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.

he has now walked a total of what?

if i know what this is maybe i can

step-by-step explanation:

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).