$$ABCD$$ is a rectangle region. $$P$$ is a point on the segment $$BC$$. $$PQRC$$ is also a rectangle region such that $$A$$, $$Q$$ lie on the opposite sides of the line $$BC$$. $$AD = 15$$, $$DR = 18$$. The perimeter of the region $$ABPQRDA$$ is

Solution

Let $$AB = h$$, $$BP = p$$ and $$PQ = k$$. Then $$BC = AD = 15$$, $$QR = PC = 15 - p$$, and $$D$$, $$C$$, $$R$$ are collinear with $$DR = h + k = 18$$. The perimeter is $$AB + BP + PQ + QR + RD + DA = h + p + k + (15 - p) + 18 + 15 = (h + k) + 48 = 66$$.

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