Question

solve for ( x ). figures are not necessarily drawn to scale.

Explanation:

Step1: Identify Similar Triangles

The two triangles (big and small) have a common angle at \( Q \) (both \( 54^\circ \)) and the lines \( RU \) and \( US \) are parallel (implied by the right angles, though not labeled, but the corresponding angles and sides suggest similarity). So, by AA (Angle-Angle) similarity, \( \triangle QRU \sim \triangle QTS \).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. Let \( QT = x \) and \( QS = x + 7 \), \( RU = 3 \), \( US = 6 \). The ratio of \( RU \) to \( US \) should equal the ratio of \( QU \) to \( QS \)? Wait, no, correct corresponding sides: \( RU \) corresponds to \( TS \)? Wait, no, \( RU = 3 \), \( US = 6 \), so \( RS = 3 + 6 = 9 \). Wait, actually, \( QU = x \), \( QS = x + 7 \), and \( RU = 3 \), \( TS \) (wait, no, the small triangle has base \( 6 \) and the big triangle has base \( 3 + 6 = 9 \)? Wait, no, looking at the diagram: \( RU = 3 \), \( US = 6 \), so the base of the small triangle (with angle \( 54^\circ \)) is \( 6 \), and the base of the big triangle is \( 3 + 6 = 9 \). The sides on \( QS \): \( QT = x \), \( TS = 7 \), so \( QS = x + 7 \).

Since \( \triangle QRU \sim \triangle QTS \) (AA similarity, same angle at \( Q \), and the other angles are equal because the lines are parallel, so corresponding angles are equal), the ratio of \( RU \) to \( US \) is \( 3:6 = 1:2 \), but wait, no, the ratio of \( QU \) to \( QS \) should be equal to the ratio of \( RU \) to \( TS \)? Wait, no, let's label the triangles: \( \triangle QRU \) has sides \( QU = x \), \( RU = 3 \), and \( \triangle QTS \) has sides \( QS = x + 7 \), \( TS = 6 \)? Wait, no, I think I mixed up. Wait, the small triangle (with base \( 6 \)) has the side on \( QS \) as \( 7 \), and the big triangle (with base \( 3 + 6 = 9 \)) has the side on \( QS \) as \( x + 7 \)? No, wait, the diagram: \( RU \) is length \( 3 \), \( US \) is length \( 6 \), so the two triangles are \( \triangle QRU \) (base \( 3 \), side \( QU = x \)) and \( \triangle QTS \) (base \( 6 \), side \( QS = x + 7 \))? No, that can't be. Wait, actually, the correct corresponding sides: \( RU = 3 \), \( TS = 6 \) (wait, no, the small triangle has base \( 6 \), big has base \( 3 + 6 = 9 \))? Wait, no, let's do it properly.

Wait, the two triangles: \( \triangle QRU \) and \( \triangle QTS \). Angle at \( Q \) is common (\( 54^\circ \)), and the angle at \( R \) and \( T \) are equal (since \( RU \) and \( TS \) are parallel, so corresponding angles are equal). So by AA similarity, \( \triangle QRU \sim \triangle QTS \). Therefore, the ratio of \( QU \) to \( QS \) is equal to the ratio of \( RU \) to \( TS \). Wait, \( QU = x \), \( QS = x + 7 \), \( RU = 3 \), \( TS = 6 \). So:

\[ \frac{QU}{QS} = \frac{RU}{TS} \]

Substituting the values:

\[ \frac{x}{x + 7} = \frac{3}{6} \]

Simplify \( \frac{3}{6} = \frac{1}{2} \), so:

\[ \frac{x}{x + 7} = \frac{1}{2} \]

Cross-multiplying:

\[ 2x = x + 7 \]

Subtract \( x \) from both sides:

\[ x = 7 \]

