Solution

(a) Given:
The perimeter of parallelogram $$ \mathrm{ABCD}=42 \mathrm{cm} $$
To find:
Lengths of the sides of the parallelogram ABCD. From fig (1) We know that, $$ A B=P $$
Then, perimeter of $$ \| \mathrm{gm} \mathrm{ABCD}=2(\mathrm{AB}+\mathrm{BC}) $$
$$\begin{array}{l}42=2(\mathrm{P}+\mathrm{BC}) \\42 / 2=\mathrm{P}+\mathrm{BC} \\21=\mathrm{P}+\mathrm{BC} \\\mathrm{BC}=21-\mathrm{P} \\\text { So, ar }(\| \mathrm{gm} \mathrm{ABCD})=\mathrm{AB} \times \mathrm{DM} \\=\mathrm{P} \times 6 \\=6 \mathrm{P} \ldots \ldots \ldots(1)\end{array}$$
Again, ar $$ (\| \mathrm{gm} \mathrm{ABCD})=\mathrm{BC} \times \mathrm{DN} $$
$$ =(21-\mathrm{P}) \times 8 $$
$$ =8(21-\mathrm{P}) \ldots \ldots \ldots $$(2)
From ( 1 ) and $$ (2), $$ we get
$$ 6 P=8(21-P) $$
$$ 6 P=168-8 P $$
$$ 6 P+8 P=168 $$
$$ 14 P=168 $$
$$ P=168 / 14 $$
$$ =12 $$
Hence, sides of $$ \| \mathrm{gm} $$ are
$$ \mathrm{AB}=12 \mathrm{cm} $$ and $$ \mathrm{BC}=(21-12) \mathrm{cm}=9 \mathrm{cm} $$