CS = x, and since AB = BC = CD = DA = 1cm -> SD = 1 - x
We see that angle DSP = angle APQ (because they simultaneously equal to 90 when added angle DPS)
So, triangle DSP = triangle APQ (equal hypotenuse - square side and equal angle.)
Similarly, we can prove triangle SCE and QBE in the same way
So, all 4 of the triangles are equal to each other.
-> SD = CE = BQ = AP = 1-x
Calculate the area of 4 triangles : [ x(1-x) : 2 ] x 4 = x(1-x) x 2
Area (PQRS) = 1 - [ x(1-x)] x 2 = 1 - 2x + 2x(squared) = [tex]2x^{2} -2x + 1[/tex]
ii) Half area of PQRS = [tex]x^{2} -x+1[/tex]
= [tex]x^{2} -\frac{1}{2} x-\frac{1}{2} x + \frac{1}{2} = x(x-\frac{1}{2} )- \frac{1}{2}(x-\frac{1}{2} ) + \frac{1}{4} = (x-\frac{1}{2})^{2} + \frac{1}{4}[/tex]
We have : [tex](x-\frac{1}{2}) ^{2}[/tex]
has the smallest value of 0 -> The smallest half value of PQRS = 1/4 -> The smallest value of PQRS = 1/2