Wait, that can't be right. Wait, maybe I mixed up the sides. Let's check again. Maybe the ratio is \( \frac{RU}{US} = \frac{QU}{TS} \). Wait, \( RU = 3 \), \( US = 6 \), so \( \frac{3}{6} = \frac{1}{2} \), and \( QU = x \), \( TS = 7 \), so \( \frac{x}{7} = \frac{1}{2} \)? No, that would give \( x = 3.5 \), which is wrong. Wait, maybe the correct ratio is \( \frac{QU}{QS} = \frac{RU}{RS} \), where \( RS = 3 + 6 = 9 \), \( RU = 3 \), so \( \frac{x}{x + 7} = \frac{3}{9} = \frac{1}{3} \). Then:

\[ 3x = x + 7 \\ 2x = 7 \\ x = 3.5 \]

No, that's not matching. Wait, maybe I got the similarity wrong. Let's look at the angles: both triangles have a \( 54^\circ \) angle at \( Q \), and the sides \( RU \) and \( TS \) are parallel (since they are both perpendicular to \( RS \), assuming right angles, though not labeled, but the diagram shows vertical sides). So \( \angle QRU = \angle QTS = 90^\circ \) (if they are right triangles). So by AA similarity, \( \triangle QRU \sim \triangle QTS \). Therefore, the ratio of \( QR \) to \( QT \) is equal to the ratio of \( RU \) to \( TS \) is equal to the ratio of \( QU \) to \( QS \).

Wait, \( RU = 3 \), \( TS = 6 \) (since \( US = 6 \) and \( TS \) is the base of the small triangle), \( QU = x \), \( QS = x + 7 \). So:

\[ \frac{RU}{TS} = \frac{QU}{QS} \\ \frac{3}{6} = \frac{x}{x + 7} \\ \frac{1}{2} = \frac{x}{x + 7} \\ x + 7 = 2x \\ 7 = x \]

Wait, that gives \( x = 7 \), but that would make \( QS = 14 \), and \( \frac{3}{6} = \frac{7}{14} = \frac{1}{2} \), which works. Wait, but let's check again. If \( RU = 3 \), \( TS = 6 \), then the ratio is \( 1:2 \), so \( QU \) should be half of \( QS \). If \( QU = x \), \( QS = x + 7 \), then \( x = \frac{1}{2}(x + 7) \), so \( 2x = x + 7 \), so \( x = 7 \). Yes, that works. So the value of \( x \) is \( 7 \)? Wait, but let's confirm the diagram: \( TS = 7 \), so \( QS = x + 7 \), and \( QU = x \), \( RU = 3 \), \( TS = 6 \). Wait, no, \( TS \) is not \( 6 \), \( US \) is \( 6 \). Oh! I see my mistake. \( US = 6 \) is the base of the small triangle, so the base of the small triangle is \( 6 \), and the base of the big triangle is \( RU + US = 3 + 6 = 9 \). So \( RU = 3 \), \( RS = 9 \). Then the ratio of the bases is \( 3:9 = 1:3 \)? No, \( RU = 3 \), \( US = 6 \), so the ratio of \( RU \) to \( US \) is \( 3:6 = 1:2 \), but the base of the big triangle is \( 3 + 6 = 9 \), so \( RU:RS = 3:9 = 1:3 \). Wait, I'm confused. Let's start over.

Let’s denote:

- \( \triangle QRU \): angle at \( Q = 54^\circ \), side \( RU = 3 \), side \( QU = x \) (the segment from \( Q \) to \( T \) is \( x \), and from \( T \) to \( S \) is \( 7 \), so \( QS = x + 7 \)).

- \( \triangle QTS \): angle at \( Q = 54^\circ \), side \( TS = 6 \) (wait, no, \( US = 6 \) is the base of the small triangle, so \( TS \) is the other side? No, the diagram shows \( US = 6 \), \( RU = 3 \), and the two triangles share the angle at \( Q \), with \( T \) on \( QS \) and \( U \) on \( RS \). So \( RU \) and \( TU \) are parallel (vertical sides), so \( \triangle QRU \) and \( \triangle QTS \) are similar by AA (angle at \( Q \) and right angles at \( R \) and \( T \)).

