Chapter 5.3, Problem 6WE

a.

To determine

To name: triangles congruent to $\Delta TNL$

Explanation of Solution

Given information:The given figure is.

  

  

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,

  $\Delta TNL\cong \Delta MLN$

Similarly,

  $\Delta LMK\cong \Delta MLN,\Delta MLN\cong \Delta NOM$

So,

  $\Delta MLN,\Delta LMK,\Delta NOM$ are congruent to $\Delta TNL$ .

b.

To determine

To calculate: Area of $\Delta TKO$

Answer to Problem 6WE

  $69.28{\text{cm}}^{\text{2}}$

Explanation of Solution

Calculation:

From part a.,

  $\Delta KLM,\Delta NML,\Delta ONM$ and $\Delta MLN$ are congruent to $\Delta TNL$ .

Also, congruent triangles have equal area.

So,

  $ar\left(\Delta KLM\right)=ar\left(\Delta NML\right)=ar\left(\Delta ONM\right)=ar\left(\Delta TNL\right)$

That is $ar\left(\Delta TKO\right)=4ar\left(\Delta NLM\right)$

Put $ar\left(\Delta NLM\right)=17.32{\text{cm}}^{\text{2}}$

So,

  $ar\left(\Delta TKO\right)=4\left(17.32\right)=69.28{\text{cm}}^{\text{2}}$