Question

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 the 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.

Answer 1

Face 1 = One hemisphere

Face 2 = Two hemispheres

Face 3 − Three hemispheres

Side of cube = 5 cm

Surface area of cube = 6a²

= 6 × 5 × 5

= 150 cm²

Total holes = 1 + 2 + 3 + 4 + 5 + 6

= 21

Radius of hemisphere = `(1.4)/2`

= 0.7 cm

∴ Total number of depressions = 21

Surface area of one hemisphere = 2πr²

Total Surface area of 21 hemispheres = 21 × 2π (0.7)²

= `42 xx 22/7 xx 0.7 xx 0.7`

= 64.68 cm²

The combined area of all the circles removed from the cube = 21 × πr²

= `21 × 22/7 xx 0.7 xx 0.7`

= 32.34 cm²

Total Surface area of the die = Area of the cube + Surface area of the hemisphere − Area of the circle

= 150 + 64.68 − 32.34

= 150 + 32.34

= 182.34 cm²

Hence, the total surface area of the die is 182.34 cm².