Math, asked by sweetmehak4237, 10 months ago

find the cost of digging a cuboidal pit 8m long , 6m board and 3m deep at the rate of rs 30 per m cube

Answers

Answered by aliasgar47
2

Answer:

area of cuboid l ×b ×h

8×6×3 =144

cost = area /per m cost

144/30

=4.8 rs

5 rs approximate

Answered by iTzShInNy
29

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\underline { \sf Given:-}

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  •  \small \sf \: Length  \: of  \: the \: Cuboidal \: pit, \: l = \boxed{ \bf \pink{ 8  \: m}} \\
  • \small \sf \: Breadth \: of  \: the \: Cuboidal \: pit, \: b = \boxed{ \bf \pink{ 6 \: m}} \\
  •  \small\sf \: Height \: of  \: the \: Cuboidal \: pit, \: h = \boxed{ \bf \pink{ 3  \: m}}

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  \small\therefore \sf{Volume \: of \: the \: Cuboidal \: pit \: ➺ \: l \times b \times h\: } \\

  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: \:  \:  \:  \:  \: \small \sf ➺ \: 8 \times 6 \times 3 \\

 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: \:  \:  \:  \:  \: \small \sf ➺ \: 8  \times 18 \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:   \:  \:  \:  \: \:  \:  \:  \:  \: \small \sf ➺ \: 144 \:  {m}^{3}  \\

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 \small \sf \: {Cost  \: of  \: digging  \: 1 {m}^{3}  \: of  \: the  \: pit \:  \longrightarrow  \: ₹ \: 30} \\

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\small \sf \: \therefore {Cost  \: of  \: digging  \: 144 \:  {m}^{3}  \: of  \: the  \: pit \:  \longrightarrow  \: ₹ \: 30 \times 144}

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 \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: \small \sf \: { \longrightarrow  \: ₹  \: 4320}

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 \underline{ \sf \: More \: Information:-}

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  •  \small \sf \: Volume  \: of \: a \: Cuboid  \large\leadsto  \small \boxed{ \bf l \times b \times h}

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  •  \small \sf \: Volume  \: of \: a \: Cube \large\leadsto  \small \boxed{ \bf  {a}^{3} }

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  •  \small \sf \: Volume  \: of \: a \: Cylinder  \large\leadsto  \small \boxed{ \bf \pi  {r}^{2}h }

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  •  \small \sf \: Volume  \: of \: a \:Cone\large\leadsto  \small \boxed{ \bf  \frac{1}{3}\pi r {}^{2} h }

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  •  \small \sf \: Volume  \: of \: a \: Sphere \large\leadsto  \small \boxed{ \bf  \frac{4}{3} \pi r {}^{3} }

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  •  \small \sf \: Volume  \: of \: a \: Hemisphere \large\leadsto  \small \boxed{ \bf  \frac{2}{3}\pi  {r}^{3}  }

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