In triangle ABC, P is mid-point of AB and Q is mid-point of AC. If AB = 9.6 cm, BC = 11 cm and AC = 11.2 cm; find the perimeter of the trapezium PBCQ.

Given: P is midpoint of AB, so AP = PB.

Q is midpoint of AC, so AQ = QC.

AB = 9.6 cm, BC = 11 cm and AC = 11.2 cm.

Since P is the midpoint of AB:

⇒ PB = AP = $\dfrac{AB}{2}$ = $\dfrac{9.6}{2}$ = 4.8 cm

Since Q is the midpoint of AC:

⇒ AQ = QC = $\dfrac{AC}{2}$ = $\dfrac{11.2}{2}$ = 5.6 cm

Using mid point theorem, PQ is parallel to BC, so:

PQ = $\dfrac{1}{2}$ BC

PQ = $\dfrac{1}{2}$ x 11 = 5.5 cm

The perimeter of the trapezium PBCQ = PB + BC + CQ + QP

= 4.8 + 11 + 5.6 + 5.5 cm

= 26.9 cm

Hence, the perimeter of the trapezium PBCQ = 26.9 cm.