Question

in the figure, e, g, h are on a vertical line, g and f are on a horizontal line with a right angle at g. eg and gh are marked as equal. the length of ef is 4a - 15, and the length of hf is a + 12.

Explanation:

Step1: Identify Equal Segments

Since \( FG \) is perpendicular to \( EH \) and \( G \) is the midpoint of \( EH \) (marked by the equal segments), triangles \( EGF \) and \( HGF \) are congruent by SAS (right angle, common side \( FG \), equal \( EG \) and \( HG \)). Thus, \( EF = HF \).
\[ 4a - 15 = a + 12 \]

Step2: Solve for \( a \)

Subtract \( a \) from both sides:
\[ 4a - a - 15 = 12 \]
\[ 3a - 15 = 12 \]
Add 15 to both sides:
\[ 3a = 12 + 15 \]
\[ 3a = 27 \]
Divide by 3:
\[ a = \frac{27}{3} = 9 \]

Answer:

\( a = 9 \)