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(1) As shown in the figure,

(2) a' ma is an isosceles right triangle, AM=A'M=a-X,

A'A=√2AM=√2(a-X),

sδa'ma= 1/2(a-x)^2,

C'C=√2CN=√2(b-X),

Trapezoidal height aa' c' c:1/2a' a+√ 2bm+1/2c' c

= 1/2(a-X)+ì2X+ 1/2(b-X)

= 1/2(a+b)+(√2- 1)X,

S trapezoid =1/2 [√ 2 (a-x)+√ 2 (b-x)] [1/2 (a+b)+(√ 2-1) x]

= 1/4[√2(a+b)-2√2X][(a+b)+2(√2- 1)X]

=√2/4[a+b-2X][a+b-2(√2- 1)X].