Please help with this question :)

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) =

ii) Half area of PQRS =

=

We have :

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

ad = ae + ed = y + 7

similar triangles therefor

ac/ab = ad/ae

6/4 = (y + 7) / y

cross multiply

6y = 4(y + 7)

6y = 4y + 28

2y = 28

y = 14

step-by-step explanation: