Mensuration
- The radius and the height of a cone are in the ratio 4 : 3. The ratio of the curved surface area and total surface area of the cone is
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r = 4 ⇒ r = h = k h 3 4 3
⇒ r = 4k; h = 3k
∴ l = √r² + k² = √16k² + 3k²
= √25k² = 5k∴ Curved surface area Total surface area = πrl πr(r + l) = l = 5k = 5 r + l 4k + 5k 9
or 5 : 9Correct Option: A
r = 4 ⇒ r = h = k h 3 4 3
⇒ r = 4k; h = 3k
∴ l = √r² + k² = √16k² + 3k²
= √25k² = 5k∴ Curved surface area Total surface area = πrl πr(r + l) = l = 5k = 5 r + l 4k + 5k 9
or 5 : 9
- The maximum length of a pencil that can be kept in a rectangular box of dimensions 8cm × 6cm × 2cm is
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Maximum length of the pencil = √8² + 6² + 2² = √64 + 36 + 4
= √104 = 2√26 cmCorrect Option: C
Maximum length of the pencil = √8² + 6² + 2² = √64 + 36 + 4
= √104 = 2√26 cm
- The volume of a cubical box is 3.375 cubic metres. The length of edge of the box is
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Length of the edge of the box = ³√3.375 . metre
= ³√1.5 × 1.5 × 1.5metre = 1.5 metreCorrect Option: B
Length of the edge of the box = ³√3.375 . metre
= ³√1.5 × 1.5 × 1.5metre = 1.5 metre
- Diagonal of a cube is 63cm. Ratio of its total surface area and volume (numerically) is
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Diagonal of the cube = 63cm
∴ 3 × edge = 63cm
⇒ Edge = 6 cm
∴ Total surface area : Volume = 6 × 6² : 6³ = 1 : 1Correct Option: C
Diagonal of the cube = 63cm
∴ 3 × edge = 63cm
⇒ Edge = 6 cm
∴ Total surface area : Volume = 6 × 6² : 6³ = 1 : 1
- The length of the largest possible rod that can be placed in a cubical room is 35 3 m. The surface area of the largest possible sphere that fit within the cubical room (assuming π = 22/7 ) (in squarem)is
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Diagonal of cubical room = 35√3metre
∴ 3 × edge = 35√3
⇒ Edge = 35 metre
→ Diameter of sphere = 35 metre
⇒ Surface area of the sphere = 4πr²= 4 × 22 × 35 × 35 = 3850 sq.metre 7 4 Correct Option: B
Diagonal of cubical room = 35√3metre
∴ 3 × edge = 35√3
⇒ Edge = 35 metre
→ Diameter of sphere = 35 metre
⇒ Surface area of the sphere = 4πr²= 4 × 22 × 35 × 35 = 3850 sq.metre 7 4
