Consider parallelogram ABCD. The shorter sides are AB and CD, so [tex]AB=CD=\sqrt{82}[/tex] cm.
The shorter altitudes are BH=CF=9 cm.
1. Consider right triangle FCD:
[tex]CD^2=DF^2+CF^2,[/tex]
[tex]\sqrt{82}^2=9^2+FD^2,\\ \\FD^2=82-81=1,\\ \\FD=1 cm.[/tex]
2. Consider right triangle FCA:
[tex]CA^2=DF^2+FA^2,[/tex]
[tex]15^2=9^2+FA^2,\\ \\FA^2=225-81=144,\\ \\FA=12 cm.[/tex]
3.
[tex]FA=AD+FD,\\ \\AD=12-1=11 cm.[/tex]
4. Therefore, the area of parallelogram ABCD is
[tex]A_{ABCD}=AD\cdot BH=11\cdot 9=99[/tex] sq. cm.
Answer: 99 sq. cm.
