Question
Download Solution PDF∠BAC = 90° గా ఉన్న ఒక లంబకోణ త్రిభుజం ABC లో, AD అనేది BC కి లంబంగా ఉంది. ΔABC యొక్క వైశాల్యం 63 cm2, ΔACD యొక్క వైశాల్యం 7 cm2 మరియు AC = 5 cm అయితే, BC యొక్క పొడవు ఎంత?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFఇచ్చినవి:
ΔABC యొక్క వైశాల్యం = 63 cm2
ΔACD యొక్క వైశాల్యం = 7 cm2
AC = 5 cm
ఉపయోగించిన సూత్రం:
ΔABC ∼ ΔXYZ అయితే
ar(ΔABC) / ar(ΔXYZ) = (AB/XY)2 = (BC/YZ)2 = (AC/XZ)2
గణన:
ΔABC మరియు ΔACD లో
∠BAC = ∠ADC = 90°
∠C = ∠C (రెండు త్రిభుజాలలో ఉమ్మడిగా)
కాబట్టి, ΔABC ∼ ΔDAC (AA సామ్యం ద్వారా)
ఇప్పుడు,
ar(ΔABC) / ar(ΔACD) = (BC/AC)2
⇒ 63/7 = (BC/5)2
⇒ 9/1 = (BC/5)2
⇒ √(9/1) = BC/5
⇒ 3 = BC/5
⇒ BC = 5 x 3 = 15 cm
∴ BC యొక్క పొడవు 15 cm
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