(1) When k=4, the area of the quadrangle OECF is 8. ⑵∵E﹙k/3,3﹚F﹙4,k/4﹚。 Delta △CEF area =1/2 ~ 4-k/3 ~× 3-k/4 ~ =1/24k? Area ÷ k+6, △ OEF =12 ÷ k-1/24 k? ÷ k+6), the areas of △OEF and △CEF are equal, so 1/24 k? ÷k+6= 12÷k÷﹙ 1/24·k？ ÷ k+6 モ The solution is k = 12. At this time, e and f overlap ∴ there is no k value, which makes the areas of △OEF and △CEF equal.