The radius of circle inscribed in a rhombus with diagonals of length 10 and 24, is ?

ABCD is a rhombus. AC and BD are the diagonals intersecting at O.

We know that, diagonals of rhombus bisect each other at 90°.

∠AOD = 90°

In ΔAOD,

AD² = OA² + OD²

∴ AD² = (12 cm)² + (5 cm)² = 144 cm² + 25 cm² = 169 cm²

⇒ AD = 13 cm

Let the radius of the inscribed circle be r cm.

∠OPD = ∠OPA = 90°       (Radius is perpendicular to the tangent at the point of contact)

Suppose PD = x cm

∴ AP = AD – PD = (13 – x) cm

In ΔOPD,

OP² + PD² = OD²

∴ r² + x² = (5)² = 25     ...(1)

In ΔOAP,

OP² + AP² = OA²

∴ r² + (13 – x)² = (12)² = 144    ...(2)

Subtracting (2) from (1), we get

r² + x² – r² – (13 – x)² = 25 – 144 = –  119

∴ x² – 169 – x² + 26 x = – 119

⇒ 26x = (169 – 119) = 50

From (1), we have

Thus, the radius of  inscribed circle is .