Step 1: Given data.
Radius, \( r = 3.5 \, \text{cm} \)
Height of the cylinder, \( h = 10 \, \text{cm} \)
Step 2: Total surface area (TSA) of the item.
Since hemispheres are taken out from both ends, there are no circular bases. Thus, \[ \text{TSA} = \text{Curved surface area of cylinder} + 2 \times \text{Curved surface area of hemisphere} \] Step 3: Write the formulas.
\[ \text{CSA of cylinder} = 2\pi rh \] \[ \text{CSA of one hemisphere} = 2\pi r^2 \] Therefore, \[ \text{TSA} = 2\pi rh + 2(2\pi r^2) = 2\pi r(h + 2r) \] Step 4: Substitute the given values.
\[ \text{TSA} = 2 \times 3.14 \times 3.5 (10 + 2 \times 3.5) \] \[ = 6.28 \times 3.5 \times 17 = 6.28 \times 59.5 = 373.66 \, \text{cm}^2 \] Step 5: Conclusion.
Hence, the total surface area of the item is approximately 373.66 cm$^2$.
The product of $\sqrt{2}$ and $(2-\sqrt{2})$ will be:
If a tangent $PQ$ at a point $P$ of a circle of radius $5 \,\text{cm}$ meets a line through the centre $O$ at a point $Q$ so that $OQ = 12 \,\text{cm}$, then length of $PQ$ will be:
In the figure $DE \parallel BC$. If $AD = 3\,\text{cm}$, $DE = 4\,\text{cm}$ and $DB = 1.5\,\text{cm}$, then the measure of $BC$ will be: