The correct option is A True
Let a,b,c be the length of the sides of the triangle, s be the semi perimeter and Δ be the area.
r1=Δ(s−a)=24(12−a)r2=Δ(s−b)=Δ(12−b)r3=Δs−c=24(12−c)
Since r1,r2,r3 are in H.P.
⇒1r1,1r2,1r3 are in A.P.
⇒2r1=1r1+1r3 or 2×12−b24=12−a24+12−c24
Or a+c=2b .....(1)
Also, a+b+c=24 ....(2)
From (1) and (2), we get b=8cm and a+c=16cm
Now Δ√s(s−a)(s−b)(s−c)
⇒24=√12(12−a)(12−b)(12−c)⇒24×24=12(12−a)×4×(12−16+a)⇒12=(12−a)(a−4)⇒a2−16a+60=0⇒a=6 or 10.
When a=6,c=10 and when a=10,c=6
Hence sides are 6,8,10 or 10,8,6