Question:
In triangle PQR, the straight line parallel to side QR meets PQ and PR at points ‘A’ and ‘B’, respectively. If PA = BR, PQ = 36 cm and PB = 6 cm, then BR = ?
In triangle PQR, the straight line parallel to side QR meets PQ and PR at points ‘A’ and ‘B’, respectively. If PA = BR, PQ = 36 cm and PB = 6 cm, then BR = ?
Updated On: Nov 12, 2025
- 12 cm
- 10 cm
- 16 cm
- 8 cm
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The Correct Option is A
Solution and Explanation
According to the question,
In triangle PAB and triangle PQR,
∠PAB = ∠PQR (corresponding angles)
∠PBA = ∠PRQ (corresponding angles)
Therefore, triangle PAB ~ triangle PQR (AA similarity)
Therefore, PA/PQ = PB/PR
Or, (PQ/PA) = (PR/PB)
Or, (PQ/PA) - 1 = (PR/PB) - 1
Or, (PQ/PA) - 1 = {(PR - PB)/PB}
Or, (PQ/PA) - 1 = (BR/PB)
Or, (36/BR) - 1 = (BR/PB) (Since, PA = BR, given)
Let BR = ‘a’
Therefore, (36/a) - 1 = (a/6)
Or, 6(36 - a) = a2
Or, a2 + 6a - 216 = 0
Or, a² – 12a + 18a – 216 = 0
Or, a(a – 12) + 18(a – 12) = 0
Or, (a + 18)(a – 12) = 0
Or, a = 12 (Since, ‘a’ cannot be negative)
Therefore, BR = 12 cm
So, the correct option is (A) : 12 cm.
In triangle PAB and triangle PQR,
∠PAB = ∠PQR (corresponding angles)
∠PBA = ∠PRQ (corresponding angles)
Therefore, triangle PAB ~ triangle PQR (AA similarity)
Therefore, PA/PQ = PB/PR
Or, (PQ/PA) = (PR/PB)
Or, (PQ/PA) - 1 = (PR/PB) - 1
Or, (PQ/PA) - 1 = {(PR - PB)/PB}
Or, (PQ/PA) - 1 = (BR/PB)
Or, (36/BR) - 1 = (BR/PB) (Since, PA = BR, given)
Let BR = ‘a’
Therefore, (36/a) - 1 = (a/6)
Or, 6(36 - a) = a2
Or, a2 + 6a - 216 = 0
Or, a² – 12a + 18a – 216 = 0
Or, a(a – 12) + 18(a – 12) = 0
Or, (a + 18)(a – 12) = 0
Or, a = 12 (Since, ‘a’ cannot be negative)
Therefore, BR = 12 cm
So, the correct option is (A) : 12 cm.
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