Therefore, the ratio of \( RU \) to \( TS \) is equal to the ratio of \( QU \) to \( QS \). Wait, \( RU = 3 \), \( TS = 6 \) (since \( US = 6 \) is the base of the small triangle, so \( TS \) is the side opposite? No, \( TS \) is not \( 6 \), \( US \) is \( 6 \). \( US \) is the horizontal side of the small triangle, so the horizontal side of the small triangle is \( 6 \), and the horizontal side of the big triangle is \( 3 + 6 = 9 \). The vertical sides: \( RU = 3 \) (vertical side of big triangle), \( TU \) (vertical side of small triangle) is unknown, but the sides on \( QS \): \( QU = x \), \( TS = 7 \) (wait, \( TS \) is labeled \( 7 \)). Oh! \( TS = 7 \), not \( 6 \). I misread the diagram. \( US = 6 \) (horizontal), \( TS = 7 \) (the side from \( T \) to \( S \)). So:

- \( \triangle QRU \): horizontal side \( RU = 3 \), hypotenuse \( QU = x \)

- \( \triangle QTS \): horizontal side \( US = 6 \), hypotenuse \( TS = 7 \)

Wait, no, hypotenuse is \( QS \) for the big triangle, and \( QT \) for the small? No, \( QS \) is the hypotenuse of the big triangle, with length \( x + 7 \), and \( QT \) is the hypotenuse of the small triangle? No, \( QT = x \), \( TS = 7 \), so \( QS = x + 7 \). The horizontal sides: \( RU = 3 \) (big triangle), \( US = 6 \) (small triangle). So by AA similarity (angle at \( Q \) is \( 54^\circ \), and the horizontal sides are parallel, so corresponding angles are equal), \( \triangle QRU \sim \triangle QTS \). Therefore, the ratio of horizontal sides is \( RU:US = 3:6 = 1:2 \), and the ratio of hypotenuses is \( QU:QS = x:(x + 7) \). Wait, no, that would be \( 3:6 = x:(x + 7) \), so \( 1:2 = x:(x + 7) \), so \( x + 7 = 2x \), so \( x = 7 \). But that would make \( QS = 14 \), and \( 3:6 = 7:14 = 1:2 \), which works. So the value of \( x \) is \( 7 \)? Wait, but let's check the sides. If \( x = 7 \), then \( QT = 7 \), \( TS = 7 \), so \( QS = 14 \). The horizontal sides: \( RU = 3 \), \( US = 6 \), ratio \( 1:2 \), and the hypotenuses: \( 7:14 = 1:2 \), so that's consistent. So the similarity ratio is \( 1:2 \), so all corresponding sides are in ratio \( 1:2 \). Therefore, \( x = 7 \).

Wait, but I think the correct proportion is \( \frac{RU}{US} = \frac{QT}{TS} \), so \( \frac{3}{6} = \frac{x}{7} \)? No, that would be \( \frac{1}{2} = \frac{x}{7} \), so \( x = 3.5 \). Wait, now I'm really confused. Which is correct?

Wait, let's use the basic proportionality theorem (Thales' theorem), which states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally.

In \( \triangle QRS \), line \( TU \) is parallel to \( QR \) (since \( RU \) and \( TU \) are both vertical, so \( TU \parallel QR \)). Therefore, by Thales' theorem, \( \frac{QU}{US} = \frac{RU}{TS} \)? No, Thales' theorem: if a line is parallel to \( RS \) (the base), then it divides \( QR \) and \( QS \) proportionally. Wait, no, \( TU \) is parallel to \( QR \), so it divides \( QS \) and \( RS \) proportionally. Wait, \( RS \) is the base, length \( 3 + 6 = 9 \), and \( TU \) is parallel to \( QR \), so \( \frac{QT}{QS} = \frac{US}{RS} \)? No, Thales' theorem: in \( \triangle QRS \), if \( TU \parallel QR \), then \( \frac{QT}{TS} = \frac{RU}{US} \). Wait, \( QT = x \), \( TS = 7 \), \( RU = 3 \), \( US = 6 \). So \( \frac{x}{7} = \frac{3}{6} \), so \( \frac{x}{7} = \frac{1}{2} \), so \( x = \frac{7}{2} = 3.5 \). Ah! Here's the mistake. I had the corresponding sides wrong. Thales' theorem: the line parallel to one side (here, \( TU \parallel QR \)) intersects the other two sides (\( QS \) and \( RS \)) and divides them proportionally. So \( \frac{QT}{TS} = \frac{RU}{US} \). Wait, \( QT = x \), \( TS = 7 \), \( RU = 3 \), \( US = 6 \). So \( \frac{x}{7} = \frac{3}{6} \), so \( \frac{x}{7} = \frac{1}{2} \), so \( x = \frac{7}{2} = 3.5 \). Wait, now this is different. Which is correct?

