Question
Download Solution PDFఒకవేళ పొడవు = 4 సెం.మీ, వెడల్పు = 6 సెం.మీ, ఎత్తు = 6 సెం.మీ కొలత ఉన్న ఘనాభాసంను అర్ధగోళంలో ఉంచండి. అర్ధగోళం యొక్క వ్యాసార్థాన్ని కనుగొనండి?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFఇచ్చినది:
ఘనాభాసం కలిగి ఉన్న కొలతలు:
పొడవు = 4 సెం.మీ., వెడల్పు = 6 సెం.మీ., ఎత్తు = 6 సెం.మీ
ఉపయోగించిన సూత్రం:
అర్ధగోళం యొక్క వ్యాసార్థం = \(1\over2\) x (√4h2 + l2 + b2)
గణన:
అర్ధగోళం యొక్క 2D చిత్రం:-
అర్ధగోళం యొక్క వ్యాసార్థం = \(\sqrt{(l/2)^2 + (b/2)^2 + h ^2}\)
అర్ధగోళం యొక్క వ్యాసార్థం = \(1\over2\) x ( √4h 2 + l 2 + b 2 )
⇒ \(1\over2\) x √{4 x (6)2 } + (4)2 + (6)2
⇒ \(1\over2\) x √144 + 16 + 36
⇒ \(1\over2\) x √ 196 = \(1\over2\) x 14 = 7 సెం.మీ.
∴ అర్ధగోళం యొక్క వ్యాసార్థం = 7 సెం.మీ
Last updated on Jun 12, 2025
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