Find the common difference of the sequence shown.
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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:
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
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
