Question

Find the area of metal-sheet required to
make an open tank of length = 10 m,
breadth = 7.5 m and depth = 3.8 m.​

Answer 1

Answer:

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Length of the tank = 10 m

Length of the tank = 10 mBreadth of the tank = 7.5 m

Length of the tank = 10 mBreadth of the tank = 7.5 mDepth of the tank = 3.8 m

Length of the tank = 10 mBreadth of the tank = 7.5 mDepth of the tank = 3.8 mArea of four walls = 2[L+B] x H = 2(10 + 7.5) x 3.8

Length of the tank = 10 mBreadth of the tank = 7.5 mDepth of the tank = 3.8 mArea of four walls = 2[L+B] x H = 2(10 + 7.5) x 3.8= 2 x 17.5 x 3.8 = 35 x 3.8 = 133 m2

Length of the tank = 10 mBreadth of the tank = 7.5 mDepth of the tank = 3.8 mArea of four walls = 2[L+B] x H = 2(10 + 7.5) x 3.8= 2 x 17.5 x 3.8 = 35 x 3.8 = 133 m2Area of the floor = L x B = 10 x 7.5 = 75 m

Length of the tank = 10 mBreadth of the tank = 7.5 mDepth of the tank = 3.8 mArea of four walls = 2[L+B] x H = 2(10 + 7.5) x 3.8= 2 x 17.5 x 3.8 = 35 x 3.8 = 133 m2Area of the floor = L x B = 10 x 7.5 = 75 mArea of metal sheet required to make the tank =Area of four walls + Area of floor = 133 m2 + 75 m2 = 208 m2

Step-by-step explanation:

I hope it helps u ✌️

Answer 2

Given :-

$\sf{Length \: of \: the \: tank = 10 m} \:$

$\sf{Breadth \: of \: the \: tank = 7.5 m} \:$

$\sf{Depth \: of \: the \: tank = 3.8 m} \:$

To find :-

$\sf{Area \: of \: metal \: sheet} \:$

Solution :-

$\sf{Area \: of \: four \: walls = 2[L+B] \: x \: H} \:$

$\sf{ ⟹ 2(10 + 7.5) \: x \: 3.8} \:$

$\sf{⟹ 2 \: x \: 17.5 \: x \: 3.8} \:$

$\sf{⟹ 5 \: x \: 3.8} \:$

$\sf{⟹ {133m}^{2} }$

$\sf{Area \: of \: the \: floor = L x B } \:$

$\sf{⟹ 10 \: x \: 7.5} \:$

$\sf{⟹ 75 m} \:$

$\sf{Area \: of \: metal \: sheet \: required \: to \: make \: the \: tank} \:$

$\sf{⟹ Area \: of \: four \: walls + Area \: of \: floor} \:$

$\sf{⟹ {133m}^{2} + {75m}^{2} }$

$\sf{⟹ {128m}^{2} }$

✰ Hence , the area of metal-sheet required to make an open tank is 128m^2. ✔︎