In a rhombus of side 10 cm, one of the diagonals is 12 cm long. The length of the second diagonals is

Explanation

Let ABCD be the given rhomus whose diagonals intersect at O. then, `AB=10 cm`
Let `AC=12 cm and BD=2xcm`.
We know that the diagonals of a rhombus bisect each other at right angles.
`:. OA=(1)/(2) AC=6 cm, OB=(1)/(2) BD= x cm, and angle AOB=90^(@)`
From right `Delta AOB`, we have
`AB^(2)=OA^(2)+OB^(2)`
`rArr OB^(2)=AB^(2)-OA^(2)`
`={(10)^(2)-(6)^(2)} cm^(2)=(100-36) cm^(2)=64 cm^(2)`
`rArr x^(2)=64 rArr x= sqrt(64)=8`
`:. OB=8 cm`
`:. BD=2xxOB=2xx8cm=16cm`
Hence, the length of the second diagonal is 16cm.