Question

solve for x. figures are not necessarily drawn to scale. image of a geometric figure with angles 54° at q and t, segments ru=3, us=6, ts=7, and qs with segment x

Explanation:

Step1: Identify Similar Triangles

The triangles \( \triangle QRU \) and \( \triangle STU \) are similar by AA (Angle - Angle) similarity criterion since \( \angle Q=\angle T = 54^{\circ}\) and \( \angle R=\angle U = 90^{\circ}\) (implied by the right - angled appearance, and the lines \( RU \) and \( SU \) are on the same base, and \( QU \) and \( TU \) are parallel in a sense due to the equal angles). For similar triangles, the ratios of corresponding sides are equal. The ratio of \( RU \) to \( SU \) is \( \frac{3}{6}=\frac{1}{2}\). Let the length of \( QT=x \) and \( TS = 7\). The ratio of \( QT\) to \( QS\) (where \( QS=QT + TS=x + 7\)) should be equal to the ratio of \( RU\) to \( RS\) (where \( RS=RU+US = 3 + 6=9\)). Wait, actually, the correct corresponding sides: since \( \triangle QRU\sim\triangle STU\), the ratio of \( QR\) to \( ST\) is equal to the ratio of \( QU\) to \( TU\) is equal to the ratio of \( RU\) to \( SU\). The ratio of \( RU\) to \( SU\) is \( \frac{3}{6}=\frac{1}{2}\), so the ratio of \( QU\) to \( TU\) is also \( \frac{1}{2}\). Let \( QU=x + 7\) and \( TU = 7\)? No, wait, let's re - establish. Let \( QT=x\) and \( QS=x + 7\). The ratio of \( RU\) to \( RS=\frac{3}{3 + 6}=\frac{3}{9}=\frac{1}{3}\)? No, \( RU = 3\), \( US=6\), so \( RS=3 + 6 = 9\), and \( RU/US=\frac{3}{6}=\frac{1}{2}\). So the ratio of similarity is \( \frac{1}{2}\) (since \( RU\) and \( US\) are corresponding sides). So the ratio of \( QU\) to \( TU\) is \( \frac{1}{2}\). Let \( QU=x + 7\) and \( TU = 7\)? No, \( QU=QT + TS=x + 7\), \( TU = 7\)? No, actually, the two triangles: \( \triangle QRU\) and \( \triangle STU\), with \( RU = 3\), \( SU=6\), so the scale factor from \( \triangle STU\) to \( \triangle QRU\) is \( \frac{RU}{SU}=\frac{3}{6}=\frac{1}{2}\). So the ratio of \( QU\) (which is \( QT + TS=x + 7\)) to \( TU\) (which is \( 7\)) should be equal to the scale factor? Wait, no. 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. Here, \( TU\parallel QR\) (since \( \angle Q=\angle T = 54^{\circ}\) and \( \angle R=\angle U = 90^{\circ}\)), so by Thales' theorem, \( \frac{RU}{US}=\frac{QT}{TS}\). We know that \( RU = 3\), \( US=6\), \( TS = 7\), and \( QT=x\). So \( \frac{3}{6}=\frac{x}{7}\).

Step2: Solve for \( x\)

From \( \frac{3}{6}=\frac{x}{7}\), we can cross - multiply. Cross - multiplying gives us \( 6x=3\times7\). So \( 6x = 21\). Then \( x=\frac{21}{6}=\frac{7}{2}=3.5\)? Wait, no, that's wrong. Wait, Thales' theorem: if a line is parallel to one side of a triangle, then it divides the other two sides proportionally. The line \( TU\) is parallel to \( QR\) (because \( \angle Q=\angle T\) and \( \angle R=\angle U\), so \( TU\parallel QR\)). So in \( \triangle QRS\), line \( TU\parallel QR\), so \( \frac{RU}{US}=\frac{QT}{TS}\). Wait, \( RU = 3\), \( US=6\), \( QT=x\), \( TS = 7\). So \( \frac{3}{6}=\frac{x}{7}\)? No, \( RU\) and \( US\) are on the base \( RS\), and \( QT\) and \( TS\) are on the side \( QS\). So \( \frac{RU}{RS}=\frac{QT}{QS}\)? \( RS=3 + 6 = 9\), \( QS=x + 7\), so \( \frac{3}{9}=\frac{x}{x + 7}\). Cross - multiplying: \( 3(x + 7)=9x\), \( 3x+21 = 9x\), \( 21=6x\), \( x=\frac{21}{6}=3.5\)? No, that's not right. Wait, the correct application of Thales' theorem: in \( \triangle QRS\), \( TU\parallel QR\), so \( \frac{US}{RU}=\frac{TS}{QT}\). Because \( US = 6\), \( RU = 3\), \( TS = 7\), \( QT=x\). So \( \frac{6}{3}=\frac{7}{x}\). Ah, that's the mistake earlier. The ratio is \( \frac{US}{RU}=\frac{TS}{QT}\) because the line \( TU\) divides \( RS\) into \( RU\) and \( US\) and \( QS\) into \( QT\) and \( TS\). So \( \frac{US}{RU}=\frac{TS}{QT}\), substituting the values: \( \frac{6}{3}=\frac{7}{x}\).

