Let the area of elapsed time t be 8 cm.

s= 1/2*PB*QB

PB = 6t QB = 2t

s=(6-t)*2t=8

Solution: t=4 or 2

The second question:

According to the meaning of the question, let t seconds make the PCQ area of the triangle equal to 12.6 square centimeters.

Point P starts from point A and moves along the AB side to B at the speed of 1cm/s, and the total time spent on the AB side is 6s.

Point Q starts from point B and moves along BC to point C at a speed of 2 cm/s, and the total time spent by BC is 4s,

Therefore, we divide the time into three sections:

1, between 1-4s.

At this time, P is on AB and Q is on BC.

At this time, the area of triangle PCQ = (6-1* t) * (2 * t)/2 =12.6.

That is: t 2-6t+ 12.6 = 0 (where t 2 refers to the square of t).

You will find that the delta of the equation is negative when you solve it! (no solution! )

After 2.6s seconds.

This is, point P is on BC, point Q is on CA,

The ratio of the height of the triangle PCQ corresponding to the bottom of PC to the length of QC is 3: 5 (Pythagorean theorem).

Then, PC = (6+8)-1* t =14-t.

QC=2*t-8=2t-8

The triangle PCQ corresponds to the height h of the bottom of PC = (2t-8) * (3/5).

The area of triangle PCQ is:

( 14-t)*[(2t-8)*(3/5)]/2 = 12.6

That is: t 2- 18t+85 = 0 (where t 2 refers to the square of t).

You will find that the delta of the equation is negative when you solve it! (no solution! )

Between 3, 4-6 s

At this time, point P is on AB, point Q is on CA, and triangle PQC is an oblique triangle! Can't directly calculate its area! Only through triangle APQ and triangle PBC can we find its area indirectly!

Triangle area PQC = triangle area ABC- triangle area PQC triangle area PBC

The area of (1) triangle ABC =6×8/2=24.

(2) Area of triangle APQ:

The ratio of the height of the triangle APQ corresponding to the base of AQ to the length of AP is 4: 5 (Pythagorean theorem).

AQ=( 10+8)-2×t (10 is the length of AC, which is also calculated by Pythagorean theorem).

= 18-2t

AP= 1*t=t

The triangle APQ corresponds to the height of the AQ base =t*4/5.

The area of triangle APQ =( 18-2t)*t*(4/5)/2.

=(36/5)*t-(4/5)*t^2

(3) Area of triangular PBC:

PB*BC/2=(6-t)*8/2=24-4t

Then, the area of the triangle pqc = 48-[(36/5) * t-(4/5) * t 2]-(24-4t).

=24-( 16/5)×t-(4/5)*t^2= 12.6

Namely: (4/5) * T2+(16/5) * t-11.4 = 0.

Solution, t= root number 73/4-2 or -2- root number 73/4 (time cannot be negative! The second one is Zeng Gen! )

So, the answer is the root number 73/4-2.