91Ó°ÊÓ

In the $\mathrm{Δ}TNL$,

$\begin{array}{l}TL=8\\ TN=5\\ NL=7\end{array}$

In quadrilateral $TLMN$,

$\begin{array}{l}TL=MN=8\\ TN=LM=5\end{array}$

So, $TLMN$ is a parallelogram (quadrilateral in which opposite sides are equal is a parallelogram).

$LN$is a diagonal in parallelogram $TLMN$ and diagonal of a parallelogram divides it into two congruent triangles.

Therefore,

$\mathrm{Δ}TNLâ‰\dots \mathrm{Δ}MLN$

Similarly,

$\mathrm{Δ}LMKâ‰\dots \mathrm{Δ}MLN$,$\mathrm{Δ}MLNâ‰\dots \mathrm{Δ}NOM$ .

So,$\mathrm{Δ}MLN$ , $\mathrm{Δ}LMK$, $\mathrm{Δ}NOM$ are congruent to $\mathrm{Δ}TNL$.

Therefore, the triangles$\mathrm{Δ}MLN$ , $\mathrm{Δ}LMK$, $\mathrm{Δ}NOM$ are congruent to $\mathrm{Δ}TNL$.

Here, area of $\mathrm{Δ}NLM$ is $17.32{\text{cm}}^2$.

From part a.,

$\mathrm{Δ}KLM$ , $\mathrm{Δ}NML$, $\mathrm{Δ}ONM$and $\mathrm{Δ}MLN$are congruent to$\mathrm{Δ}TNL$.

Also, congruent triangles have equal area.

So,

$ar\left(\mathrm{Δ}KLM\right)=ar\left(\mathrm{Δ}NML\right)=ar\left(\mathrm{Δ}ONM\right)=ar\left(\mathrm{Δ}TNL\right)$

That is$ar\left(\mathrm{Δ}TKO\right)=4ar\left(\mathrm{Δ}NLM\right)$

Put $ar\left(\mathrm{Δ}NLM\right)=17.32c{m}^2$.

So,

$\begin{array}{c}ar\left(\mathrm{Δ}TKO\right)=4\left(17.32\right)\\ =69.28c{m}^2\end{array}$

Therefore, area of $\mathrm{Δ}TKO$ is $69.28c{m}^2$.