Chapter 5.3, Problem 1WE

To determine

To fill the blanks: with the measure of sides for the given figure .

Answer to Problem 1WE

  $\overline{AB}=12$

  $\overline{EF}=12$

Explanation of Solution

Given Information:

Two triangles XCD and YCD having the same base, with A, B, E and F are the midpoints of sides $\overline{XC},\overline{XD},\overline{YC}and\overline{YD}$ respectively.

Formula used:

The following theorems are used.

1) Theorem 5-11 - In a triangle the line joining the midpoints of two legs will be parallel to the third leg and will be equal to half its measure.

In the triangle ABC given below, M and N are midpoints of sides AB and AC respectively. Therefore, by the theorem,

  $\begin{array}{l}MN\left|\right|BC\\ MN=\frac{1}{2}BC\end{array}$

  

Calculation: Consider the figure below.

  

Here, A, B, E and F are the midpoints of sides $\overline{XC},\overline{XD},\overline{YC}and\overline{YD}$ respectively.

Using the theorem presented above, in triangles XCD and YCD the sides $\overline{AB}\text{ and }\overline{EF}$ can be expressed as,

  $\overline{AB}\text{ =}\frac{1}{2}\overline{CD}$

  $\overline{EF}\text{ =}\frac{1}{2}\overline{CD}$

It is given that $\overline{CD}=24$ . Substituting this in the above equation, the measures of the sides $\overline{AB}\text{ and }\overline{EF}$ can be obtained as,

  $\begin{array}{c}\overline{AB}\text{ =}\frac{1}{2}\overline{CD}\\ =\frac{1}{2}\times 24\\ =12\end{array}$

  $\begin{array}{c}\overline{EF}\text{ =}\frac{1}{2}\overline{CD}\\ =\frac{1}{2}\times 24\\ =12\end{array}$