Height of hypotenuse: 3× 4 ÷ 5 = 2.4

Make two high AE, DF,

Let BE=X, then CF=4.5-0.5-X=4-X,

In rt Δ Abe, AE 2 = 3 2-x 2,

In rt δ DCF, df 3 = 5 2-(4-x) 2,

AE = DF，

∴9-X^2=25- 16+8X-X^2

8X=0,AE^2=9,AE=3，

This trapezoid is a right-angled trapezoid with a height of 3.

The circle of 2.8㎝ has no holes.