Question

The internal measures of a cuboidal room are 15m x 10m x 8m . Find the cost of whitewashing all four walls and the ceiling of a room, if the cost of whitewashing is Rs 5 per square meter.​

Answer 1

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The internal measures of a cuboidal room are 15m x 10m x 8m . Find the cost of whitewashing all four walls and the ceiling of a room, if the cost of whitewashing is Rs 5 per square meter.

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➡Cost of white washing of walls and ceiling is Rs 2750m²

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• Length of wall = 15m
• Breadth of wall = 10m
• Height of wall = 8m

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• Area of four walls = L.S.A. of room and area of ceiling
• Total cost of whitewashing

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L.S.A of cuboid =

$\bf \pink{2h(l + b)}$

Area of rectangle =

$\bf \pink{length \times breadth}$

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Total Area used for white washing = L.S.A of room + Area of ceiling

First we will find L.S.A of room

$\bf {L.S.A = 2h(l + b)}$

$\bf = 2 \times 8(15 + 10)$

$\bf = 16 \times 25$

$\bf = {400m}^{2}$

Area of ceiling =

$\bf= l \times b$

$\bf = 15 \times 10$

$\bf = {150m}^{2}$

Total Area =

$\bf = 400 + 150$

$\bf = 550 {m}^{2}$

Total Cost of whitewashing at the rate of Rs 5 per square meter

$\bf = total \: area \times rate$

$\bf = 550 \times 5$

$\bf = Rs2750$

Therefore, Cost of white washing of walls and ceiling is Rs 2750m².

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Answer 2

Given:

• Length of wall (l) = 15 m
• Breadth of wall (b) = 10 m
• Height of wall (h) = 8 m
• Cost of whitewashing = Rs. 5 per m²

To Find:

• The cost of whitewashing all four walls and the celing of a room.

Solution:

★ Lateral surface area of room,

⇒ LSA = 2h(l + b)

⇒ LSA = 2 × 8 (15 + 10)

⇒ LSA = 16 × 25

⇒ LSA = 400 m²

★ Area of ceiling,

⇒ A = l × b

⇒ A = 15 × 10

⇒ A = 150 m²

Now,

Total Area used for white washing = LSA of room + Area of ceiling.

⇒ Total area = 400 + 150

⇒ Total area = 550 m²

So,

Total Cost of whitewashing at the rate of Rs. 5 per m².

⇒ Total area × Rate

⇒ 550 × 5

⇒ Rs. 2750

Hence:

The cost of white washing of walls and ceiling is Rs 2750.