In the given figure PT : TS : SR = 2 : 1 : 1 and SU is parallel to TQ. If RU = 10 cm. RS = 8 cm and SU = 6 cm, then what is the value (in cm) of (PQ + QR)?
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e. 40.
By BPT theorem:
RU/QU = RS/ST
10/QU = 1/1
∴ QU = 10 cm
QR = QU + UR = 20 cm
RT = SR + ST = 8 + 8 = 16 cm [∵ ST = SR = 8 cm]
By BPT theorem, RS/RT = SU/TQ
½ = 6/TQ
∴ TQ = 12 cm
In triangle TQR, TQ = 12 cm, QR = 20 cm and TR = 16 cm
∴ ∠QTR = 90° [∵ The sides forms Pythagorean triplet]
∴ Triangle PQT is a right triangle
QT = 12 cm and PT = 2 × SR = 16 cm
PQ = √(QT2 + PT2) = 20 cm
Thus,
The value of (PQ + QR) = 20 + 20 = 40 cm