The diagonals of a convex quadrilateral are mutually perpendicular. The sum of the lengths of the diagonals is 12. We want to find the maximum possible area of such a quadrilateral. Let us denote by x and y the lengths of the two diagonals. Then the area of the quadrilateral is the following function of x and y:

Maximum area =18

The area of the quadrilateral as a function of x and y = xy/2

Step-by-step explanation:

As the quadrilateral has mutually perpendicular diagonals, it is a rhombus. The area of a rhombus is denoted by the formula xy/2.

So, the area of the quadrilateral is xy/2.

As the sum of the sides is 12

x + y = 12

y = 12- x

So, the area, as a function of x alone, becomes x(12- x)/2

To find the maximum area, we find the derivative of the area function with respect to x and equate it to 0.

d/dx(x(12 - x)/2) = 0

d/dx(12x - x^2) = 0

12- 2x = 0

x = 6

The maximum area will then be 6(12- 6)/2 = 36/2 = 18

So, the maximum area is 18.

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150

step-by-step explanation:

this is not possible to solve with the info given.   you need at least an one angle somewhere.