In the figure, ACB is a right triangle. CD is the altitude. The circles are inscribed within the triangles ACD and BCD. P and Q are the centres of the circles. Find the distance PQ.
Explanation:
∆ABC is a right triangle, hence AB = 25 cm (15, 20 and 25 is a pythagorean triplet)
∴ CD = AC×BCAB = 15×2025 = 12 cm
In right ∆ACD AD = 9 cm (9, 12 and 15 is a pythagorean triplet)
Inradius = AD+CD-AC2 = 9+12-152 = 3 cm
In right ∆ACD BD = 25 - 9 = 16 cm.
Inradius = CD+BD-BC2 = 12+16-202 = 4 cm
In ∆PQR PR = 3 + 4 = 7 QR = 4 - 3 = 1
PQ2 = PR2 + QR2 ⇒ PQ = √50 = 5√2
Hence, option (c).
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