Question

A cuboid has dimensions 4cm, 3cm and 1cm.
what is the length of diagonal?​

Answer 1

Given :

• A cuboid has dimensions 4 cm, 3 cm and 1 cm.

To find :

• Length of diagonal.

Solution :

• Length = 4 cm.
• Breadth = 3 cm.
• Height = 1 cm.

We know,

${\boxed{\green{\bold{Diagonal\: of\: cuboid=\sqrt{Length^2+Breadth^2+Height^2}}}}}$

$\sf{Diagonal=\sqrt{Length^2+Breadth^2+Height^2}}$

$\implies\sf{Diagonal=\sqrt{4^2+3^2+1^2}\:cm}$

$\implies\sf{Diagonal=\sqrt{16+9+1}\:cm}$

$\implies\sf{Diagonal=\sqrt{26}\:cm}$

${\boxed{\bold{Diagonal\:of\: cuboid=\sqrt{26}\:cm}}}$

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More information :

★ Volume of cuboid = length × breadth × height

★ TSA of cuboid = 2(l×b + l×h + b×h )

★ CSA of cuboid = 2( l+b )h

[ Where l = length , b = breadth and h = height ]

★ Volume of cube = a³

★ TSA of cube = 6a²

[ where a = side ]

_________________________

Answer 2

AnsWer :

√26 Cm.

To Find :

Diagonal of Cuboid.

Solution :

We have,

• A Cuboid, Sides 4 Cm , 3 Cm and 1 Cm.

$\tt\dagger \: \: \: Diagonal \: Of \: Cuboid = \sqrt{ {l}^{2} + {b}^{2} + {h}^{2} }$

Here,

• L Length of Cuboid.
• B Breadth of Cuboid.
• H Height of Cuboid.

$\tt \longmapsto \sqrt{ {4}^{2} + {3}^{2} + {1}^{2} }$

$\tt \longmapsto \sqrt{9 + 16+ 1}$

$\tt \longmapsto \sqrt{26} \: cm.$

Therefore, the Diagonal of Cuboid be √26 cm.

Some information :

Cuboid,

• TSA => 2( lb + bh + hl )
• CSA => 2( l + b )h
• Volume => l * b * h
• Diagonal² => l² + b² + h²