Mensuration
- The base of a right pyramid is an equilateral triangle of side 4 cm each. Each slant edge is 5 cm long. The volume of the pyramid is
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Area of the base of pyramid = √3 × (side)² 4 = √3 × 4 × 4 = 4√3sq.cm. 4
Length of median on the base (AD) = √4² - 2²= √12
= 2√3 cm∴ OD = 1 × 2√3 = 2 cm. 3 √3 Height of pyramid = √5² - 
2 
² √3 Height of pyramid = √25 - 4 = √71 cm 3 √3 ∴ Volume of pyramid = 1 × area of base × height 3 ∴ Volume of pyramid = 1 × area of base × height 3 = 4√71 cu.cm. 3 Correct Option: E
Area of the base of pyramid = √3 × (side)² 4 = √3 × 4 × 4 = 4√3sq.cm. 4
Length of median on the base (AD) = √4² - 2²= √12
= 2√3 cm∴ OD = 1 × 2√3 = 2 cm. 3 √3 Height of pyramid = √5² - 2 ² √3 Height of pyramid = √25 - 4 = √71 cm 3 √3 ∴ Volume of pyramid = 1 × area of base × height 3 ∴ Volume of pyramid = 1 × area of base × height 3 = 4√71 cu.cm. 3
- If the radius of the base of a cone be 7 cm and its curved surface area be 550 sq. cm, then the volume of the cone is
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Curved surface area of cone = 550 sq. cm.
⇒ πrl = 550⇒ 22 × 7 × √r² + h² = 550 7 ⇒ 22 × 7 × √49 + h² = 550 7 ⇒ √49 + h² = 550 = 25 22
⇒ 49 + h² = (25)² = 625
→ h² = 625 – 49 = 576
⇒ h = 576 = 24 cm.∴ Volume of cone = 1 × 22 × 7 × 7 × 24 3 7
= 1232 cu. cm.Correct Option: A
Curved surface area of cone = 550 sq. cm.
⇒ πrl = 550⇒ 22 × 7 × √r² + h² = 550 7 ⇒ 22 × 7 × √49 + h² = 550 7 ⇒ √49 + h² = 550 = 25 22
⇒ 49 + h² = (25)² = 625
→ h² = 625 – 49 = 576
⇒ h = 576 = 24 cm.∴ Volume of cone = 1 × 22 × 7 × 7 × 24 3 7
= 1232 cu. cm.
- A hemisphere of iron is melted and recast in the shape of a right circular cylinder of diameter 18 cm and height 162 cm. The radius of the hemisphere is
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Volume of cylinder = πr²h = (π × 92 × 162) cu.cm.
∴ Volume of hemisphere = (π × 9² × 162) cu. cm.∴ 2 π × R³ = π × 9² × 162 3 ⇒ R³ = 9² × 162 × 3 = 9² × 8 × 3 2
∴ R = ³√ × 9² × 9² × 3;
= 9 × 3 = 27 cm.Correct Option: A
Volume of cylinder = πr²h = (π × 92 × 162) cu.cm.
∴ Volume of hemisphere = (π × 9² × 162) cu. cm.∴ 2 π × R³ = π × 9² × 162 3 ⇒ R³ = 9² × 162 × 3 = 9² × 8 × 3 2
∴ R = ³√ × 9² × 9² × 3;
= 9 × 3 = 27 cm.
- An iron sphere of radius 27 cm is melted to form a wire of length 729 cm. The radius of wire is
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Volume of iron sphere = 4 πr³ 3 Volume of metallic sphere = 4 × π × 27 × 27 × 27 3
= (36 × 27 × 27)π cu. cm.
∴ If the radius of wire be R cm., then π × R² × 729 = 36 × 27 × 27π [∵ V = πR²H]⇒ R² = 36 × 27 × 27 = 36 729
∴ R = 36 = 6 cm.Correct Option: A
Volume of iron sphere = 4 πr³ 3 Volume of metallic sphere = 4 × π × 27 × 27 × 27 3
= (36 × 27 × 27)π cu. cm.
∴ If the radius of wire be R cm., then π × R² × 729 = 36 × 27 × 27π [∵ V = πR²H]⇒ R² = 36 × 27 × 27 = 36 729
∴ R = 36 = 6 cm.
- The ratio of weights of two spheres of different materials is 8 : 17 and the ratio of weights per 1 cc of materials of each is 289 : 64. The ratio of radii of the two spheres is
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Ratio of the volumes of spheres = 8 × 64 289 × 17 ⇒ 4 πr1³ = 3 8 × 8 × 8 4 πr2³ 17 × 17 × 17 3 ⇒ r1³ = 8 ³ r2³ 17 ⇒ r1 = 8 r2 17 Correct Option: A
Ratio of the volumes of spheres = 8 × 64 289 × 17 ⇒ 4 πr1³ = 3 8 × 8 × 8 4 πr2³ 17 × 17 × 17 3 ⇒ r1³ = 8 ³ r2³ 17 ⇒ r1 = 8 r2 17
