A heap of wheat is in the form of a cone, the diameter of whose base is 14m and height is 3m .fine it's volume. the heap it to be covered by canvas to protect it from rain. fine the are area of canvas required
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Answered by
11
Hii friend,
DIAMETER = 14 M
RADIUS = D/2 = 14/2 = 7 M
HEIGHT OF THE HEAP = 3 M
THEREFORE,
VOLUME OF THE HEAP = 1/3 πr² h = 1/3 × 22/7 × 7× 7 × 3 = 22 × 7 × 3/3 = 154 M³
AREA OF THE CANVAS REQUIRED FOR COVER THE TOP OF THE HEAP = πr² = 22/7 × 7 × 7 = 22×7 = 154 M².
HOPE IT WILL HELP YOU...... :-)
DIAMETER = 14 M
RADIUS = D/2 = 14/2 = 7 M
HEIGHT OF THE HEAP = 3 M
THEREFORE,
VOLUME OF THE HEAP = 1/3 πr² h = 1/3 × 22/7 × 7× 7 × 3 = 22 × 7 × 3/3 = 154 M³
AREA OF THE CANVAS REQUIRED FOR COVER THE TOP OF THE HEAP = πr² = 22/7 × 7 × 7 = 22×7 = 154 M².
HOPE IT WILL HELP YOU...... :-)
sivaprasath:
Area of canvas required to cover the heap of wheat is CSA of cone, not area of it's base,. We can't put an umbrella-like structure to protect it from rain ,.(lol),. So, you should have used πrl units and find CSA,.
Answered by
6
Diameter of Base = 14m
.°. Radius = 14/2 = 7cm
Height of cone = 3m
Latent height = root(3²+7²= 9 +49) =
.°. Volume of cone = ( πr²h ) / 3
= (3.14 * 7 * 7 * 3) / 3
= (3.14 * 49)
=153.86m³
CSA of cone = πrl
= 3.14 * 7 * 7.14
= 156.93 cm²
Hence, the area of canvas required is 156.93cm²
.°. Radius = 14/2 = 7cm
Height of cone = 3m
Latent height = root(3²+7²= 9 +49) =
.°. Volume of cone = ( πr²h ) / 3
= (3.14 * 7 * 7 * 3) / 3
= (3.14 * 49)
=153.86m³
CSA of cone = πrl
= 3.14 * 7 * 7.14
= 156.93 cm²
Hence, the area of canvas required is 156.93cm²
hi bro
pls give me a answer this question
60 circular plates of each radius 7cm and Thikenss 1.3cm are placed and above the another the form a solid circular cylinder. fine the volume of cycling so formed
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