Chapter 5.5, Problem 7PSA

To determine

To find: Thevalues of EJ, JG and HJ.

Answer to Problem 7PSA

The values of $EJ=8,JG=5\text{and}HJ=5.$ .

Explanation of Solution

Given Information:

  $EJ=x+5$

  $JG=2x-1$

  $HF=13$

Formula used:

Congruent sides of similar triangles are in proportion.

Calculation:

We know that, in an isosceles trapezium, the diagonals are congruent.

In trapezium $EFGH,EG\text{and}HF$ are the diagonals and so they will be equal

As $HF=13$

  $\begin{array}{l}\Rightarrow EG=HF\\ \Rightarrow EJ+JG=13\end{array}$

As $EJ=x+5\text{and}JG=2x-1$

  $\begin{array}{l}\Rightarrow x+5+2x-1=13\\ \Rightarrow 3x=9\\ \Rightarrow x=3\\ \therefore EJ=x+5=8\\ JG=2x-1=5\end{array}$

In $\Delta GHJ\text{and}\Delta EFJ,$

  $\angle GHJ=\angle EFJ$ … (Alternate angles)

  $\angle HGJ=\angle FEJ$ … (Alternate angles)

  $\angle GJH=\angle EJF$ … (Vertically opposite angles)

  $\therefore \Delta GHJ\sim \Delta EFJ$ … (By AAA similarity test)

As the corresponding sides of similar triangles are in proportion

  $\Rightarrow \frac{FJ}{HJ}=\frac{EJ}{GJ}$

  $\begin{array}{l}\Rightarrow \frac{FJ}{HJ}+1=\frac{EJ}{GJ}+1\\ \Rightarrow \frac{FJ+HJ}{HJ}=\frac{EJ+GJ}{GJ}\\ \Rightarrow \frac{FH}{HJ}=\frac{EG}{GJ}\\ \Rightarrow \frac{13}{HJ}=\frac{13}{GJ}\\ \Rightarrow HJ=GJ\\ \Rightarrow HJ=5\mathrm{...}\left(\because GJ=5\right)\end{array}$

Hence, $EJ=8,JG=5\text{and}HJ=5.$