Hello,

to #13:

let x and y be the lengthes of the diagonals. Then the area of the quadrilateral is calculated by:

a = ½ * x * y

From the given conditions you know: x + y = 6 ==> y = 6 - x

Plug in the term for y into the first equation:

a(x) = ½*x*(6 - x) = -½x² + 3x

The graph of this function is a parabola opening downwards. The maximum value is at it's vertex: V(3, 4.5)

You get: x = 3, y = 3 the maximum area is 4.5 cm²

The quadrilateral is a square with diagonals which are 3 cm long.