एक समकोण त्रिभुज की भुजाएँ (सेमी में) (x − 13), (x − 26) और x हैं। इसका क्षेत्रफल (सेमी2 में) है:

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RRB NTPC Graduate Level CBT-I Official Paper (Held On: 06 Jun, 2025 Shift 1)
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  1. 1012
  2. 999
  3. 1010
  4. 1014

Answer (Detailed Solution Below)

Option 4 : 1014
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दिया गया है:

समकोण त्रिभुज की भुजाएँ: (x − 13), (x − 26), और x सेमी

प्रयुक्त सूत्र:

पाइथागोरस प्रमेय: कर्ण2 = आधार2 + ऊँचाई2

क्षेत्रफल = (1/2) × आधार × ऊँचाई

गणना:

कर्ण = x, आधार = (x - 26), ऊँचाई = (x - 13)

मान लीजिए कर्ण = x

⇒ x2 = (x − 13)2 + (x − 26)2

⇒ x2 = (x2 − 26x + 169) + (x2 − 52x + 676)

⇒ x2 = 2x2 − 78x + 845

⇒ सभी पदों को एक तरफ ले जाएँ:

⇒ x2 − 2x2 + 78x − 845 = 0

⇒ −x2 + 78x − 845 = 0

⇒ x2 − 78x + 845 = 0

द्विघात सूत्र का उपयोग करके हल करें:

x = [78 ± √(782 − 4 × 1 × 845)] ÷ 2

x = [78 ± √(6084 − 3380)] ÷ 2

x = [78 ± √2704] ÷ 2

x = [78 ± 52] ÷ 2

⇒ x = (78 + 52)/2 = 130 ÷ 2 = 65 (मान्य)

⇒ x = (78 − 52)/2 = 26 ÷ 2 = 13 (अमान्य क्योंकि भुजाएँ 0 हो जाती हैं)

यदि x = 13 है, तो एक भुजा (x - 13) = (13 - 13) = 0 है, जो एक त्रिभुज के लिए संभव नहीं है। इसलिए, x ≠ 13 है। 

इसलिए, x = 65

अब, भुजाओं की लंबाई ज्ञात कीजिए:

भुजा 1 = x - 13 = 65 - 13 = 52 सेमी

भुजा 2 = x - 26 = 65 - 26 = 39 सेमी

कर्ण = x = 65 सेमी

क्षेत्रफल = (1/2) × 39 × 52 = 1014 सेमी2

∴ क्षेत्रफल = 1014 सेमी2

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