Question

$\Large \mathtt \green{Question :}$
The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V, prove that V2 = xyz.

$\red{note - }$
❎No spam
❎No copied answer ​

Answer 1

$\large \dag$ Question :-

The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V, prove that V² =xyz

$\large \dag$ Step by step Solution :-

We know that all faces of a cuboid are rectangle, and area of rectangle is a × b where a and b are its sides.

Let us Assume that dimensions of the cuboid are $\large\rm l \times b \times h$

And here we have the areas of three adjacent faces of a cuboid are x, y, and z.

Therefore,

• $\rm l \times b = x \: - - - - (1)$

• $\rm b \times h = y \: \: - - - - (2)$

• $\rm l \times h = z \: \: - - - - (3)$

$\large \bigstar$ Multiplying all the three equations ;

$:\longmapsto \rm l \times b \times b \times h \times l \times h = xyz$

$:\longmapsto \rm l {}^{2} b {}^{2} h {}^{2} = xyz$

$:\longmapsto \rm ( {lbh)}^{2} = xyz$

As Volume of cuboid is lenght × breadth × height

$:\longmapsto \rm {(Volume)}^{2} = xyz$

$\purple{ \large :\longmapsto  \underline {\boxed{{\bf V {}^{2} = xyz} }}}$

$\Large\red \maltese \: \: \: \underline{\pink{\underline{\frak{\pmb{\text Hence\:\:Proved }}}}}$

Answer 2

QUESTION-:

The areas of three adjacent faces of a cuboid are x, y, and z. If the volume is V, prove that V²= xyz.

EXPLANATION-:

Since the given figure is cuboid so the length and breadth and heigh of each face will be different.

Find area of each face.

→x= length×Breadth

→y=length × Height

→z=length × Height

→xyz=length×Breadth×length×Height×length×height

→xyz=(length)²×(breadth)²×(height)²

→xyz=(length×breadth×height)²

We know that-:

→Volume=length×breadth×height

→(volume)²=xyz

Hence proved