Wait, let's label the triangle properly:

- \( \triangle QRS \): vertices \( Q \), \( R \), \( S \), with \( R \) and \( S \) on the base, \( R \) to \( U \) is \( 3 \), \( U \) to \( S \) is \( 6 \), so \( RS = 9 \).

- Line \( TU \) is drawn parallel to \( QR \), intersecting \( QS \) at \( T \) and \( RS \) at \( U \).

By Thales' theorem (Basic Proportionality Theorem), \( \frac{QT}{TS} = \frac{RU}{US} \).

So \( QT = x \), \( TS = 7 \), \( RU = 3 \), \( US = 6 \).

Therefore, \( \frac{x}{7} = \frac{3}{6} \)

Simplify \( \frac{3}{6} = \frac{1}{2} \), so \( \frac{x}{7} = \frac{1}{2} \)

Multiply both sides by \( 7 \): \( x = \frac{7}{2} = 3.5 \)

Ah! This is the correct application of Thales' theorem. I had the corresponding sides wrong earlier. The line \( TU \) is parallel to \( QR \), so it divides \( QS \) and \( RS \) proportionally. So \( \frac{QT}{TS} = \frac{RU}{US} \), not \( \frac{QU}{QS} \). So that's the mistake. So \( x = 3.5 \), which is \( \frac{7}{2} \) or \( 3.5 \).

Wait, let's verify with similar triangles. \( \triangle QTU \sim \triangle QRS \) (

Answer:

Step1: Identify Similar Triangles

The two triangles (big and small) have a common angle at \( Q \) (both \( 54^\circ \)) and the lines \( RU \) and \( US \) are parallel (implied by the right angles, though not labeled, but the corresponding angles and sides suggest similarity). So, by AA (Angle-Angle) similarity, \( \triangle QRU \sim \triangle QTS \).

Step2: Set Up Proportion

For similar triangles, the ratios of corresponding sides are equal. Let \( QT = x \) and \( QS = x + 7 \), \( RU = 3 \), \( US = 6 \). The ratio of \( RU \) to \( US \) should equal the ratio of \( QU \) to \( QS \)? Wait, no, correct corresponding sides: \( RU \) corresponds to \( TS \)? Wait, no, \( RU = 3 \), \( US = 6 \), so \( RS = 3 + 6 = 9 \). Wait, actually, \( QU = x \), \( QS = x + 7 \), and \( RU = 3 \), \( TS \) (wait, no, the small triangle has base \( 6 \) and the big triangle has base \( 3 + 6 = 9 \)? Wait, no, looking at the diagram: \( RU = 3 \), \( US = 6 \), so the base of the small triangle (with angle \( 54^\circ \)) is \( 6 \), and the base of the big triangle is \( 3 + 6 = 9 \). The sides on \( QS \): \( QT = x \), \( TS = 7 \), so \( QS = x + 7 \).

Since \( \triangle QRU \sim \triangle QTS \) (AA similarity, same angle at \( Q \), and the other angles are equal because the lines are parallel, so corresponding angles are equal), the ratio of \( RU \) to \( US \) is \( 3:6 = 1:2 \), but wait, no, the ratio of \( QU \) to \( QS \) should be equal to the ratio of \( RU \) to \( TS \)? Wait, no, let's label the triangles: \( \triangle QRU \) has sides \( QU = x \), \( RU = 3 \), and \( \triangle QTS \) has sides \( QS = x + 7 \), \( TS = 6 \)? Wait, no, I think I mixed up. Wait, the small triangle (with base \( 6 \)) has the side on \( QS \) as \( 7 \), and the big triangle (with base \( 3 + 6 = 9 \)) has the side on \( QS \) as \( x + 7 \)? No, wait, the diagram: \( RU \) is length \( 3 \), \( US \) is length \( 6 \), so the two triangles are \( \triangle QRU \) (base \( 3 \), side \( QU = x \)) and \( \triangle QTS \) (base \( 6 \), side \( QS = x + 7 \))? No, that can't be. Wait, actually, the correct corresponding sides: \( RU = 3 \), \( TS = 6 \) (wait, no, the small triangle has base \( 6 \), big has base \( 3 + 6 = 9 \))? Wait, no, let's do it properly.

