Question

find qs.
s ----42---- r ---- q
\\ /
\\ /
\\ /
\\ /
\\ /
t
|
| 27
|
u
|
| 18
|
t
qs =
submit

Explanation:

Step1: Identify Similar Triangles

From the diagram, \( RU \parallel ST \) (implied by the congruent marks), so \( \triangle QRU \sim \triangle QST \) by the Basic Proportionality Theorem (Thales' theorem).

Step2: Set Up Proportion

Let \( QS = x \). The ratio of corresponding sides in similar triangles is equal. So, \( \frac{QU}{QT}=\frac{QR}{QS} \)? Wait, no, correct proportion: \( \frac{QU}{QT}=\frac{QR}{QS} \)? Wait, actually, \( QU = 18 \), \( QT = 27 \), \( SR = 42 \), and \( QR = x - 42 \)? Wait, no, let's re - examine. The sides: \( QU = 18 \), \( QT = 27 \), \( SR = 42 \), and we need to find \( QS \). Since \( RU \parallel ST \), \( \triangle QRU \sim \triangle QST \), so \( \frac{QU}{QT}=\frac{QR}{QS} \). Wait, \( QR = QS - SR=QS - 42 \), \( QU = 18 \), \( QT = 27 \). So \( \frac{18}{27}=\frac{QS - 42}{QS} \).

Step3: Solve the Proportion

Cross - multiply: \( 18\times QS=27\times(QS - 42) \)
Expand: \( 18QS = 27QS-27\times42 \)
\( 27\times42=1134 \), so \( 18QS=27QS - 1134 \)
Subtract \( 18QS \) from both sides: \( 0 = 9QS-1134 \)
Add 1134 to both sides: \( 9QS = 1134 \)
Divide both sides by 9: \( QS=\frac{1134}{9}=126 \)? Wait, no, that's wrong. Wait, maybe the correct proportion is \( \frac{QU}{QT}=\frac{QR}{SR} \)? No, let's start over.

Wait, the correct approach: Since \( RU \) is parallel to \( ST \), \( \triangle QRU \sim \triangle QST \) (AA similarity, as \( \angle Q \) is common and \( \angle QRU=\angle QST \) because of parallel lines). So the ratio of corresponding sides: \( \frac{QU}{QT}=\frac{QR}{QS} \). Wait, \( QU = 18 \), \( QT = 27 \), \( SR = 42 \), and \( QS=QR + RS \), let \( QR = y \), then \( QS=y + 42 \). So \( \frac{18}{27}=\frac{y}{y + 42} \)
Cross - multiply: \( 18(y + 42)=27y \)
\( 18y+756 = 27y \)
\( 756=27y - 18y \)
\( 9y = 756 \)
\( y = 84 \)
Then \( QS=y + 42=84 + 42 = 126 \)? No, that's not right. Wait, maybe the proportion is \( \frac{QU}{QT}=\frac{SR}{QS} \)? Wait, no, let's look at the lengths again. The segment \( QU = 18 \), \( QT = 27 \), \( SR = 42 \). The correct similarity ratio: \( \frac{QU}{QT}=\frac{QR}{QS} \), but \( QR=QS - SR \). Wait, another way: The ratio of \( QU \) to \( QT \) is \( \frac{18}{27}=\frac{2}{3} \). So the ratio of \( QR \) to \( QS \) is also \( \frac{2}{3} \). Let \( QS=x \), then \( QR=\frac{2}{3}x \). But \( QS=QR + RS \), so \( x=\frac{2}{3}x + 42 \)
Subtract \( \frac{2}{3}x \) from both sides: \( x-\frac{2}{3}x=42 \)
\( \frac{1}{3}x = 42 \)
Multiply both sides by 3: \( x = 126 \). Wait, but that seems large. Wait, maybe I mixed up the sides. Let's try again. If \( RU \parallel ST \), then \( \frac{QU}{QT}=\frac{QR}{QS} \), where \( QU = 18 \), \( QT = 27 \), \( QR=QS - SR=QS - 42 \). So \( \frac{18}{27}=\frac{QS - 42}{QS} \)
\( 18QS=27QS-27\times42 \)
\( 27\times42 = 1134 \)
\( 18QS=27QS - 1134 \)
\( 1134 = 27QS-18QS \)
\( 9QS=1134 \)
\( QS = 126 \). Wait, but let's check with the ratio. \( \frac{18}{27}=\frac{2}{3} \), \( \frac{QS - 42}{QS}=\frac{126 - 42}{126}=\frac{84}{126}=\frac{2}{3} \), which matches. So the length of \( QS \) is 126.

Answer:

126