Mensuration
- Surface areas of three adjacent faces of a cuboid are p, q, r. Its volume is
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Let the length, breadth and height of a cuboid be l, b and h units respectively, then
p = lb; q = bh, r = hl
⇒ pqr = l²b²h²
∴ Volume of the cuboid = lbh = √pqrCorrect Option: D
Let the length, breadth and height of a cuboid be l, b and h units respectively, then
p = lb; q = bh, r = hl
⇒ pqr = l²b²h²
∴ Volume of the cuboid = lbh = √pqr
- A godown is 15 m long and 12 m broad. The sum of the area of the floor and the ceiling is equal to the sum of areas of the four walls. The volume (in m³) of the godownis:
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If the height of the godown be h metre, then
2(15 × 12) = 2 × h(15 + 12)
⇒ 27h = 15 × 12h = 15 × 12 = 20 metre 27 3 h = 15 × 12 = 20 metre 27 3 Correct Option: B
If the height of the godown be h metre, then
2(15 × 12) = 2 × h(15 + 12)
⇒ 27h = 15 × 12h = 15 × 12 = 20 metre 27 3 h = 15 × 12 = 20 metre 27 3
- If the total surface area of a cube is 96 cm², its volume is
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Let Edge of cube = x cm
∴ 6x² = 96 ⇒ x² = 96 = 16 6
⇒ x = √16 = 4cm
Volume of cube = (edge)³ = (4)³ = 64 cu. cmCorrect Option: C
Let Edge of cube = x cm
∴ 6x² = 96 ⇒ x² = 96 = 16 6
⇒ x = √16 = 4cm
Volume of cube = (edge)³ = (4)³ = 64 cu. cm
- A parallelogram has sides 15 cm and 7 cm long. The length of one of the diagonals is 20 cm. The area of the parallelogram is
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Using Rule 2 and 3,
Area of parallelogram ABCD
= Area of 2 ∆ ABC
Semi-perimeter of ∆ ABCS = 20 + 7 + 15 = 42 = 21 cm 2 2
∴ area of ∆ ABC
= √s(s - a)(s - b)(s - c)
= √21(21 - 7)(21 - 20)(21 - 15)
= √21 × 14 × 6 = 42 sq.cm.
∴ Area of parallelogram = 2 × 42 = 84 sq. cm.Correct Option: C
Using Rule 2 and 3,
Area of parallelogram ABCD
= Area of 2 ∆ ABC
Semi-perimeter of ∆ ABCS = 20 + 7 + 15 = 42 = 21 cm 2 2
∴ area of ∆ ABC
= √s(s - a)(s - b)(s - c)
= √21(21 - 7)(21 - 20)(21 - 15)
= √21 × 14 × 6 = 42 sq.cm.
∴ Area of parallelogram = 2 × 42 = 84 sq. cm.
- The sides of a triangle are 16 cm, 12 cm and 20 cm. Find the area.
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Using Rule 2 and 3,
Semi–perimeterS = 16 + 12 + 20 2 = 48 = 24 cm 2
Area of triangle
= √s(s - a)(s -b)(s - c)
= √24(24 - 16)(24 - 12)(24 - 20)
= √24 × 8 × 12 × 4 = 96 sq.cmCorrect Option: C
Using Rule 2 and 3,
Semi–perimeterS = 16 + 12 + 20 2 = 48 = 24 cm 2
Area of triangle
= √s(s - a)(s -b)(s - c)
= √24(24 - 16)(24 - 12)(24 - 20)
= √24 × 8 × 12 × 4 = 96 sq.cm
