91Ó°ÊÓ

The perimeter of a parallelogram STOP is given by

$\text{Perimeterof}â–\pm STOP=ST+TO+PO+SP$

The opposite sides of a parallelogram are congruent.

In $â–\pm STOP$, $\stackrel{¯}{ST},\stackrel{¯}{PO}$and $\stackrel{¯}{SP},\stackrel{¯}{TO}$are opposite sides. Therefore, $\stackrel{¯}{ST}â‰\dots \stackrel{¯}{PO}$and $\stackrel{¯}{SP}â‰\dots \stackrel{¯}{TO}$, that is, $ST=PO$and $SP=TO$.

Substitute $ST$for $PO$and $SP$for $TO$into the equation of perimeter of a parallelogram STOP and simplify.

$\begin{array}{c}\text{Perimeterof}â–\pm STOP=ST+TO+PO+SP\\ =ST+SP+ST+SP\\ =2ST+2SP\\ =2\left(ST+SP\right)\end{array}$

It is given that STis $1\mathrm{cm}$ longer than SP. Let SP be $x\mathrm{cm}$then $ST=\left(x+1\right)\mathrm{cm}$.

Substitute x for SP, $x+1$ for ST and 54 for perimeter of $â–\pm STOP$.

$\begin{array}{c}\text{Perimeterof}â–\pm STOP=2\left(ST+SP\right)\\ 54=2\left(x+1+x\right)\\ 54=4x+2\\ 4x=52\\ x=13\end{array}$

Substitute 13 for x into $ST=\left(x+1\right)\mathrm{cm}$ to find the value of ST.

$\begin{array}{c}ST=x+1\\ =13+1\\ =14\mathrm{cm}\end{array}$

Therefore, the value of ST and SP is $13\mathrm{cm}$and $14\mathrm{cm}$respectively.