Question:
A circle is inscribed in triangle PQR, touching PQ, QR and PR at the points X, Y and Z, respectively. If PQ – QR = 3 cm, PQ – PR = 1 cm and the perimeter of triangle PQR = 32 cm, then QY + PZ in cm is:
त्रिभुज PQR के अंतर्गत एक वृत्त खींचा गया है, जो भुजाओं PQ, QR और PR को क्रमशः बिंदुओं X, Y और Z पर स्पर्श करता है। यदि PQ – QR = 3 सेमी, PQ – PR = 1 सेमी और त्रिभुज PQR का परिमाप = 32 सेमी है, तो (QY + PZ) का मान सेमी में क्या है?
Options:
- 10
- 12
- 14
- 15
- 10
- 12
- 14
- 15
Correct Answer: b)
Explanation:
Let \(PQ = c, QR = a, PR = b\).
\(a = c – 3\) and \(b = c – 1\).
Perimeter \(a + b + c = 32 \implies (c – 3) + (c – 1) + c = 32 \implies 3c – 4 = 32 \implies c = 12\) cm.
Sides: \(PQ = 12\), \(QR = 9\), \(PR = 11\) cm. Semi-perimeter \(s = 16\) cm.
By tangent segments: \(QY = s – PR = 16 – 11 = 5\) cm.
\(PZ = s – QR = 16 – 9 = 7\) cm.
\(QY + PZ = 5 + 7 = 12\) cm.
माना \(PQ = c\), तब \(QR = c – 3\) और \(PR = c – 1\)।
परिमाप \(= 3c – 4 = 32 \implies c = 12\) सेमी।
भुजाएँ: \(PQ = 12\), \(QR = 9\), \(PR = 11\) सेमी। अर्ध-परिमाप \(s = 16\) सेमी।
स्पर्श रेखा खंड: \(QY = s – PR = 16 – 11 = 5\) सेमी, \(PZ = s – QR = 16 – 9 = 7\) सेमी।
\(QY + PZ = 5 + 7 = 12\) सेमी।