Question
Download Solution PDFC वर काटकोन असलेल्या काटकोन त्रिकोण ABC मध्ये 4 सेमी त्रिज्या असलेले एक वर्तुळ कोरलेले आहे. जर AC = 12 सेमी, तर CB चे मूल्य आहे:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिलेल्याप्रमाणे:
वर्तुळाची त्रिज्या = 4 सेमी
AC = 12 सेमी;
∠ACB = 90°
वापरलेली संकल्पना:
स्पर्शिका वर्तुळाच्या त्रिज्येसोबत स्पर्श बिंदूवर काटकोन बनवते.
जर वर्तुळाच्या बाह्य बिंदूपासून दोन स्पर्शिका काढल्या तर त्या लांबीने समान असतात.
वापरलेला सूत्र:
पायथागोरस प्रमेय:
H2 = P2 + B2
येथे, P = लंब; B = पाया;
H = कर्ण
(a + b)2 = a2 + b2 + 2ab
गणना:
चतुर्भुज PCRO हे एक चौरस आहे.
PC = CR = 4 सेमी
AR = (AC - CR) = (12 - 4) = 8 सेमी
AR = AQ = 8 सेमी
△ABC मध्ये
(AB)2 = (AC)2 + (BC)2
⇒ (8 + x)2 = 122 + (4 + x)2
⇒ 64 + x2 + 16x = 144 + 16 + x2 + 8x
⇒ 64 + 16x = 160 + 8x
⇒ 8x = (160 - 64) = 96
⇒ x = 96/8 = 12
BC = (4 + x) = 4 + 12 = 16 सेमी
म्हणूनच, योग्य उत्तर 16 सेमी आहे.
Last updated on Jun 13, 2025
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