Hello,
i propose this exercise if anyone can answer me.
let abc be a right triangle, of hypotenuse bc = a, where .
let a' be the symmetric of a with respect to bc.
determine x so that:
-the area of the triangle aa'c is maximum
show and justify all the steps (take the picture).

you.

Step-by-step explanation:

Call the point of intersection of AA' and BC point X. Then ...

  CX = AC·cos(90°-x) = AC·sin(x)

and the area of AA'C is ...

  area = AC·CX·sin(90°-x) = AC²·sin(x)cos(x) = (1/2)AC²·sin(2x)

Obviously, area is maximized for 2x = π/2, or x = π/4 when AC is fixed.

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On the other hand, ...

  AC = AB·tan(x)

so the area of the triangle is ...

  area = (1/2)AC²·sin(2x) = (1/2)(AB·tan(x))²·sin(2x) = AB²·sin(x)³/cos(x)

For fixed AB, area approaches infinity as x approaches 90°.

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Comment on the attachments

The attached diagrams show AC=1 and B free to move. Values of x around 45° are shown. The number in the middle of the figure is the approximate area of ΔAA'C.

the answer is 1030 it's very simple math.

x=-2

step-by-step explanation:

i began by flipping the problem to give me x= on the left (keeps it clearer for me).   so my problem simply became 10x-3=4x-6.  

1)   subtract 4x from the right side, which equals 0.   you do the same to the left of the equal as the right; therefore, subtract 4x from 10x, leaving 6x on the left.

              10x-3=4x-6   :   4x-4x on the right leaves only the -6        

2)             6x-3= 0-6 :    

                6x+3=   -6

3) divide