Question

John has built a cuboidal water tank in his house.The top of the water tank is covered with an iron lid.he wants to cover the inner surface of the tank including the base with tiles of size 10cm by 8cm.if the dimensions of the water tank are 180cm×120cm×60cm and cost of tiles is rs 480 per dozen, the find the total amount required for tiles.

Answer 1

what we have here is a cuboidal tank john built. Here the area where we are suppose to co er with tiles is whole area of the cuboid leaving the upper surface area since it is covered with iron lid.

so what we have to do is:-

total surface area of cuboid -the upper surface are.
given:- length=180
breast=120
height= 60

first the total surface area:-

$2(lb + bh +hl)$
2(180×120+120×60+60×180)=79200cm^2 -{1}

Upper surface area = length×breadth
= 180×120
= 21600cm^2 -{2}


So here we subtract as we earlier saw
{1} - {2} we will get
79200-21600=57600 cm^2 -{3}

Now 57600cm^2 is what john 8s foi g to put the tiles on.

Now the area of the tiles is length ×breadth
= 10×8cm^2
= 80 cm^2 - {4}



Now we want number of tiles requires to co er the surface so we will divide total surface e area by surface area of one tile:-
$\frac{ < 3 > }{ \ < 4 > }$
$\frac{57600}{80} = 720tiles$
given 480 rs per dozen.
1dozen= 12 pieces

1 tile=
$\frac{480}{12} = 40rupees$The amount required =number of tiles× cost of per tiles.

=720×40
=28800Rs

Therefore we need 28800Rs to cover the cuboidal tank with tiles.