Chapter 5.3, Problem 6CE

a.

To determine

To Find: the lengths of three sides of $\Delta TNZ$ shown in figure such that M, N and T are midpoints of the sides of $\Delta XYZ$ .

Answer to Problem 6CE

  $NM=4$ , $TM=7$ and $TN=5$ .

Explanation of Solution

Given:

Figure is shown below.

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

  $XY=10,YZ=14\text{ and }XZ=8$ .

Calculation:

In $\Delta XYZ$ and $\Delta TNZ$,

  $\angle Z=\angle Z\therefore \left(\text{Common}\right)$

  $ZT=\frac{1}{2}ZX\therefore \left(T\text{ is the midpoint of ZX}\right)$

  $ZN=\frac{1}{2}YZ\therefore \left(\text{N is the midpoint of YZ}\right)$

  $\Delta XYZ\cong \Delta TNZ$ By SAS postulate.

Therefore, by congruent parts of congruent triangles $TN=\frac{1}{2}XY$ .

Similarly, $TM=\frac{1}{2}ZY$ and $NM=\frac{1}{2}ZX$ .

Length of TN is.

  $\begin{array}{l}TN=\frac{1}{2}XY\\ TN=\frac{1}{2}\times 10\\ TN=5\end{array}$

Length of TM is.

  $\begin{array}{l}TM=\frac{1}{2}ZY\\ TM=\frac{1}{2}\times 14\\ TM=7\end{array}$

Length of NM is.

  $\begin{array}{l}NM=\frac{1}{2}ZX\\ NM=\frac{1}{2}\times 8\\ NM=4\end{array}$

Therefore, the answer is $NM=4$ , $TM=7$ and $TN=5$ .

b.

To determine

To Find: the lengths of three sides of $\Delta MYN$ shown in figure such that M, N and T are midpoints of the sides of $\Delta XYZ$ .

Answer to Problem 6CE

  $MY=5$ , $NY=7$ and $NM=4$ .

Explanation of Solution

Given:

Figure is shown below.

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

  $XY=10,YZ=14\text{ and }XZ=8$ .

Calculation:

In $\Delta YZX$ and $\Delta YMN$.

  $\angle Z=\angle Z\therefore \left(\text{Common}\right)$

  $ZT=\frac{1}{2}ZX\therefore \left(T\text{ is the midpoint of ZX}\right)$

  $ZN=\frac{1}{2}YZ\therefore \left(\text{N is the midpoint of YZ}\right)$

  $\Delta XYZ\cong \Delta TNZ$ By Side-angle-Side postulate.

Therefore, by congruent parts of congruent triangles $TN=\frac{1}{2}XY$ .

Similarly, $TM=\frac{1}{2}ZY$ and $NM=\frac{1}{2}ZX$ .

Length of NM is.

  $\begin{array}{l}NM=\frac{1}{2}ZX\\ NM=\frac{1}{2}\times 8\\ NM=4\end{array}$

Length of MY is.

  $\begin{array}{l}MY=\frac{1}{2}XY\\ MY=\frac{1}{2}\times 10\\ MY=5\end{array}$

Length of NY is.

  $\begin{array}{l}NY=\frac{1}{2}YZ\\ NY=\frac{1}{2}\times 14\\ NY=7\end{array}$

Therefore, the answer is $MY=5$ , $NY=7$ and $NM=4$ .

(c)

To determine

To Find: the lengths of three sides of $\Delta XMT$ shown in figure such that M,N and T are midpoints of the sides of $\Delta XYZ$ .

Answer to Problem 6CE

  $XT=4$ , $MY=5$ and $NY=7$ .

Explanation of Solution

Given:

Figure is shown below.

that M,N and T are midpoints of the sides of $\Delta XYZ$ .

  $XY=10,YZ=14\text{ and }XZ=8$ .

Calculation:

In$\Delta YZX$ and $\Delta NTZ$.

  $\angle Z=\angle Z\therefore \left(\text{Common}\right)$

  $ZT=\frac{1}{2}ZX\therefore \left(T\text{ is the midpoint of ZX}\right)$

  $ZN=\frac{1}{2}YZ\therefore \left(\text{N is the midpoint of YZ}\right)$

  $\Delta XYZ\cong \Delta TNZ$ By SAS postulate.

Therefore, by CPCT $TN=\frac{1}{2}XY$ .

Similarly, $TM=\frac{1}{2}ZY$ and $NM=\frac{1}{2}ZX$ .

Length of XT is.

  $\begin{array}{l}XT=\frac{1}{2}ZX\\ XT=\frac{1}{2}\times 8\\ XT=4\end{array}$

Length of XM is.

  $\begin{array}{l}XM=\frac{1}{2}XY\\ MY=\frac{1}{2}\times 10\\ MY=5\end{array}$

Length of NY is.

  $\begin{array}{l}NY=\frac{1}{2}YZ\\ NY=\frac{1}{2}\times 14\\ NY=7\end{array}$

Therefore, the answer is $XT=4$ , $MY=5$ and $NY=7$ .

(d)

To determine

To Find: the lengths of three sides of $\Delta NMT$ shown in figure such that M,N and T are midpoints of the sides of $\Delta XYZ$ .

Answer to Problem 6CE

  $XT=4$ , $MY=5$ and $NY=7$ .

Explanation of Solution

Given:

Figure is shown below.

that M,N and T are midpoints of the sides of $\Delta XYZ$ .

  $XY=10,YZ=14\text{ and }XZ=8$ .

Calculation:

In$\Delta XYZ$ and $\Delta TNZ$.

  $\angle Z=\angle Z\therefore \left(\text{Common}\right)$

  $ZT=\frac{1}{2}ZX\therefore \left(T\text{ is the midpoint of ZX}\right)$

  $ZN=\frac{1}{2}YZ\therefore \left(\text{N is the midpoint of YZ}\right)$

  $\Delta XYZ\cong \Delta TNZ$ By SAS postulate.

Therefore, by CPCT $TN=\frac{1}{2}XY$ .

Similarly, $TM=\frac{1}{2}ZY$ and $NM=\frac{1}{2}ZX$ .

Length of NM is.

  $\begin{array}{l}NM=\frac{1}{2}ZX\\ NM=\frac{1}{2}\times 8\\ NM=4\end{array}$

Length of TM is.

  $\begin{array}{l}TM=\frac{1}{2}ZY\\ TM=\frac{1}{2}\times 14\\ TM=7\end{array}$

Length of TN is.

  $\begin{array}{l}TN=\frac{1}{2}XY\\ TN=\frac{1}{2}\times 10\\ TN=5\end{array}$

Therefore, the answer is $NM=4$ , $TM=7$ and $TN=5$ .