Answer:

Let a=8 cm , b =6 cm and ANGLE between

sides a and b is C. Area of triangle ABC =

1/2.a.b.sinC.

1/2.8 cm.6 cm.sinC = 12 cm^2

24 cm^2.sinC=12 cm^2

sinC= 12cm^2/24 cm^2= 1/2

C= 30° ,150° are only possible (as sum of all

interior angles is 180°).

Thus, only two TRIANGLES are possible in which

in 1st triangle angle between given sides is 30°

and in 2nd triangle angle between sides is

150°. , Answer.

Second method:-

Let third side is x cm long.

a= 8 cm , b = 6 cm and c = x cm.

s=(a+b+c)/2=(8+6+x)/2= 7+(x/2).

Area of triangle =[s (s-a) (s-b) (s-c)]^1/2

12 =[{7+(x/2)}.{x/2–1}.{x/2+1}.{7-x/2}]^1/2

12 =[{49-x^2/4}.{x^2/4–1}]^1/2

On squaring both sides

144 =49x^2/4–49-x^4/16+x^/4

193 =50x^2/4-x^4/16

x^4/16–50x^2/4+193 =0

(x^2/4)^2–2×(x^2/4)×25+(25)^2–625+193=0

(x^2/4 -25)^2= 432.

(x^2/4-25)= +/-12.(3^1/2) = +/-20.784

x^2/4 = 25 +/-20.784

x^2/4= 45.784 , 4.216

x^2= 183.136 , 16.864

x= 13.53 cm , 4.11 cm.

Hare both the length of 3RD side are possible

to construct the triangle, so that two triangles

are possible. , Answer.