A math problem in the ninth grade

The intersection of the straight line y=-x+6 with the X axis A (6,0) and the Y axis (0,6).

The straight line n is y=-x+t, its intersection with the X axis is D(t, 0), and its intersection with the Y axis is (0, t).

The point p of CD is (t/2, t/2).

So: CP=DP=√2t/2.

So: the center p is on the straight line y = X.

Because: OP is perpendicular to the lines n and l.

Therefore, the tangent point of the straight line l must be on the straight line y = X.

So: the tangent point e is (3,3)

So: PE=CP=DP=√2t/2.

So: (3-t/2) 2+(3-t/2) 2 = (√ 2t/2) 2.

So: (3-t/2) 2 = (t/2) 2.

So: 3-t/2=t/2.

Solution: t=3

3. 1) center P(t/2, t/2), radius R=√2t/2.

So the semicircle area:

S=π(R^2)/2=π(t^2)/4

s=π(t^2)/4,0<； = t & lt=6

3.2)CD=√2t，AB=6√2

Height of trapezoid h=AD/√2=(6-t)/√2.

Trapezoidal area S=(√2t+6√2)*(6-t)/√2/2.

=(t+6)(6-t)/2

= 18-(t^2)/2

According to the meaning, π (T2)/4 = (π/4) * [18-(T2)/2]

So: t 2 = 18-(t 2)/2.

So: t 2 = 12.

Solution: t=2√3