Question

$\overline{rs} \parallel \overline{qt}$. find $ps$.
(image of a triangle with points r, s, p, q, t; lengths: rq=32, qp=48, st=14)

Explanation:

Step1: Identify Similar Triangles

Since \( \overline{RS} \parallel \overline{QT} \), by the Basic Proportionality Theorem (Thales' theorem), \( \triangle RPS \sim \triangle QPT \). So, the ratios of corresponding sides are equal: \( \frac{RP}{QP} = \frac{PS}{PT} \). Wait, actually, let's correct the segment labels. \( RP = RQ + QP = 32 + 48 = 80 \)? No, wait, \( RQ = 32 \), \( QP = 48 \), so \( RP = RQ + QP = 32 + 48 = 80 \)? Wait, no, the sides: \( RQ = 32 \), \( QP = 48 \), so the ratio of \( RQ \) to \( QP \) is \( 32:48 = 2:3 \). Wait, actually, the triangles are similar, so \( \frac{RQ}{QP} = \frac{ST}{TP} \)? Wait, no, \( RS \parallel QT \), so \( \triangle RST \) and \( \triangle QPT \)? Wait, maybe better to set up the proportion as \( \frac{PS}{PS - 14} = \frac{32 + 48}{48} \)? Wait, no. Let's define \( PT = x \), so \( PS = PT + ST = x + 14 \)? Wait, no, \( ST = 14 \), \( PT = x \), so \( PS = x + 14 \). The sides \( RQ = 32 \), \( QP = 48 \), so the ratio of \( RP \) to \( QP \) is \( (32 + 48)/48 = 80/48 = 5/3 \). Wait, no, similar triangles: \( \triangle RPS \sim \triangle QPT \), so \( \frac{RP}{QP} = \frac{PS}{PT} \). \( RP = RQ + QP = 32 + 48 = 80 \), \( QP = 48 \), \( PS = PT + 14 \) (since \( ST = 14 \), \( PT = PS - 14 \)). So \( \frac{80}{48} = \frac{PS}{PS - 14} \). Simplify \( 80/48 = 5/3 \), so \( \frac{5}{3} = \frac{PS}{PS - 14} \). Cross-multiplying: \( 5(PS - 14) = 3PS \) → \( 5PS - 70 = 3PS \) → \( 2PS = 70 \) → \( PS = 35 \)? Wait, that can't be. Wait, maybe I mixed up the segments. Let's re-express: \( RQ = 32 \), \( QP = 48 \), so the ratio of \( RQ \) to \( QP \) is \( 32:48 = 2:3 \). So the ratio of \( RS \) to \( QT \) is 2:3, but actually, the segments on the base: \( ST = 14 \), \( PT = x \), so \( PS = 14 + x \). The triangles \( \triangle RST \) and \( \triangle QPT \) are similar? Wait, no, \( RS \parallel QT \), so \( \angle R = \angle Q \) (common angle? No, \( \angle R \) is common to \( \triangle RPS \) and \( \triangle RQT \)? Wait, maybe \( \triangle RQT \sim \triangle RPS \). So \( \frac{RQ}{RP} = \frac{QT}{PS} \)? No, \( RQ = 32 \), \( RP = 32 + 48 = 80 \), \( QT = PT \) (no, \( ST = 14 \), \( PT = x \), so \( QT \) is parallel to \( RS \), so \( \frac{RQ}{RP} = \frac{ST}{PS} \)? Wait, \( \frac{RQ}{RP} = \frac{32}{80} = \frac{2}{5} \), and \( \frac{ST}{PS} = \frac{14}{PS} \). So \( \frac{2}{5} = \frac{14}{PS} \)? No, that gives \( PS = 35 \), but that seems low. Wait, maybe the correct proportion is \( \frac{RQ}{QP} = \frac{ST}{PT} \). \( RQ = 32 \), \( QP = 48 \), \( ST = 14 \), \( PT = x \). So \( \frac{32}{48} = \frac{14}{x} \). Simplify \( 32/48 = 2/3 \), so \( 2/3 = 14/x \) → \( 2x = 42 \) → \( x = 21 \). Then \( PS = PT + ST = 21 + 14 = 35 \)? Wait, no, \( ST = 14 \), \( PT = x \), so \( PS = PT + ST = x + 14 \)? Wait, no, \( S---T---P \), so \( ST = 14 \), \( TP = x \), so \( PS = ST + TP = 14 + x \). The ratio of \( RQ \) to \( QP \) is \( 32:48 = 2:3 \), so \( \frac{RQ}{QP} = \frac{ST}{TP} \) → \( \frac{32}{48} = \frac{14}{x} \) → \( x = (14 \times 48)/32 = (14 \times 3)/2 = 21 \). Then \( PS = 14 + 21 = 35 \)? Wait, but let's check the similar triangles again. \( RS \parallel QT \), so \( \angle R = \angle Q \) (common angle? No, \( \angle R \) is in \( \triangle RRS \) and \( \triangle RQT \). Wait, actually, \( \triangle RPS \) and \( \triangle QPT \) are similar because \( RS \parallel QT \), so corresponding angles are equal. So \( \frac{RP}{QP} = \frac{PS}{PT} \). \( RP = 32 + 48 = 80 \), \( QP = 48 \), \( PS = PT + 14 \) (since \( ST = 14 \), \( PT = PS - 14 \)). So \( \frac{80}{48} = \frac{PS}{PS - 14} \). Simplify \( 80/48 = 5/3 \), so \( 5/3 = PS/(PS - 14) \). Cross-multiplying: \( 5(PS - 14) = 3PS \) → \( 5PS - 70 = 3PS \) → \( 2PS = 70 \) → \( PS = 35 \). Yes, that matches. So the length of \( PS \) is 35.

Step1: Set Up Proportion

Since \( \overline{RS} \parallel \overline{QT} \), \( \triangle RPS \sim \triangle QPT \) (by AA similarity, as \( \angle R \) is common and \( \angle RSP = \angle QTP \) due to parallel lines). Thus, the ratios of corresponding sides are equal:
\[ \frac{RP}{QP} = \frac{PS}{PT} \]
Here, \( RP = RQ + QP = 32 + 48 = 80 \), \( QP = 48 \), and \( PT = PS - 14 \) (since \( ST = 14 \), \( PT = PS - ST \)). Substitute these into the proportion:
\[ \frac{80}{48} = \frac{PS}{PS - 14} \]

Step2: Simplify and Solve

Simplify \( \frac{80}{48} = \frac{5}{3} \), so:
\[ \frac{5}{3} = \frac{PS}{PS - 14} \]
Cross-multiply to eliminate the fraction:
\[ 5(PS - 14) = 3 \cdot PS \]
Expand and solve for \( PS \):
\[ 5PS - 70 = 3PS \\ 5PS - 3PS = 70 \\ 2PS = 70 \\ PS = \frac{70}{2} = 35 \]

Answer:

\( \boxed{35} \)