textbooks.
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/
See lessIf all the people in the world are crowded together in one location, how much area would it cover?
Assuming the current world population is approximately 8 billion people. If tightly packed in a crowd, one standing person occupies roughly 0.25 square metres (4 people per square metre). The total area required would be 8,000,000,000 × 0.25 = 2,000,000,000 m². Converting to square kilometres, thisRead more
Assuming the current world population is approximately 8 billion people. If tightly packed in a crowd, one standing person occupies roughly 0.25 square metres (4 people per square metre). The total area required would be 8,000,000,000 × 0.25 = 2,000,000,000 m². Converting to square kilometres, this equals 2,000 km², which is approximately comparable to the area of a large metropolitan region like Tokyo or Greater London.
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/
See less