Question

A cistern 6m long and 4 m wide contains water up to a depth of 1 m 25 cm. The total area of the wet surface is:

A) 49
B) 50
C) 53.5
D) 55

Don't attend if you don't know!!

Answer 1

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$here \: is \: your \: \: answer↪length = 6m↪ breadth = 4m \:so \: using \: the \: formula ,↪Area \: of \: the \: wet \: surface,= [2(lb + bh + lh) - lb]= 2(bh + lh) + lb= [2 (4 x 1.25 + 6 x 1.25) + 6 x 4]m^2 = 49m^2.the \: total \: area \: of \: the \: wet \:surface \: is \: = 49 {m}^{2}the \: total \:area \: of \: the \: wet \: surface \: is=49m2\:hope \: it \: helps\beta e \: \beta \gamma\alpha inlyβeβγαinly$

Answer 2

Given

• Length = 6 m
• Breadth = 4m
• Depth = 1m 25 cm = 1.25 m

Explanation

We know that all the surfaces are wet (except of upper side area) so as we know that:-

$\bullet \ {\purple{\underline{\underline{\boxed{\sf{ Total \ surface \ Area _{(Cuboid)} = 2(lb+bh+hl) }}}}}}$

We know that All areas are wet only except of upper side so, we can remove upper side area from Total Surface Area as:

$\colon\implies{\sf{ 2(lb+bh+hl) - lb }} \\ \\ \\ \colon\implies{\sf{ 2(6 \times 4+4 \times 1.25+1.25 \times 6) - 6 \times 4 }} \\ \\ \\ \colon\implies{\sf{ 2(24+5+7.5) - 24 }} \\ \\ \\ \colon\implies{\sf{ 2(36.5) - 24 }} \\ \\ \\ \colon\implies{\sf{ 73 - 24 }} \\ \\ \\ \colon\implies{\underline{\underline{\sf{ 49 \ m^2 }}}}$

Hence,

• The total area of the wet surface of cistern is 49 m².