Connect PT and GT and extend TG to PQ, cross PQ to s, do TU⊥PQ to u, TV⊥QR to v,

∵T is the heart of △PQR, and g is the center of gravity of △PQT.

∴TU=TV=2, PS=GQ=5 (center of gravity, inner nature)

∫PQ = 10，QR=8，PR=6

∴△PQR is RT△, ∠ PRQ = 90 (Pythagorean theorem).

∵T is the heart of △PQR.

∴TR bisection ∠PRQ (intrinsic property)

∴∠PRT=∠QRT=45

∵TV⊥QR

∴TV=RV=2

∴QV=8-2=6

TU = TV = 2，QT=QT

∴△QTU≌△QTV(HL)

Qu = qv = 6 (congruence), Pu = 10-60 = 4, Su = 1 (line length).

∵TU⊥PQ

∴ PQ = 2 √ 5, ST = √ 5 (vertical)

∴△PST is RT△, ∠ pts = 90 (Pythagorean theorem).

∵G is the center of gravity of △ △PQT.

∴SG∶GT= 1∶2

∴ gt = (2 √ 5)/3 (the nature of the center of gravity)

∴ pg = (10 √ 2)/3 (Pythagorean theorem)

To solve the above problems, we must know what is the center (the center of the inscribed circle of a triangle), that is, the bisector of three internal angles, and the distance to each side is equal. One theorem is that the ratio of the distance from the center of gravity to the vertex to the distance from the center of gravity to the midpoint of the opposite side is 2: 1. Of course, there are Pythagorean theorems, congruence properties and so on.

I also graduated from grade three, which is quite predestined. Glad to answer your question, I hope you can adopt it!