Question

M, N and T are the midpoints of the sides of $\Delta \text{XYZ}$.

If $\text{XY}=\text{k}$, then$\text{TN}=\underset{¯}{?}$

Short answer

The length of TN is $\frac{\text{k}}{\mathrm{2}}$ units.

Step-by-step solution

Step 1. Consider the diagram.

$\Delta XYZ$is:

Here, M, N and T are midpoints of $\Delta XYZ$. And $XY=k$.

Step 2. Show the calculation.

Consider in triangles TXM and ZXY .

$\angle TXM=\angle ZXY$ (Common).

$\frac{TX}{ZX}=\frac{XM}{XY}=\frac{1}{2}$ (Midpoints).

Therefore, $\Delta TXM~\Delta ZXY$(SAS).

$\begin{array}{l}\angle XMT=\angle XYZ\\ \angle MTX=\angle YZX\end{array}$

Angle XMT and XYZ are the corresponding angles. Hence, line TM and $ZY$ or ZN are parallel to each other.

Consider in triangles NMY and ZXY .

$\angle NYM=\angle ZYX$ (Common).

$\frac{NY}{ZY}=\frac{YM}{YX}=\frac{1}{2}$ (Midpoints).

Therefore, $\Delta NMY~\Delta ZXY$(SAS).

$\begin{array}{l}\angle YNM=\angle YZX\\ \angle YMN=\angle YXZ\end{array}$

Angle YNM and YZX are the corresponding angles. Hence, line NM and ZX or ZT are parallel to each other.

Thus, MNZT is parallelogram.

Similarly, TNXM is also a parallelogram

$TN=XM$.

The side TN is calculated as follows:

$\begin{array}{c}TN=XM\\ =\frac{XY}{2}\\ TN=\frac{k}{2}\text{units}\end{array}$

Step 3. State the conclusion.

Therefore, the length of TN is $\frac{k}{2}$units.