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We are given an isosceles trapezium \( EFGH \), where we are asked to find \( \angle EFH \). 

### Known Information:
- This is an isosceles trapezium (trapezoid).
- The angle \( \angle FGH = 38^\circ \).
- Since it is an isosceles trapezium, the non-parallel sides \( EF \) and \( HG \) are congruent, meaning \( EF = GH \).
  
### Approach:
1. In any trapezium, the consecutive angles along each leg (i.e., on the same side) are supplementary (they add up to \( 180^\circ \)).
   - Here, \( \angle EFH \) and \( \angle FGH \) are consecutive angles along leg \( FH \).
   - Therefore, \( \angle EFH + \angle FGH = 180^\circ \).

2. Substituting the given value for \( \angle FGH = 38^\circ \), we get:
   \[
   \angle EFH + 38^\circ = 180^\circ
   \]
   \[
   \angle EFH = 180^\circ - 38^\circ = 142^\circ
   \]

Thus, \( \angle EFH = 142^\circ \).

### Explanation:
In an isosceles trapezium, the two angles on either side of a leg are supplementary. This property helped us solve the problem directly by using basic angle relationships.

Would you like further details or have any additional questions?

Here are some related questions to deepen your understanding:
1. How do you calculate angles in different types of quadrilaterals?
2. What properties define an isosceles trapezium?
3. How do supplementary angles help in geometry problems involving polygons?
4. Could the same principle apply to non-isosceles trapezoids?
5. What are the conditions for congruent sides in isosceles trapeziums?

**Tip**: Always look for symmetry and supplementary angles when dealing with trapeziums. They often provide the key to solving problems!

In this isosceles trapezium, find ∠EFH.

Solve the problem of finding ∠EFH in an isosceles trapezium where ∠FGH is 38°. This solution covers the properties of trapeziums, including supplementary angles along the legs, and is suitable for students in Grades 7-9.

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