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A wooden cubical die is formed by forming hemispherical depressions on each face of the cube such that face $1$ has one depression, face $2$ has two depressions and so on. The sum of number of hemispherical depressions on opposite faces is always $7$. If the edge of the cubical die measures $5$ cm and each hemispherical depression is of diameter $1.4$ cm, find the total surface area of the die so formed.
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Number of hemispherical depressions = $21$
Total surface area of the die formed
= TSA of cube + CSA of $21$ hemispheres – Base area of $21$ hemispherical depressions
$= 6 \times 5^2 + 21 \times 2 \times \frac{22}{7} \times (0.7)^2 - 21 \times \frac{22}{7} \times (0.7)^2$
$= 182.34$ cm$^2$
Total surface area of the die formed
= TSA of cube + CSA of $21$ hemispheres – Base area of $21$ hemispherical depressions
$= 6 \times 5^2 + 21 \times 2 \times \frac{22}{7} \times (0.7)^2 - 21 \times \frac{22}{7} \times (0.7)^2$
$= 182.34$ cm$^2$