Question

the surface area of three faces of a cuboidal room with integral dimensions are 35m², 40m² and 56m² what will be the volume in (m³) of the room​

Answer 1

Answer:

So the Volume of the room will be 280 m^3.

Step-by-step explanation:

I hope my answer will help you a lot.

Answer 2

Given:

The surface area of three faces of a cuboidal room with integral dimensions are 35m², 40m² and 56m² what will be the volume in (m³) of the room​?

To find:

The volume in (m³) of the room

Solution:

Let "l", "b" & "h" be the length, breadth and height of the cuboidal room.

The surface area of three faces of a cuboidal room with integral dimensions are 35m², 40m² and 56m²

So, we can write the equations as,

$A_1 = l \times b = 35 \:m^2$ . . . (1)

$A_2 = b\times h = 40 \:m^2$ . . . (2)

$A_3 = h\times l = 56 \:m^2$ . . . (3)

On multiplying equations (1), (2) and (3), we get

$l \times b \times b\times h \times h\times l = 35 \times 40 \times56$

$\implies l^2 \times b^2 \times h^2 = 78400$

$\implies \sqrt{l^2 \times b^2 \times h^2} = \sqrt{78400}$

$\implies l \times b \times h = 280$

since $Volume \:of\: a\: cuboid = length \times breadth \times height$

$\implies \bold{Volume = 280\:m^3}$

Thus, the volume of the cuboidal-shaped room is → 280 m³.

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