Question

similar triangle measures worksheet

1. the triangles are similar, solve for x.
2. the triangles are similar
3. δabc ~ δdef. solve for x and y.
4. δjkl ~ δmno. solve fo
5. δpqr has been dilated by a scale factor of 1.76 to form δpqr. find x and y.
6. δdef has been dilated by a scale factor of 1/2 to form δdef.
7. δhij ~ δpqr. solve for x, y, and z.
8. δabc ~ δdef

Explanation:

Step1: Solve for \( x \) in the first similar triangles problem

We have two similar triangles. The sides of the first triangle are 8 and 12, and the corresponding sides of the second triangle are \( x \) and 9. Since the triangles are similar, the ratios of corresponding sides are equal. So, \(\frac{8}{x} = \frac{12}{9}\). Cross - multiplying gives \( 12x = 8\times9 = 72 \). Then, \( x=\frac{72}{12}=6 \).

Step2: Solve for \( x \) and \( y \) in \( \triangle ABC\sim\triangle DEF \)

For side lengths: In \( \triangle ABC \), side \( BC = 7 \), and in \( \triangle DEF \), side \( EF=10.5 \), and side \( DE = 9 \). Let the side \( AB=x \). Since the triangles are similar, \(\frac{AB}{DE}=\frac{BC}{EF}\), so \(\frac{x}{9}=\frac{7}{10.5}\). Cross - multiplying, \( 10.5x=9\times7 = 63 \), and \( x=\frac{63}{10.5}=6 \). For angles: In similar triangles, corresponding angles are equal. So \( y = 61^{\circ} \) (since \( \angle C = 61^{\circ} \) and \( \angle F=y^{\circ} \), and they are corresponding angles).

Step3: Solve for \( x \) and \( y \) in the dilated triangle problem (\( \triangle PQR \) dilated to \( \triangle P'Q'R' \) with scale factor 1.76)

For side lengths: The scale factor \( k = 1.76 \). We know that \( P'Q' = 12 \) and \( PQ=x \), and \( \frac{P'Q'}{PQ}=k \), so \( \frac{12}{x}=1.76 \), then \( x=\frac{12}{1.76}\approx6.82 \) (or if we consider the other side, \( P'R' = 15 \) and \( PR = 8.5 \), and \( \frac{P'R'}{PR}=1.76 \) (since \( 15\div8.5\approx1.76 \)). For angles: In similar triangles (dilation preserves similarity), corresponding angles are equal. So \( y = 55^{\circ} \) (since \( \angle R'=55^{\circ} \) and \( \angle R = y^{\circ} \)).

Step4: Solve for \( x \), \( y \), and \( z \) in \( \triangle HIJ\sim\triangle PQR \)

For angles: In \( \triangle HIJ \), \( \angle I = 90^{\circ} \), \( \angle J \) can be found using the fact that the sum of angles in a triangle is \( 180^{\circ} \). In \( \triangle PQR \), \( \angle R = 47^{\circ} \), \( \angle P=43^{\circ} \), so \( \angle Q=x^{\circ}=180-(47 + 43)=90^{\circ} \). In \( \triangle HIJ \), \( \angle H=y^{\circ} \), and since \( \triangle HIJ\sim\triangle PQR \), \( \angle H=\angle P = 43^{\circ} \) (or we can calculate \( y = 180-(90 + \angle J) \), and \( \angle J \) corresponds to \( \angle R = 47^{\circ} \), so \( y=43^{\circ} \)). For side lengths: The ratio of sides of \( \triangle HIJ \) is \( \frac{HI}{IJ}=\frac{2}{3.4}\approx0.588 \), and in \( \triangle PQR \), \( \frac{PQ}{QR}=\frac{x}{4.1} \) (but \( x = 90^{\circ} \) for angle, for side \( z \): \( \frac{IJ}{QR}=\frac{3.4}{4.1} \), and \( \frac{HI}{PQ}=\frac{2}{x_{side}} \), but maybe a better way is to use the ratio of sides. Since \( \triangle HIJ\sim\triangle PQR \), \( \frac{HI}{PQ}=\frac{IJ}{QR}=\frac{HJ}{PR} \). We know \( HI = 2 \), \( IJ = 3.4 \), \( QR = 4.1 \). Let's find the ratio \( \frac{IJ}{QR}=\frac{3.4}{4.1}\approx0.829 \). Then, if we assume \( PQ \) corresponds to \( HI \), and \( PR = z \), \( HJ \) corresponds to \( PR \). First, find the angle \( x \): In \( \triangle PQR \), \( x=180-(47 + 43)=90^{\circ} \). Angle \( y \): In \( \triangle HIJ \), \( y = 43^{\circ} \) (since \( \angle P = 43^{\circ} \) and \( \angle H=y^{\circ} \) are corresponding angles). For side \( z \): The ratio of sides \( \frac{HJ}{PR}=\frac{IJ}{QR} \). First, find \( HJ \) using Pythagoras in \( \triangle HIJ \): \( HJ=\sqrt{HI^{2}+IJ^{2}}=\sqrt{2^{2}+3.4^{2}}=\sqrt{4 + 11.56}=\sqrt{15.56}\approx3.94 \). Then \( \frac{3.94}{z}=\frac{3.4}{4.1} \), cross - multiplying gives \( 3.4z=3.94\times4.1\approx16.154 \), so \( z=\frac{16.154}{3.4}\approx4.75 \).

Answer:

1. \( x = 6 \)
2. \( x = 6,y = 61^{\circ} \)
3. \( x\approx6.82,y = 55^{\circ} \)
4. \( x = 90^{\circ},y = 43^{\circ},z\approx4.75 \) (answers may vary slightly depending on the method of calculation for the dilated and similar triangle side - length problems)