Question

The Lengths of the shorter altitude and the shorter side of parallelogram are 9cm and root 82 cm, respectively. The Length of a longer diagonal is 15 cm. What is the area of this parallelogram

Answer 1

Consider parallelogram ABCD. The shorter sides are AB and CD, so $AB=CD=\sqrt{82}$ cm.

The shorter altitudes are BH=CF=9 cm.

1. Consider right triangle FCD:

$CD^2=DF^2+CF^2,$

$\sqrt{82}^2=9^2+FD^2,\\ \\FD^2=82-81=1,\\ \\FD=1 cm.$

2. Consider right triangle FCA:

$CA^2=DF^2+FA^2,$

$15^2=9^2+FA^2,\\ \\FA^2=225-81=144,\\ \\FA=12 cm.$

3.

$FA=AD+FD,\\ \\AD=12-1=11 cm.$

4. Therefore, the area of parallelogram ABCD is

$A_{ABCD}=AD\cdot BH=11\cdot 9=99$ sq. cm.

Answer: 99 sq. cm.