A line moving along the hypotenuse of a right triangle ABC is a triangle with sides $AB = 6 m$, $BC = 8m$, and $AC = 10m$. A line $k$ in the plane of the triangle $ABC$ moves along the segment $AC$ at the rate of $1cm$ per sec. The line starts at A and ends at $C$ and is always perpendicular to $AC$.


*

*How long does it take the line to reach the point B?

*How long does it take the line to bisect the area of the triangle?

*What is the area of the region that the line sweeps in its movement after $6$ min and $45$ sec?

*What is the area of the region that the line sweeps at any give moment from the start of its movement?


I have noticed that $(6,8,10)$ are Pythagorean triple and  have solved the first question (6 mints is the answer), but the last three seems a bit tedious, could somebody tell me which could be the most easiest approach to solve these three?
 A: Let $D_1$ be the point on $AC$ such that $D_1B \perp AC$. You have that by similar triangles, $AD_1 = 3.6m$, so you are correct that part 1 is 360 secs. 
Observe that the area $ABD_1 = \frac12 \times 3.6 \times 4.8 m^2$, and $BCD_1 = \frac12 \times 4.8 \times 6.4 m^2$. Let $D_2$ be the point on $AC$ such that the perpendicular line through $D_2$ to $AC$ bisects the triangle in area, notice that $D_2$ is between $C$ and $D_1$. The area of the right triangle cut out by $D_2$ and the corner $C$ is going to be $\frac12 (CD_2)^2 \times \frac{6}{8}$ using similar triangles. You set this to equal to half the total area $\frac12 \times 48$ and you get 
$$ (CD_2)^2 = 24 \times \frac{8}{6} = 32 $$
so $CD_2 = 4\sqrt{2}$. So the time to get there is $(10 - 4\sqrt{2})\times 100$ seconds which is roughly 435 secs. 
Question 3 you already solved as part of Question 1. 
Question 4 you have to treat the times before and after passing through $D_1$ separately. Before passing through $D_1$, you get a similar triangle to a $(6,8,10)$ triangle with the $6$ side replaced by $AD = $ the time passed in seconds multiplied by 1 centimeter per second. So the area is 
$$ \frac12 \times \frac{8}{6} \times \left(\frac{t}{100}\right)^2 $$
in meter squared. 
After passing through point $D_1$, the area swept is the total area minus the area remaining. This time you compute the area of a $(6,8,10)$ triangle with the $8$ side replaced by $1000 - AD$ centimeters, where $AD$ is equal to the time elapsed in seconds. 
A: Here is a hint. (Hint as in: it's quite a lot of work to find the precise answer, but this may help).
I claim $\triangle ABC \sim \triangle AYX $: note that the angles $\hat A$ are the same and the angles $A\hat BC$ and $A\hat YX$ are both $90^\circ$. So, if $XY$ is the sweeping line, then the area of $\triangle ABC$ is equal to that of $\triangle AYX$ multiplied with the similarity-factor, squared. In particular
$$ A(\triangle ABC)  \frac{ |AY|^2}{|AB|^2} = A(\triangle AYX)$$

(This allows you to determine the area of the sweeped region, before the line $XY$ passes the point $B$. After $B$ is passed, one needs a similar consideration in the other half of the triangle to have a complete description.)