Wait, the two triangles: \( \triangle QRU \) and \( \triangle QTS \). Angle at \( Q \) is common (\( 54^\circ \)), and the angle at \( R \) and \( T \) are equal (since \( RU \) and \( TS \) are parallel, so corresponding angles are equal). So by AA similarity, \( \triangle QRU \sim \triangle QTS \). Therefore, the ratio of \( QU \) to \( QS \) is equal to the ratio of \( RU \) to \( TS \). Wait, \( QU = x \), \( QS = x + 7 \), \( RU = 3 \), \( TS = 6 \). So:

$$ \frac{QU}{QS} = \frac{RU}{TS} $$

Substituting the values:

$$ \frac{x}{x + 7} = \frac{3}{6} $$

Simplify \( \frac{3}{6} = \frac{1}{2} \), so:

$$ \frac{x}{x + 7} = \frac{1}{2} $$

Cross-multiplying:

$$ 2x = x + 7 $$

Subtract \( x \) from both sides:

$$ x = 7 $$

Wait, that can't be right. Wait, maybe I mixed up the sides. Let's check again. Maybe the ratio is \( \frac{RU}{US} = \frac{QU}{TS} \). Wait, \( RU = 3 \), \( US = 6 \), so \( \frac{3}{6} = \frac{1}{2} \), and \( QU = x \), \( TS = 7 \), so \( \frac{x}{7} = \frac{1}{2} \)? No, that would give \( x = 3.5 \), which is wrong. Wait, maybe the correct ratio is \( \frac{QU}{QS} = \frac{RU}{RS} \), where \( RS = 3 + 6 = 9 \), \( RU = 3 \), so \( \frac{x}{x + 7} = \frac{3}{9} = \frac{1}{3} \). Then:

$$ 3x = x + 7 \\ 2x = 7 \\ x = 3.5 $$

No, that's not matching. Wait, maybe I got the similarity wrong. Let's look at the angles: both triangles have a \( 54^\circ \) angle at \( Q \), and the sides \( RU \) and \( TS \) are parallel (since they are both perpendicular to \( RS \), assuming right angles, though not labeled, but the diagram shows vertical sides). So \( \angle QRU = \angle QTS = 90^\circ \) (if they are right triangles). So by AA similarity, \( \triangle QRU \sim \triangle QTS \). Therefore, the ratio of \( QR \) to \( QT \) is equal to the ratio of \( RU \) to \( TS \) is equal to the ratio of \( QU \) to \( QS \).

Wait, \( RU = 3 \), \( TS = 6 \) (since \( US = 6 \) and \( TS \) is the base of the small triangle), \( QU = x \), \( QS = x + 7 \). So:

$$ \frac{RU}{TS} = \frac{QU}{QS} \\ \frac{3}{6} = \frac{x}{x + 7} \\ \frac{1}{2} = \frac{x}{x + 7} \\ x + 7 = 2x \\ 7 = x $$

