Mensuration
- The whole surface area of a pyramid whose base is a regular polygon is 340 cm² and area of its base is 100 cm². Area of each lateral face is 30 cm². Then the number of lateral faces is
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Total surface area = Lateral surface area + area of base
⇒ 340 = Lateral surface area + 100
⇒ Lateral surface area = 340 – 100 = 240 sq. cm.
Area of each lateral surface = 30 sq. cm.
∴ Number of lateral surfaces = 240/30 = 8Correct Option: A
Total surface area = Lateral surface area + area of base
⇒ 340 = Lateral surface area + 100
⇒ Lateral surface area = 340 – 100 = 240 sq. cm.
Area of each lateral surface = 30 sq. cm.
∴ Number of lateral surfaces = 240/30 = 8
- A sphere has the same curved surface area as a cone of vertical height 40 cm and radius 30 cm. The radius of the sphere is
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Slant height of cone (l) = √h² + r²
= √30² + 40² = √900 + 1600
= √2500 = 50 cm.
∴ Curved surface area of cone = πrl = (π × 30 × 50) sq. cm.
= 1500π sq. cm.
If the radius of sphere be R cm, then 4πR² = 1500π⇒ R² = 1500 = 375 4
⇒ R = √375 = √5 × 5 × 15
= 5√15 cm.Correct Option: C
Slant height of cone (l) = √h² + r²
= √30² + 40² = √900 + 1600
= √2500 = 50 cm.
∴ Curved surface area of cone = πrl = (π × 30 × 50) sq. cm.
= 1500π sq. cm.
If the radius of sphere be R cm, then 4πR² = 1500π⇒ R² = 1500 = 375 4
⇒ R = √375 = √5 × 5 × 15
= 5√15 cm.
- The diameter of a sphere is twice the diameter of another sphere. The curved surface area of the first and the volume of the second are numerically equal. The numerical value of the radius of the first sphere is
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Radius of first sphere = 2r units (let).
∴ Radius of second sphere = r units
Curved surface of first sphere = 4πR² = 4π(2r)² = 16πr² sq. units.Volume of second sphere = 4 πr³ cu. units 3
According to the question,= 4 πr³ = 16πr² 3
⇒ 4r = 16 × 3⇒ r = 16 × 3 = 12 units 4
∴ Radius of first sphere = 24 unitsCorrect Option: B
Radius of first sphere = 2r units (let).
∴ Radius of second sphere = r units
Curved surface of first sphere = 4πR² = 4π(2r)² = 16πr² sq. units.Volume of second sphere = 4 πr³ cu. units 3
According to the question,= 4 πr³ = 16πr² 3
⇒ 4r = 16 × 3⇒ r = 16 × 3 = 12 units 4
∴ Radius of first sphere = 24 units
- A well of diameter 3m is dug 14m deep. The earth taken out of it has been spread evenly all around it in the shape of a circular ring of width 4m to form an embankment. Find the height of the embankment.
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Volume of the earth taken out
= πr²h = π × 
3 
² × 14 2 = 63 π cubic metre 2 Ex–radius of embankment = 3 + 4 = 11 metre 2 2
∴Volume of embankment = π(R² – r²) × h1= π 11 ² - 3 ² × h1 2 2 = π 11 + 3 11 - 3 × h1 2 2 2 2
= π × 7 × 4h1
= 28πh1 cu. metre∴ 28πh1 = 63 π 2 &rArr4; h1 = 63 = 1.125 metre 2 × 28 Correct Option: C
Volume of the earth taken out
= πr²h = π × 3 ² × 14 2 = 63 π cubic metre 2 Ex–radius of embankment = 3 + 4 = 11 metre 2 2
∴Volume of embankment = π(R² – r²) × h1= π 11 ² - 3 ² × h1 2 2 = π 11 + 3 11 - 3 × h1 2 2 2 2
= π × 7 × 4h1
= 28πh1 cu. metre∴ 28πh1 = 63 π 2 &rArr4; h1 = 63 = 1.125 metre 2 × 28
- The diagonal of a cuboid of length 5 cm, width 4 cm and height 3 cm is
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Diagonal of the cuboid = √l² + b² + h² = √5² + 4²² + 3²
= √25 + 16 + 9 = √50 = 5√2 cm.Correct Option: A
Diagonal of the cuboid = √l² + b² + h² = √5² + 4²² + 3²
= √25 + 16 + 9 = √50 = 5√2 cm.
