Question

& From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the
same height and same diameter is hollowed out. Find the total surface area of the
remaining solid to the nearest cm.
​

Answer 1

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Your correct answer is in attachment. ✔

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.

$\huge \sf\red{learn \: more}$

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0 \end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}$

Answer 2

Answer:

Refer to attachment

Step-by-step explanation:

Hope it will help you ....