Question

A cuboidal hall measures 10ft×8ft×8ft the wall and ceiling needs to be plastered . find area to be plastered ​

Answer 1

• Area to be plastered is 368 ft².

Step-by-step explanation:

Given:-

• Length of cuboidal hall is 10 ft.
• Breadth of cuboidal hall is 8 ft.
• Height of cuboidal hall is 8 ft.

To find:-

• Area to be plastered.

Solution:-

• We will use lateral surface area for area of walls.

Lateral surface area of cuboid = 2h(l + b)

• Where, l, b and h are length, breadth and height of cuboid.

Put l, b and h in formula:

$\longrightarrow$ 2 × 8(10 + 8)

$\longrightarrow$ 16 × (10 + 8)

$\longrightarrow$ 128 + 160

$\longrightarrow$ 288

Area of walls of cuboidal hall is 288 ft².

The ceiling of hall is of rectangular shape.

Area of rectangle = l×b

$\longrightarrow$ 10 × 8

$\longrightarrow$ 80

Area of ceiling of cuboidal hall is 80 ft².

________________________________

Area to be plastered = Area of walls + Area of ceiling.

$\longrightarrow$ 288 + 80

$\longrightarrow$ 368

Therefore,

Area to be plastered is 368 ft².

Answer 2

Area to be Plastered: lb + 2h(l+b)

[As only ceiling and walls need to be plastered.]

Dimensions: 10 ft × 8 ft × 8 ft

∴ Area:-

(10 ft)(8 ft) + 2(8 ft)(10 ft + 8 ft)

= 80 ft² + 16(18) ft²

= 80 ft² + 288 ft²

= 368 ft²

∴ 368 ft² area is to be plastered.