Question

desemal. 0.6 et 23) a 9 ft c h b d 7 ft e 15 ft 25) x x y 9m z w 8m z

Explanation:

Problem 23 (Triangle Similarity for Height \( h \))

Step 1: Identify Similar Triangles

Triangles \( ABE \) and \( CDE \) are similar (right triangles with shared angle at \( E \), so AA similarity). Thus, their corresponding sides are proportional. The base of \( ABE \) is \( BE = 15 \) ft, and the base of \( CDE \) is \( DE = 7 \) ft? Wait, no—wait, \( BD \) is \( 15 - 7 = 8 \) ft? Wait, no, \( BE = 15 \) ft, \( DE = 7 \) ft, so \( BD = 15 - 7 = 8 \) ft? Wait, no, the height of \( ABE \) is \( AB = 9 \) ft, height of \( CDE \) is \( h \). The bases: \( BE = 15 \) ft, \( DE = 7 \) ft? Wait, no, actually, \( CD \) is parallel to \( AB \), so triangles \( ABE \) and \( CDE \) are similar. So \( \frac{AB}{BE} = \frac{CD}{DE} \)? Wait, no, \( BE \) is \( 15 \) ft, \( DE \) is \( 7 \) ft? Wait, no, \( BD \) is \( 15 - 7 = 8 \) ft? Wait, maybe I misread. Wait, \( BE \) is \( 15 \) ft, \( DE \) is \( 7 \) ft, so the horizontal segment from \( B \) to \( D \) is \( 15 - 7 = 8 \) ft. But the height of \( ABE \) is \( AB = 9 \) ft, height of \( CDE \) is \( h \). So the ratio of heights should equal the ratio of their bases. Wait, actually, the triangles are similar, so \( \frac{AB}{BE} = \frac{h}{DE} \)? Wait, no, \( BE \) is the entire base (15 ft), \( DE \) is the base of the smaller triangle (7 ft)? Wait, no, maybe \( BD \) is \( 8 \) ft, but \( DE \) is \( 7 \) ft, so \( BE = BD + DE = 8 + 7 = 15 \) ft (correct). So triangle \( ABE \) has height \( AB = 9 \) ft, base \( BE = 15 \) ft. Triangle \( CDE \) has height \( h \), base \( DE = 7 \) ft? Wait, no, that can't be, because \( D \) is between \( B \) and \( E \), so the base of \( CDE \) is \( DE = 7 \) ft, and the base of \( ABE \) is \( BE = 15 \) ft. So the similarity ratio is \( \frac{DE}{BE} = \frac{7}{15} \), but wait, no—wait, \( AB \) is vertical, \( CD \) is vertical, so \( AB \parallel CD \), so angle at \( E \) is common, so triangles \( ECD \) and \( EAB \) are similar. So \( \frac{CD}{AB} = \frac{DE}{BE} \). So \( \frac{h}{9} = \frac{7}{15} \)? Wait, no, \( BE \) is \( 15 \) ft, \( DE \) is \( 7 \) ft? Wait, that would make \( BD = 8 \) ft, but maybe the base of \( ABE \) is \( BE = 15 \) ft, and the base of \( CDE \) is \( DE = 7 \) ft. Then:

\( \frac{h}{9} = \frac{7}{15} \)? Wait, no, that would give \( h = \frac{63}{15} = 4.2 \) ft? Wait, but maybe I mixed up the bases. Wait, maybe \( BD \) is \( 8 \) ft, so the base of \( ABE \) is \( BE = 15 \) ft, and the base of \( CDE \) is \( DE = 7 \) ft, but actually, the horizontal distance from \( B \) to \( E \) is \( 15 \) ft, and from \( D \) to \( E \) is \( 7 \) ft, so the horizontal distance from \( B \) to \( D \) is \( 15 - 7 = 8 \) ft. But the height of \( ABE \) is \( 9 \) ft, height of \( CDE \) is \( h \). So the ratio of heights should equal the ratio of their corresponding bases. Wait, maybe the triangles are \( ABE \) (height 9, base 15) and \( CDE \) (height \( h \), base 7). So:

\( \frac{h}{9} = \frac{7}{15} \)

Step 2: Solve for \( h \)

Multiply both sides by 9: \( h = 9 \times \frac{7}{15} = \frac{63}{15} = 4.2 \) ft. Wait, but let's check again. Wait, maybe the base of \( ABE \) is \( BE = 15 \) ft, and the base of \( CDE \) is \( DE = 7 \) ft, but actually, \( CD \) is at distance \( 7 \) ft from \( E \), and \( AB \) is at distance \( 15 \) ft from \( E \)? No, \( B \) is at the left end, \( E \) at the right. So \( AB \) is at \( B \), \( CD \) is at \( D \), so the distance from \( B \) to \( E \) is \( 15 \) ft, from \( D \) to \( E \) is \( 7 \) ft, so from \( B \) to \( D \) is \( 8 \) ft. So the two triangles: \( ABE \) (height 9, base 15) an…

