In a right-angled triangle PQR, ∠Q = 90°. A and B are the mid-points of PQ and PR, respectively. If PQ = 16 cm, QR = 30 cm and PR = 34 cm, what is perimeter (in cm) of the trapezium ABRQ?

70

Considering \(\Delta \)PQR,

\( {PQ }^{2 } \) = \( {QR }^{2 } \) + \( {PR }^{2 } \)

\( {16 }^{2 } \) = \( {30 }^{2 } \) + \( {PR }^{2 } \)

PR = \(\sqrt {1156 }\)

PR = 34

According to the question,

PA = AQ = \(\frac{16}{2}\) =  8 cm

PB = BR = \(\frac{34}{2}\) = 17 cm

Now, \(\Delta \)PAB is a right angled triangle at A.

Considering \(\Delta \)PAB ,

\( {PA }^{2 } \) = \( {AB }^{2 } \) + \( {PB }^{2 } \)

\( {8 }^{2 } \) = \( {AB }^{2 } \) + \( {17 }^{2 } \)

\( {AB }^{2 } \) = 289 - 64

AB = \(\sqrt {225 }\)

AB = 15

Now, the perimeter of the trapezium ABRQ = 15 + 17 + 30 + 8 = 70cm

Therefore, the perimeter of trapezium ABRQ is 70 cm.