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4.
(i)
By Pythagoras’ Theorem, $(2x)^{2} + (y-2)^{2} = (2\sqrt{65})^{2}$

$\therefore, 5x^{2} + y^{2}-2xy = 260$

(ii)

Perimeter $= 6x+2y = 60$

$\Rightarrow 5x^{2} + (30-3x)^{2}-2x(30-3x) = 260$

$x^{2} -12x +32 = 0$

$x = 4 ~\mathrm{or}~ x= 8$

When $x = 4, y = 18$

When $latex x = 8, y = 6 < x$, which is not possible given the question. Thus, rejected. $latex \therefore, x=4, y=18$