Question

6.

a. Name all triangles congruent to$\Delta \text{TNL}$

b. Suppose you are told that the area of$\Delta \text{NLM}$ is$\mathbf{17}\text{.}\mathbf{32}{\text{cm}}^2$ . What is the area of $\Delta \text{TKO}$.

Short answer

a. Triangles congruent to $\Delta \text{TNL}$ are $\Delta \text{MLN}$, $\Delta \text{LMK}$, $\Delta \text{NOM}$.

b. Area of $\Delta \text{TKO}$ is $\mathbf{69}.\mathbf{28}{\text{cm}}^2$.

Step-by-step solution

a. Step 1 - Consider the diagram.

In the $\Delta TNL$,

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

- State the explanation.

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 $\Delta TNL$.

- State the conclusion.

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

b. Step 1 - Consider the diagram.

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

- Show the 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 $ar\left(\Delta NLM\right)=17.32c{m}^2$.

So,

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

- State the conclusion.

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