Question

A tank 10m long and 5m wide has water up to a depth of 3m. What is the total area of the wet surface?

Answer 1

Area=CSA+area of lb

=2h(l+b)+lb
=2*3(10+5)+10*5
=6(15)+50
=90+50
=140

Answer 2

A tank which is of 10 m long and 5 m wide has different length and breadth, so it must take the form of a rectangle. Hence, we denote it as a cuboid

To find out the area of the wet surface, the following steps are to be taken  

Step: 1

Calculate the total surface area of the cuboid (area of all the six sides)  

Surface area of a cuboid = 2(lb + bh + lh)

    = 2 (10×5 + 5× 3 + 3× 10)

    = 2×(50 + 15 + 30)

    = 2×95

    = 190 m

Step 2:

Find the surface area of the top portion (which is of rectangular)

Since the single side of a tank represents a rectangle. Then,

Surface area of the top portion =10×5

=50  

(c) Subtract the surface area of the top portion from the surface area of the cuboid to get the surface area of the wet portion

Surface area of the wet portion = Surface area of the cuboid – surface area of the top portion

             = 190 – 50

                                                     = 140 $m^2$  

∴ The total area of the wet surface = 140$m^2$