Wait, that gives \( x = 7 \), but that would make \( QS = 14 \), and \( \frac{3}{6} = \frac{7}{14} = \frac{1}{2} \), which works. Wait, but let's check again. If \( RU = 3 \), \( TS = 6 \), then the ratio is \( 1:2 \), so \( QU \) should be half of \( QS \). If \( QU = x \), \( QS = x + 7 \), then \( x = \frac{1}{2}(x + 7) \), so \( 2x = x + 7 \), so \( x = 7 \). Yes, that works. So the value of \( x \) is \( 7 \)? Wait, but let's confirm the diagram: \( TS = 7 \), so \( QS = x + 7 \), and \( QU = x \), \( RU = 3 \), \( TS = 6 \). Wait, no, \( TS \) is not \( 6 \), \( US \) is \( 6 \). Oh! I see my mistake. \( US = 6 \) is the base of the small triangle, so the base of the small triangle is \( 6 \), and the base of the big triangle is \( RU + US = 3 + 6 = 9 \). So \( RU = 3 \), \( RS = 9 \). Then the ratio of the bases is \( 3:9 = 1:3 \)? No, \( RU = 3 \), \( US = 6 \), so the ratio of \( RU \) to \( US \) is \( 3:6 = 1:2 \), but the base of the big triangle is \( 3 + 6 = 9 \), so \( RU:RS = 3:9 = 1:3 \). Wait, I'm confused. Let's start over.

Let’s denote:

• \( \triangle QRU \): angle at \( Q = 54^\circ \), side \( RU = 3 \), side \( QU = x \) (the segment from \( Q \) to \( T \) is \( x \), and from \( T \) to \( S \) is \( 7 \), so \( QS = x + 7 \)).

• \( \triangle QTS \): angle at \( Q = 54^\circ \), side \( TS = 6 \) (wait, no, \( US = 6 \) is the base of the small triangle, so \( TS \) is the other side? No, the diagram shows \( US = 6 \), \( RU = 3 \), and the two triangles share the angle at \( Q \), with \( T \) on \( QS \) and \( U \) on \( RS \). So \( RU \) and \( TU \) are parallel (vertical sides), so \( \triangle QRU \) and \( \triangle QTS \) are similar by AA (angle at \( Q \) and right angles at \( R \) and \( T \)).

Therefore, the ratio of \( RU \) to \( TS \) is equal to the ratio of \( QU \) to \( QS \). Wait, \( RU = 3 \), \( TS = 6 \) (since \( US = 6 \) is the base of the small triangle, so \( TS \) is the side opposite? No, \( TS \) is not \( 6 \), \( US \) is \( 6 \). \( US \) is the horizontal side of the small triangle, so the horizontal side of the small triangle is \( 6 \), and the horizontal side of the big triangle is \( 3 + 6 = 9 \). The vertical sides: \( RU = 3 \) (vertical side of big triangle), \( TU \) (vertical side of small triangle) is unknown, but the sides on \( QS \): \( QU = x \), \( TS = 7 \) (wait, \( TS \) is labeled \( 7 \)). Oh! \( TS = 7 \), not \( 6 \). I misread the diagram. \( US = 6 \) (horizontal), \( TS = 7 \) (the side from \( T \) to \( S \)). So:

• \( \triangle QRU \): horizontal side \( RU = 3 \), hypotenuse \( QU = x \)

• \( \triangle QTS \): horizontal side \( US = 6 \), hypotenuse \( TS = 7 \)

Wait, no, hypotenuse is \( QS \) for the big triangle, and \( QT \) for the small? No, \( QS \) is the hypotenuse of the big triangle, with length \( x + 7 \), and \( QT \) is the hypotenuse of the small triangle? No, \( QT = x \), \( TS = 7 \), so \( QS = x + 7 \). The horizontal sides: \( RU = 3 \) (big triangle), \( US = 6 \) (small triangle). So by AA similarity (angle at \( Q \) is \( 54^\circ \), and the horizontal sides are parallel, so corresponding angles are equal), \( \triangle QRU \sim \triangle QTS \). Therefore, the ratio of horizontal sides is \( RU:US = 3:6 = 1:2 \), and the ratio of hypotenuses is \( QU:QS = x:(x + 7) \). Wait, no, that would be \( 3:6 = x:(x + 7) \), so \( 1:2 = x:(x + 7) \), so \( x + 7 = 2x \), so \( x = 7 \). But that would make \( QS = 14 \), and \( 3:6 = 7:14 = 1:2 \), which works. So the value of \( x \) is \( 7 \)? Wait, but let's check the sides. If \( x = 7 \), then \( QT = 7 \), \( TS = 7 \), so \( QS = 14 \). The horizontal sides: \( RU = 3 \), \( US = 6 \), ratio \( 1:2 \), and the hypotenuses: \( 7:14 = 1:2 \), so that's consistent. So the similarity ratio is \( 1:2 \), so all corresponding sides are in ratio \( 1:2 \). Therefore, \( x = 7 \).

