From the right angled isosceles triangle QRS of side QR = RS = a, the smaller triangle POS of side PO = OS = a/3 is cut out. The centroid of the resultant area will be

1
x \(\rm\frac{a}{3}\), y = \(\rm\frac{a}{3}\)
2
x = \(\frac{5}{18}\)a, y = \(\frac{13}{36}\)a
3
x \(\frac{5}{36}\)a, y = \(\frac{5}{18}\)a
4
x = \(\frac{13}{36}\)a, y = \(\frac{5}{18}\)a

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