हिंदी

A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed.

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प्रश्न

A hemisphere of maximum possible diameter is placed over a cuboidal block of side 7 cm. Find the surface area of the solid so formed.

योग
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उत्तर

Given: Side of cuboidal block (cube) = 7 cm.

Hemisphere placed on it has maximum diameter = 7 cm → radius r = `7/2` = 3.5 cm.

Step-wise calculation:

1. Total surface area (TSA) of the cube = 6 × (7)2 

= 6 × 49

= 294 cm2

2. Area of the circular base (face) covered by the hemisphere = πr2

= π × (3.5)2

 = 12.25π cm2

3. Curved surface area (CSA) of the hemisphere = 2πr2

= 2 × 12.25π

= 24.5π cm2

4. Surface area of the combined solid = (TSA of cube) – (Area of face covered) + (CSA of hemisphere)

= 294 – 12.25π + 24.5π

= 294 + 12.25π cm2

5. If `π = 22/7`, 12.25π = 38.5, so numeric value = 294 + 38.5 = 332.5 cm2.

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अध्याय 17: Volumes and Surface Areas of Solids - EXERCISE 17A [पृष्ठ ७८९]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 17 Volumes and Surface Areas of Solids
EXERCISE 17A | Q 29. (i) | पृष्ठ ७८९
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