Discussion

Explanation:

∆ABC is a right triangle, hence AB = 25 cm (15, 20 and 25 is a pythagorean triplet)

∴ CD = $\frac{\mathrm{AC}\times \mathrm{BC}}{\mathrm{AB}}$ = $\frac{15\times 20}{25}$ = 12 cm

In right ∆ACD
AD = 9 cm (9, 12 and 15 is a pythagorean triplet)

Inradius = $\frac{\mathrm{AD}+\mathrm{CD}-\mathrm{AC}}{2}$ = $\frac{9+12-15}{2}$ = 3 cm

In right ∆ACD
BD = 25 - 9 = 16 cm.

Inradius = $\frac{\mathrm{CD}+\mathrm{BD}-\mathrm{BC}}{2}$ = $\frac{12+16-20}{2}$ = 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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