(1) The first question. . . . The first time I read this topic, I was scared to pee. . . Goodbye. . . But the first question, I think, can still be worked out. . . . If two right-angled sides of a triangle are 6 and 8, which are also right-angled triangles, then it is easy to get that the hypotenuse is 10. . . . . Through the area, 6*8= 10*X, then the height of x as the hypotenuse is 4.8. . . . . .

(2) the second question. . . . I didn't expect such a large amount of calculation. . . Is there something wrong with my calculation method? . . . .

The BPQ area of triangle is required, and the first reaction is not directly calculated. Such a thing that only knows BQ is 2t tells me what to do! ! ! So its area must be calculated indirectly. Since D is the midpoint, you can calculate the size of Y by finding the areas of BCP and PQD, preferably because the areas of these three triangles are half the area of the whole triangle! ! ! ! !

But actually forget it, I found it is not so easy to find Bcp and triangle PQD, because they are not special, but when we calculate Bcp area, my first reaction is to take CP as the base. As long as the height is calculated according to CD, the area of BCP is easy to calculate, so we might as well make an auxiliary line with B as the vertex, make it high, and use O point to cross the extension line of CD. . . . .

But it's not hard to find! ! ! ! The BCO triangle and CBE triangle are actually congruent triangles, because. . See for yourself, so the height is 4.8. . . . Then the area of the triangle BCP is actually 0.5 * t * 4.8 = 2.4t .......

Finally figured it out! ! ! ! ! Then let's look at the area of PQD! ! ! ! Similarly, the triangle PQD should be based on QD, so QD is actually BD-BQ=5-2t! ! ! ! Isn't it amazing? ! ! !

But it seems that the height is not so easy to find, so we need to make an auxiliary line with P as the vertex and a vertical line to AB. Then through the similarity theorem of triangles, CD/PD=CE/ (the height of triangle PQD), the height can be expressed as 4.8*(5-t)/2, and the area of triangle PQD can also be expressed. . . .

I didn't calculate the relationship between y and t, I calculated the time when y is the largest, which is actually the time when the areas of the two triangles are the smallest! ! !

The relationship of easy-to-get y is 0.96t 2-4.8t+ 12 ... I don't know if there is any miscalculation, the opening of this parabola is downward, and there must be a lowest point! ! ! But I seem to forget that T has a range, between 0 and 5, and the lowest point t=2.5, so it meets the requirements! ! ! ! ! So y is the largest when t=2.5! ! ! ! ! !

The third question. . . . QD is 5-2t, PD is 5t, and the two triangles should be isosceles. . . I'm afraid it's not easy . Unless t=0. . . .