Step3: Solve the Proportion

Simplify \( \frac{6}{3}=\frac{7}{x}\), \( 2=\frac{7}{x}\). Cross - multiply: \( 2x=7\), so \( x=\frac{7}{2}=3.5\)? No, wait, \( \frac{US}{RU}=\frac{TS}{QT}\), \( US = 6\), \( RU = 3\), \( TS = 7\), \( QT=x\). So \( \frac{6}{3}=\frac{7}{x}\) implies \( 2=\frac{7}{x}\), so \( x=\frac{7}{2}=3.5\)? Wait, but let's check the angle. The two triangles have a \( 54^{\circ}\) angle, so they are similar. The ratio of the sides adjacent to the \( 54^{\circ}\) angle: in the smaller triangle, the base is \( 6\) and the side is \( 7\), in the larger triangle, the base is \( 3+6 = 9\) and the side is \( x + 7\). The ratio of the bases is \( \frac{9}{6}=\frac{3}{2}\), so the ratio of the sides should also be \( \frac{3}{2}\). So \( \frac{x + 7}{7}=\frac{9}{6}\). Simplify \( \frac{9}{6}=\frac{3}{2}\), so \( \frac{x + 7}{7}=\frac{3}{2}\). Cross - multiply: \( 2(x + 7)=21\), \( 2x+14 = 21\), \( 2x=7\), \( x=\frac{7}{2}=3.5\). Wait, that's the same result. So the correct proportion is based on similar triangles. Since \( \triangle QRU\) and \( \triangle STU\) are similar (AA similarity: \( \angle Q=\angle T = 54^{\circ}\), \( \angle R=\angle U = 90^{\circ}\)), the ratio of corresponding sides is equal. The ratio of \( RU\) to \( SU\) is \( \frac{3}{6}=\frac{1}{2}\), so the ratio of \( QU\) to \( TU\) is \( \frac{1}{2}\). Wait, \( QU=x + 7\), \( TU = 7\), so \( \frac{x + 7}{7}=\frac{3 + 6}{6}=\frac{9}{6}=\frac{3}{2}\)? No, I think I messed up the corresponding sides. Let's label the triangles properly. Let \( \triangle QRU\) with \( \angle Q = 54^{\circ}\), \( \angle R=90^{\circ}\), \( RU = 3\). Let \( \triangle STU\) with \( \angle T = 54^{\circ}\), \( \angle U=90^{\circ}\), \( SU = 6\). So the ratio of the legs (the sides adjacent to the \( 54^{\circ}\) angle) of \( \triangle QRU\) to \( \triangle STU\) is \( \frac{RU}{SU}=\frac{3}{6}=\frac{1}{2}\), so the ratio of the hypotenuses (or the other legs) should also be \( \frac{1}{2}\). The hypotenuse of \( \triangle STU\) is \( 7\), so the hypotenuse of \( \triangle QRU\) is \( \frac{7}{2}\)? No, that's not right. Wait, the correct way: the two triangles are similar, so the ratio of \( RU\) to \( SU\) is equal to the ratio of \( QT\) to \( TS\). \( RU = 3\), \( SU = 6\), \( QT=x\), \( TS = 7\). So \( \frac{3}{6}=\frac{x}{7}\) is wrong, \( \frac{3}{6}=\frac{TS}{QT}\)? No, \( \frac{RU}{SU}=\frac{QT}{TS}\) is incorrect. The correct proportion from similar triangles (AA) is that \( \frac{RU}{SU}=\frac{QR}{ST}=\frac{QU}{TU}\). Since \( RU = 3\), \( SU = 6\), the ratio \( k=\frac{RU}{SU}=\frac{3}{6}=\frac{1}{2}\). So \( QU=k\times TU\)? No, \( k\) is the scale factor from \( \triangle STU\) to \( \triangle QRU\), so \( QR=k\times ST\), \( QU=k\times TU\), \( RU=k\times SU\). Since \( RU = 3\) and \( SU = 6\), \( k=\frac{RU}{SU}=\frac{3}{6}=\frac{1}{2}\). So \( QU=\frac{1}{2}\times TU\)? No, that would mean \( TU = 2\times QU\). If \( TU = 7\), then \( QU=\frac{7}{2}\), but \( QU=x + 7\)? No, I think the correct approach is using 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 \( TU\parallel QR\), so \( \frac{RU}{US}=\frac{QT}{TS}\). Here, \( RU = 3\), \( US = 6\), \( QT=x\), \( TS = 7\). So \( \frac{3}{6}=\frac{x}{7}\) is incorrect. Wait, \( RU\) and \( US\) are segments on \( RS\), and \( QT\) and \( TS\) are segments on \( QS\). So the correct proportion is \( \frac{RU}{RS}=\frac{QT}{QS}\). \( RS=RU + US=3 + 6 = 9\), \( QS=QT + TS=x + 7\). So \( \frac{3}{9}=\frac{x}{x + 7}\). Cross - multiplying: \( 3(x + 7)=9x\), \( 3x+21 = 9x\), \( 21 = 6x\), \( x=\frac{21}{6}=\frac{7}{2}=3.5\). Yes, that's correct. So the value of \( x\) is \( \frac{7}{2}\) or \( 3.5\).

Answer:

\( \boldsymbol{\frac{7}{2}}\) (or \( 3.5\))