Question

Find the volume of the cuboid whose length =16cm, breadth = 9cm and height = 6c​

Answer 1

Volume of a cuboid = (length × breadth × height) cubic units.

= (l × b × h) cubic units.

(Since area = ℓ × b)

Volume of a cuboid = area of one surface × height cubic units

Let us look at the given cuboid.

The length of the cuboid = 5 cm

The breadth of the cuboid = 3 cm

The height of cuboid (thickness) = 2 cm

The number of 1 cm cubes in the given cuboid = 30 cubes = 5 × 3 × 2

We find that volume of the given cuboid with length 5 cm, breadth 3 cm and height 2 cm is 30 cu cm.

Therefore, volume of a cuboid = length × breadth × height

it's_Khu°°°☺️

Answer 2

Given :

• Length =16cm

• Breadth =9cm

• Height = 6cm

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ \rule{200pt}{3pt} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

To Find :

• Area of Cuboid = ?

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Solution

$✭ {\underline{\pmb{\frak{ Formula \; Used \; :- }}}}$

${\underline{\boxed{\pink{ \pmb{\sf{ Volume\small_{(Area)} =Length×Breadth×Height}}}}}}$

Where :

• L = Length = 16cm

• B = Breadth = 09cm

• H = Height = 06cm

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ \qquad{\underline{\rule{150pt}{1pt}}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$✭ {\underline{\pmb{\frak{ Calculating \; the \; Area\; :- }}}}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} \dashrightarrow \; \; \sf {Volume\small_{(Area)} = Length×Breadth×Height } \\ \end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} \dashrightarrow \; \; \sf {Volume\small_{(Area)} = 16 \times 09 \times 06 } \\ \end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} \dashrightarrow \; \; \sf {Volume\small_{(Area)} = 16 \times 09\times 06 = 864 } \\ \end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} \dashrightarrow \; \; \sf {Volume\small_{(Area)} = 864 } \\ \end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \begin{gathered} {\qquad \; \; {\therefore \; {\underline{\boxed{\orange{\pmb{\frak{ Volume\small_{(Area)}= ₹ \; 864}}}}}}}} \\ \end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ \qquad{\underline{\rule{150pt}{1pt}}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$

~Therefore

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$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} \\ {\underline{\rule{212pt}{6pt}}} \end{gathered}\end{gathered}\end{gathered}\end{gathered}$