The dimensions of a box are 12 cm X 4 cm X 3cm .Find the length of the longest rod which can be placed in this box. About the author Sadie
Given :- Dimensions of the box = 12cm × 4cm × 3cm Aim :- To find the length of the longest rod which can be placed in the box Formula to use :- The length of the longest rod which can be placed in the box must be placed diagonally in order to be the longest. To find the length of this rod, [tex]\longrightarrow \sf \sqrt{(length)^{2} + (breadth)^{2} + (height)^{2}}[/tex] Substituting the values, [tex]\implies \sf \sqrt{(12)^{2} + (4)^{2} + (3)^{2}}[/tex] [tex]\implies \sf \sqrt{144 + 16 + 9}[/tex] [tex]\implies \sf \sqrt{169}[/tex] [tex]\implies\sf 13[/tex] Hence, the length of the longest rod which can be placed in the box is 13cm. Some more formulas :- Total surface area of a cuboid = 2 ×[(length + breadth) + (breadth + height) + (height + length)] Lateral surface area of a cuboid = 2 × height + length + 2 × height × breadth ⇒ 2 × height (length + breadth) Volume of a cuboid = Length × Breadth × Height Length of the longest rod which can be placed in a cube = √3 × side Reply
Given :-
Aim :-
Formula to use :-
The length of the longest rod which can be placed in the box must be placed diagonally in order to be the longest.
To find the length of this rod,
[tex]\longrightarrow \sf \sqrt{(length)^{2} + (breadth)^{2} + (height)^{2}}[/tex]
Substituting the values,
[tex]\implies \sf \sqrt{(12)^{2} + (4)^{2} + (3)^{2}}[/tex]
[tex]\implies \sf \sqrt{144 + 16 + 9}[/tex]
[tex]\implies \sf \sqrt{169}[/tex]
[tex]\implies\sf 13[/tex]
Hence, the length of the longest rod which can be placed in the box is 13cm.
Some more formulas :-