Question

Please help with this question :)

Answer 1

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) =
`2x^(2) -2x + 1`

ii) Half area of PQRS =
`x^(2) -x+1`

=
`x^(2) -(1)/(2) x-(1)/(2) x + (1)/(2) = x(x-(1)/(2) )- (1)/(2)(x-(1)/(2) ) + (1)/(4) = (x-(1)/(2))^(2) + (1)/(4)`

We have :
`(x-(1)/(2)) ^(2)`

has the smallest value of 0 -> The smallest half value of PQRS = 1/4 -> The smallest value of PQRS = 1/2