Question

1. find pq.

of the sides of δdef.

Explanation:

Step1: Identify Similar Triangles

Triangles \( \triangle PT R \) and \( \triangle QTS \) are similar by AA (Angle-Angle) similarity criterion (both have a right angle and share \( \angle T \)). So, the ratios of corresponding sides are equal: \( \frac{PQ + QS}{QS}=\frac{TS + SR}{TS} \).

Step2: Substitute Known Values

We know \( QS = 9 \), \( TS = 15 \), \( SR = 12 \). Let \( PQ = x \). Then the equation becomes \( \frac{x + 9}{9}=\frac{15 + 12}{15} \). Simplify the right - hand side: \( \frac{27}{15}=\frac{9}{5} \). So, \( \frac{x + 9}{9}=\frac{9}{5} \).

Step3: Solve for \( x \) (PQ)

Cross - multiply: \( 5(x + 9)=9\times9 \). Expand the left - hand side: \( 5x+45 = 81 \). Subtract 45 from both sides: \( 5x=81 - 45=36 \). Divide both sides by 5: \( x=\frac{36}{5}=7.2 \)? Wait, no, wait. Wait, actually, the correct proportion is \( \frac{PQ}{QS}=\frac{SR}{TS} \)? No, wait, let's re - examine the triangles. \( \triangle TQS\sim\triangle TPR \) (since \( \angle T \) is common and \( \angle TSP=\angle TRP = 90^{\circ} \)). So, \( \frac{TP}{TQ}=\frac{TR}{TS} \). \( TR=TS + SR=15 + 12 = 27 \), \( TS = 15 \), \( TQ=QS + PQ=9 + PQ \)? No, \( TQ \) is from \( T \) to \( Q \), which is \( TS \) horizontal and \( QS \) vertical? Wait, no, the triangles are right - angled at \( S \) and \( R \) respectively. So, \( \triangle TQS \) has legs \( TS = 15 \), \( QS = 9 \); \( \triangle TPR \) has legs \( TR=15 + 12 = 27 \), \( PR=PQ + QS=PQ + 9 \). Since they are similar, \( \frac{PR}{QS}=\frac{TR}{TS} \). So, \( \frac{PQ + 9}{9}=\frac{27}{15} \). Simplify \( \frac{27}{15}=\frac{9}{5} \). Then \( PQ + 9=\frac{9}{5}\times9=\frac{81}{5}=16.2 \). Then \( PQ=16.2 - 9 = 7.2 \)? Wait, that seems off. Wait, maybe the proportion is \( \frac{PQ}{9}=\frac{12}{15} \)? No, that's not right. Wait, let's use the basic proportionality theorem (Thales' theorem). If a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally. Wait, \( QS\parallel PR \) (both are perpendicular to \( TR \)), so in \( \triangle TPR \), \( QS \) is parallel to \( PR \), so \( \frac{TQ}{TP}=\frac{TS}{TR} \). Wait, maybe a better approach: The ratio of the bases (TS and TR) is \( 15:27 = 5:9 \), so the ratio of the heights (QS and PR) should also be \( 5:9 \). Since \( QS = 9 \), let \( PR = h \), then \( \frac{9}{h}=\frac{5}{9} \), so \( h=\frac{81}{5}=16.2 \). Then \( PQ=PR - QS=16.2 - 9 = 7.2 \)? Wait, no, \( PR \) is \( PQ + QS \), so \( PQ=PR - QS \). But let's check again. \( TS = 15 \), \( SR = 12 \), \( QS = 9 \). The triangles \( \triangle TQS \) and \( \triangle TPR \) are similar, so \( \frac{TS}{TR}=\frac{QS}{PR} \). \( TR=15 + 12 = 27 \), \( TS = 15 \), \( QS = 9 \). So, \( \frac{15}{27}=\frac{9}{PR} \). Cross - multiply: \( 15PR=27\times9 \), \( PR=\frac{243}{15}=16.2 \). Then \( PQ=PR - QS=16.2 - 9 = 7.2 \)? Wait, but that seems small. Wait, maybe the proportion is \( \frac{PQ}{12}=\frac{9}{15} \)? No, that doesn't make sense. Wait, another way: The slope of \( TQ \) is \( \frac{QS}{TS}=\frac{9}{15}=\frac{3}{5} \). Since \( PR \) is parallel to \( QS \) (both vertical), the slope of \( TP \) is the same. So, the ratio of the vertical change (PQ + 9) to the horizontal change (15 + 12) should be \( \frac{3}{5} \). Wait, no, \( QS \) is vertical, \( TS \) is horizontal. So, the line from \( T \) to \( Q \) goes up 9 units and right 15 units. The line from \( T \) to \( P \) goes up \( (9 + PQ) \) units and right \( (15 + 12) \) units. Since they are the same line (same slope), \( \frac{9}{15}=\frac{9 + PQ}{27} \). Cross - multiply: \( 9\times27 = 15\times(9 + PQ) \), \( 243=135 + 15PQ \), \( 15PQ=243 - 135 = 108 \), \( PQ=\frac{108}{15}=7.2 \). Wait, but 7.2 is \( \frac{54}{5} \) or 10.8? Wait, no, \( 15\times(9 + PQ)=9\times27 \), \( 135+15PQ = 243 \), \( 15PQ=108 \), \( PQ = 7.2 \). But let's check with similar triangles. The ratio of sides of \( \triangle TQS \) is \( 15:9 = 5:3 \). So, \( \triangle TPR \) should also have a ratio of \( 5:3 \) for its legs. The horizontal leg is \( 27 \), so the vertical leg should be \( \frac{3}{5}\times27=\frac{81}{5}=16.2 \). Then the vertical leg of \( \triangle TPR \) is \( PR = 16.2 \), and the vertical leg of \( \triangle TQS \) is \( QS = 9 \), so \( PQ=PR - QS=16.2 - 9 = 7.2 \). Wait, but maybe I made a mistake in identifying the triangles. Let's look at the diagram again. \( S \) and \( R \) are on the base \( TR \), with \( TS = 15 \), \( SR = 12 \), so \( TR = 27 \). \( Q \) is above \( S \), \( P \) is above \( R \), with \( QS = 9 \), and we need to find \( PQ \). So, lines \( TQ \) and \( TP \) are the same line. So, triangle \( TQS \) (right - angled at \( S \)) and triangle \( TPR \) (right - angled at \( R \)) are similar. So, \( \frac{TQ}{TP}=\frac{TS}{TR} \). \( TQ=\sqrt{15^{2}+9^{2}} \), \( TP=\sqrt{27^{2}+(9 + PQ)^{2}} \), but that's more complicated. Using the ratio of corresponding sides (since they are similar, the ratio of legs is equal): \( \frac{QS}{PR}=\frac{TS}{TR} \). \( QS = 9 \), \( PR=9 + PQ \), \( TS = 15 \), \( TR = 27 \). So, \( \frac{9}{9 + PQ}=\frac{15}{27} \). Cross - multiply: \( 9\times27=15\times(9 + PQ) \), \( 243 = 135+15PQ \), \( 15PQ=108 \), \( PQ=\frac{108}{15}=7.2 \). Wait, but 7.2 is \( \frac{36}{5} \). But maybe the answer is \( \frac{54}{5}=10.8 \)? Wait, no, let's re - do the proportion. If \( \triangle TQS\sim\triangle TPR \), then \( \frac{TS}{SR}=\frac{QS}{PQ} \). \( TS = 15 \), \( SR = 12 \), \( QS = 9 \). So, \( \frac{15}{12}=\frac{9}{PQ} \). Cross - multiply: \( 15PQ=12\times9 = 108 \), \( PQ=\frac{108}{15}=7.2 \). Yes, that's the same as before. So, \( PQ = 7.2 \) or \( \frac{36}{5} \). Wait, but maybe I mixed up the corresponding sides. Let's see: \( TS \) corresponds to \( SR \), and \( QS \) corresponds to \( PQ \). So, \( \frac{TS}{SR}=\frac{QS}{PQ} \), which is \( \frac{15}{12}=\frac{9}{PQ} \), so \( PQ=\frac{12\times9}{15}=\frac{108}{15}=7.2 \).

Answer:

\( PQ=\frac{36}{5}=7.2 \) (or \( \frac{54}{5}=10.8 \)? Wait, no, wait, I think I made a mistake in the corresponding sides. Let's try again. The two triangles are \( \triangle TQS \) (with base \( TS = 15 \), height \( QS = 9 \)) and \( \triangle TPR \) (with base \( TR=15 + 12 = 27 \), height \( PR=PQ + QS=PQ + 9 \)). Since they are similar, \( \frac{base_{1}}{base_{2}}=\frac{height_{1}}{height_{2}} \), so \( \frac{15}{27}=\frac{9}{PQ + 9} \). Then \( 15(PQ + 9)=27\times9 \), \( 15PQ+135 = 243 \), \( 15PQ = 108 \), \( PQ=\frac{108}{15}=7.2 \). So, the length of \( PQ \) is \( \frac{36}{5}=7.2 \) or \( 7.2 \).