In a triangle PQR, PQ = PR and the perimeter of ΔPQR is 8(2 + \(\sqrt {2}\))cm. If the length of QR is \(\sqrt {2}\) times the length of PQ, then find the half of area of ΔPQR.

16

Perimeter of ΔPQR = 2 + \(\sqrt {2}\)

ATQ,

Perimete = (2 + \(\sqrt {2}\)) unit = 8(2 + \(\sqrt {2}\)) cm

1 unit = 8 cm

area of ΔPQR = \(\frac{1}{2}\) × b × h = \(\frac{1}{2}\) × 8 × 8 = 32 cm²

Half of area of  ΔPQR = \(\frac{1}{2}\) × 32 = 16 cm²