Wait, but I think the correct proportion is \( \frac{RU}{US} = \frac{QT}{TS} \), so \( \frac{3}{6} = \frac{x}{7} \)? No, that would be \( \frac{1}{2} = \frac{x}{7} \), so \( x = 3.5 \). Wait, now I'm really confused. Which is correct?

Wait, let's use the basic proportionality theorem (Thales' theorem), which states that if a line is drawn parallel to one side of a triangle, intersecting the other two sides, then it divides those sides proportionally.

In \( \triangle QRS \), line \( TU \) is parallel to \( QR \) (since \( RU \) and \( TU \) are both vertical, so \( TU \parallel QR \)). Therefore, by Thales' theorem, \( \frac{QU}{US} = \frac{RU}{TS} \)? No, Thales' theorem: if a line is parallel to \( RS \) (the base), then it divides \( QR \) and \( QS \) proportionally. Wait, no, \( TU \) is parallel to \( QR \), so it divides \( QS \) and \( RS \) proportionally. Wait, \( RS \) is the base, length \( 3 + 6 = 9 \), and \( TU \) is parallel to \( QR \), so \( \frac{QT}{QS} = \frac{US}{RS} \)? No, Thales' theorem: in \( \triangle QRS \), if \( TU \parallel QR \), then \( \frac{QT}{TS} = \frac{RU}{US} \). Wait, \( QT = x \), \( TS = 7 \), \( RU = 3 \), \( US = 6 \). So \( \frac{x}{7} = \frac{3}{6} \), so \( \frac{x}{7} = \frac{1}{2} \), so \( x = \frac{7}{2} = 3.5 \). Ah! Here's the mistake. I had the corresponding sides wrong. Thales' theorem: the line parallel to one side (here, \( TU \parallel QR \)) intersects the other two sides (\( QS \) and \( RS \)) and divides them proportionally. So \( \frac{QT}{TS} = \frac{RU}{US} \). Wait, \( QT = x \), \( TS = 7 \), \( RU = 3 \), \( US = 6 \). So \( \frac{x}{7} = \frac{3}{6} \), so \( \frac{x}{7} = \frac{1}{2} \), so \( x = \frac{7}{2} = 3.5 \). Wait, now this is different. Which is correct?

Wait, let's label the triangle properly:

• \( \triangle QRS \): vertices \( Q \), \( R \), \( S \), with \( R \) and \( S \) on the base, \( R \) to \( U \) is \( 3 \), \( U \) to \( S \) is \( 6 \), so \( RS = 9 \).

• Line \( TU \) is drawn parallel to \( QR \), intersecting \( QS \) at \( T \) and \( RS \) at \( U \).

By Thales' theorem (Basic Proportionality Theorem), \( \frac{QT}{TS} = \frac{RU}{US} \).

So \( QT = x \), \( TS = 7 \), \( RU = 3 \), \( US = 6 \).

Therefore, \( \frac{x}{7} = \frac{3}{6} \)

Simplify \( \frac{3}{6} = \frac{1}{2} \), so \( \frac{x}{7} = \frac{1}{2} \)

Multiply both sides by \( 7 \): \( x = \frac{7}{2} = 3.5 \)

Ah! This is the correct application of Thales' theorem. I had the corresponding sides wrong earlier. The line \( TU \) is parallel to \( QR \), so it divides \( QS \) and \( RS \) proportionally. So \( \frac{QT}{TS} = \frac{RU}{US} \), not \( \frac{QU}{QS} \). So that's the mistake. So \( x = 3.5 \), which is \( \frac{7}{2} \) or \( 3.5 \).

Wait, let's verify with similar triangles. \( \triangle QTU \sim \triangle QRS \) (