Question:
In a triangle PQR, points M and N lie on QR such that \(PM = PN\) and \(\angle QPM = \angle RPN\). If \(PQ = (3x + 2)\) units, \(QM = x\) units, \(PR = (2y + 5)\) units, \(RN = y\) units, find the value of \((x + y)\).
त्रिभुज PQR में, बिंदु M और N भुजा QR पर इस प्रकार स्थित हैं कि \(PM = PN\) और \(\angle QPM = \angle RPN\) है। यदि \(PQ = (3x + 2)\) इकाई, \(QM = x\) इकाई, \(PR = (2y + 5)\) इकाई, \(RN = y\) इकाई है, तो \((x + y)\) का मान ज्ञात कीजिए।
Options:
- 4
- 5
- 6
- 3
- 4
- 5
- 6
- 3
Correct Answer: c)
Explanation:
Since \(PM = PN\), \(\angle PMN = \angle PNM\), which implies their supplementary exterior angles are equal: \(\angle PMQ = \angle PNR\).
In \(\triangle PQM\) and \(\triangle PRN\):
1. \(\angle QPM = \angle RPN\) (given)
2. \(PM = PN\) (given)
3. \(\angle PMQ = \angle PNR\)
By ASA congruence criterion, \(\triangle PQM \cong \triangle PRN\).
Therefore, corresponding sides are equal:
\(QM = RN \implies x = y\)
\(PQ = PR \implies 3x + 2 = 2y + 5\).
Substitute \(y = x\):
\(3x + 2 = 2x + 5 \implies x = 3\).
Thus, \(y = 3\) and \(x + y = 3 + 3 = 6\).
चूँकि \(PM = PN\), इसलिए \(\angle PMN = \angle PNM\), जिससे \(\angle PMQ = \angle PNR\) होगा।
\(\triangle PQM\) और \(\triangle PRN\) में:
\(\angle QPM = \angle RPN\), \(PM = PN\), तथा \(\angle PMQ = \angle PNR\)।
ASA सर्वांगसमता से, \(\triangle PQM \cong \triangle PRN\)।
अतः संगत भुजाएँ बराबर होंगी:
\(QM = RN \implies x = y\)
\(PQ = PR \implies 3x + 2 = 2y + 5\)।
\(y = x\) रखने पर: \(3x + 2 = 2x + 5 \implies x = 3\)।
अतः \(y = 3\) और \(x + y = 3 + 3 = 6\)।