Question

Seema is painting the walls and ceiling of a hall whose dimensions are 10 m x 15 m x 5 m, from each can pf paint 120m2 of area is painted. How many cans of paint she need to pain the hall?​

Answer 1

$\large\underline{\sf{Solution-}}$

Dimensions of Room :-

• Length of Room = 10 m

• Breadth of Room = 15 m

• Height of Room = 5 m

We know, that

$\boxed{ \sf \: Area_{(walls)} \: = \: 2(length + breadth) \times height}$

So,

• On plugging the values, we have

$\rm :\longmapsto\:Area_{(walls)} = 2(15 + 10) \times 5$

$\rm :\longmapsto\:Area_{(walls)} = 10 \times 25$

$\bf\implies \:Area_{(walls)} = 250 \: {m}^{2}$

We know that,

$\boxed{ \sf \: Area_{(ceiling)} = length \times breadth}$

• On plugging the values, we have

$\rm :\longmapsto\:Area_{(ceiling)} = 10 \times 15$

$\bf\implies \:Area_{(ceiling)} = 150 \: {m}^{2}$

Total area to be painted,

$\rm :\longmapsto\:Area_{(to \: be \: painted)} = Area_{(walls)} + Area_{(ceiling)}$

$\rm :\longmapsto\:Area_{(to \: be \: painted)} = 250 + 150$

$\bf\implies \:Area_{(to \: be \: painted)} = 400 \: {m}^{2}$

Now,

• In one can, area to be painted = 120 square meter.

• To paint area of 400 square meter, number of cans required,

$\rm :\longmapsto\:Number_{(cans)} \: = \dfrac{400}{120} = 3.33 \:$

$\bf\implies \:Number_{(cans)} \: = \: 4 \: cans \: required.$

Additional Information :-

$\boxed{ \sf{ \: CSA{(cylinder)} = 2\pi \: rh}}$

$\boxed{ \sf{ \: Volume_{(cuboid)} = lbh}}$

$\boxed{ \sf{ \: Volume_{(cone)} = \dfrac{1}{3} \pi \: {r}^{2} h}}$

$\boxed{ \sf{ \: Volume_{(cube)} = {(edge)}^{3} }}$

$\boxed{ \sf{ \: TSA{(cone)} = {6(edge)}^{2} }}$

$\boxed{ \sf{ \: TSA{(cuboid)} = 2(lb + bh + hl)}}$

$\boxed{ \sf{ \: TSA{(cone)} = \pi \: r(l + r)}}$

$\boxed{ \sf{ \: TSA{(cylinder)} = 2\pi \: r(h + r)}}$

$\boxed{ \sf{ \: CSA{(cone)} = \pi \: rl}}$