Answer:
The answer is below
Step-by-step explanation:
A parallelogram is a quadrilateral (has 4 sides and 4 angle) with two pair of parallel and opposite sides. Opposite sides of a parallelogram are parallel and equal.
Given parallelogram ABCD:
AB = CD = 18 cm; BC = AD = 8 cm
∠P = ∠P, ∠PDA = ∠PCQ (corresponding angles are equal).
Hence ΔPCQ and ΔPDA are similar by angle-angle similarity theorem. For similar triangles, the ratio of their corresponding sides equal. Therefore:
[tex]\frac{CD}{PC}= \frac{AD}{CQ}\\\\\frac{18}{6}=\frac{8}{x} \\\\x=\frac{6*8}{18}=\frac{8}{3}\ cm[/tex]
Perimeter of CPQ = CP + CQ + PQ
15 = 6 + 8/3 + PQ
PQ = 15 - (6 + 8/3)
PQ = 6.33
∠CQP = ∠AQB (vertical angles), ∠QCP = ∠QBA (alternate angles are equal).
Hence ΔCPQ and ΔABQ are similar by angle-angle similarity theorem
[tex]\frac{AQ}{QP}=\frac{AB}{CP} \\\\\frac{AQ}{6.33} =\frac{18}{6} \\\\AQ=\frac{18}{6}*6.33\\\\AQ = 19[/tex]
[tex]\frac{BQ}{CQ}=\frac{AB}{CP} \\\\\frac{BQ}{8/3} =\frac{18}{6} \\\\BQ=\frac{18}{6}*\frac{8}{3} \\\\BQ =8[/tex]
Perimeter of BAQ = AB + BQ + AQ = 18 + 8 + 19 = 45cm
PA = AQ + PQ = 19 + 6.33 = 25.33
PD = CD + DP = 18 + 6 = 24
Perimeter of PDA = PA + PD + AD = 24 + 25.33 + 8 = 57.33 cm
