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?
Answer & explanation
Correct answer: option 4
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.