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≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.