Question

A wall 10 m long, 20 cm thick and 1.2 m high
has been constructed as a partition between two
plots of land. Find the cost of cementing it on five
sides at the rate of 25 per sq. m.​

Answer 1

Question:

A wall 10 m long, 20 cm thick and 1.2 m high has been constructed as a partition between two plots of land. Find the cost of cementing it on five sides at the rate of 25 per sq. m.

Given:

$length \: = 10m \\ breadth \: = 1.2m \\ height \: = 20cm. \\ it \: is \: in \: cm \: so \: we \: can \: change \: it \: in \: m. \\ \longmapsto \frac{20}{100} = 0.2m$

$area \: to \: be \: cemented \: the \: wall \: \longmapsto t.s.a. - area \: of \: bottom \\ \longmapsto 2(lb + bh + hl) - hl \\ \longmapsto 2(10 \times 1.2 + 1.2 \times 0.2 + 0.2 \times 10) - (0.2 \times 10) \\ \longmapsto 2(12 + 0.24 + 2) - 2 \\ \longmapsto 26.48 {m}^{2}$

So, The Area Of Cemented The Wall Is

$\longmapsto 26.48 {m}^{2}$

Now,

$cost \: of \: painting \: the \: wall\longmapsto area \: of \: cemented \: wall \: \times rate \: of \: painting \\ \longmapsto 26.48 \times 25 \\ \longmapsto ₹662$

So, The Final Answer Is ₹662.

Answer 2

Step-by-step explanation:

TSA of a cuboid=2(lw+wh+hl)

were,

l=length

w=width

h=height

therefore,

2[(10×20/100)m+(10×1.2)m+(20×1.2)m]

=2(2+12+24)

=2×38

=76m^2

now,

as one of the side is not cemented because of standing on the ground.

Therefore,

76m^2-(10×20/100)m^2

=76m^2-2m^2

=74m^2

cost of rupees 25 per Sq m.

74×25

=1850 rupees.

answer.rupees1850.

Hope it's help uh...!!!