Step 1: Identify Similar Triangles

Triangles \( XYZ \) (wait, no, triangles \( XYZ \) and \( WZ \)? Wait, the diagram shows two triangles intersecting at \( Z \), with vertical angles at \( Z \), and angle \( X \) equal to angle \( Z \), angle \( Y \) equal to angle \( W \)? Wait, the markings: angle at \( X \) and angle at \( Z \) are equal (curved marks), angle at \( Y \) and angle at \( W \) are equal (single curved mark), and vertical angles at \( Z \) are equal. So triangles \( XYZ \) and \( WZ \)? Wait, triangle \( XYZ \) and triangle \( W Z \)? Wait, the labels: \( X, Y, Z \) and \( W, Z, \dots \)? Wait, the triangle with sides \( 9m \) ( \( XZ \) ), \( x \) ( \( XY \) ), and the other triangle with sides \( 12m \) ( \( YZ \)? Wait, no, the diagram: \( X \) to \( Z \) is \( 9m \), \( Y \) to \( Z \) is \( 12m \), \( W \) to \( Z \) is \( 12m \)? Wait, no, the lower triangle has base \( 8m \) ( \( WZ \) ), side \( 12m \) ( \( YZ \)? Wait, maybe triangles \( XYZ \) and \( W Z \) are similar by AA similarity (vertical angles at \( Z \), and two pairs of equal angles). So triangle \( XYZ \sim \) triangle \( W Z \)? Wait, let's label:

• Triangle \( XYZ \): sides \( XY = x \), \( XZ = 9m \), \( YZ = 12m \) (wait, no, \( XZ = 9m \), \( YZ = 12m \), and angle at \( X \) equals angle at \( W \), angle at \( Y \) equals angle at \( Z \)? Wait, the lower triangle is \( W Z \) with base \( 8m \) ( \( WZ \) ), side \( 12m \) ( \( YZ \)? No, the sides: \( XZ = 9m \), \( YZ = 12m \), \( WZ = 8m \), \( WX \)? Wait, maybe the triangles are \( XYZ \) and \( W Z \), with \( XZ = 9m \), \( YZ = 12m \), \( WZ = 8m \), and \( XY = x \), \( WZ = 8m \), and they are similar. So the ratio of corresponding sides: \( \frac{XZ}{WZ} = \frac{XY}{WZ} \)? Wait, no, let's use similarity. Since angles at \( Z \) are vertical (equal), and angle at \( X \) equals angle at \( W \), angle at \( Y \) equals angle at \( Z \)? Wait, maybe the ratio of sides: \( \frac{XZ}{WZ} = \frac{XY}{WZ} \)? No, let's see:

If triangle \( XYZ \sim \) triangle \( W Z \) (let's say triangle \( W Z \) is \( W Z Y \)? No, the lower triangle is \( W, Z, \) and the other vertex is \( \dots \)? Wait, the lower triangle has base \( 8m \) ( \( WZ \) ), side \( 12m \) ( \( YZ \) ), and the upper triangle has side \( 9m \) ( \( XZ \) ), side \( x \) ( \( XY \) ). So by similarity, \( \frac{XZ}{WZ} = \frac{XY}{WZ} \)? No, maybe \( \frac{XZ}{WZ} = \frac{XY}{WZ} \)? Wait, no, let's set up the proportion. Since the triangles are similar, the ratio of \( XZ \) to \( WZ \) should equal the ratio of \( XY \) to \( WZ \)? Wait, no, let's list the corresponding sides:

• \( XZ = 9m \) corresponds to \( WZ = 8m \)? No, that doesn't make sense. Wait, maybe \( XZ = 9m \), \( YZ = 12m \), and the lower triangle has \( WZ = 8m \), \( YZ = 12m \)? No, the lower triangle's side is \( 12m \) ( \( YZ \) ), and the upper triangle's side is \( 9m \) ( \( XZ \) ), and the base of the lower triangle is \( 8m \) ( \( WZ \) ), base of the upper triangle is \( x \) ( \( XY \) ). So the proportion is \( \frac{XZ}{WZ} = \frac{XY}{WZ} \)? No, that's not right. Wait, maybe \( \frac{XZ}{YZ} = \frac{XY}{WZ} \). Wait, \( XZ = 9m \), \( YZ = 12m \), \( WZ = 8m \), so \( \frac{9}{12} = \frac{x}{8} \)? Wait, no, \( 9/12 = 3/4 \), so \( x = 8 \times (3/4) = 6 \)? Wait, no, \( 9/12 = x/8 \)? Wait, \( 9 \times 8 = 12x \) → \( 72 = 12x \) → \( x = 6 \). Wait, that makes sense. So the proportion is \( \frac{XZ}{YZ} = \frac{XY}{WZ} \), so \( \frac{9}{12} = \frac{x}{8} \), solving for \( x \): \( x = \frac{9 \times 8}{12} = 6 \)…

(Problem 23):

Step 1: Similar Triangles

Triangles \( ABE \) and \( CDE \) are similar (AA similarity: right angles, shared angle at \( E \)). Thus, \( \frac{AB}{BE} = \frac{CD}{DE} \).

Step 2: Substitute Values

\( AB = 9 \) ft, \( BE = 15 \) ft, \( DE = 7 \) ft? Wait, no—wait, \( DE = 7 \) ft, \( BE = 15 \) ft, so:

\( \frac{9}{15} = \frac{h}{7} \)

Step 3: Solve for \( h \)

Cross-multiply: \( 15h = 9 \times 7 \) → \( 15h = 63 \) → \( h = \frac{63}{15} = 4.2 \) ft.

Answer:

s:

• Problem 23: \( h = 4.2 \) ft (or \( \frac{21}{5} \) ft)
• Problem 25: \( x = 6 \) m

Final Answers (if focusing on Problem 23):