Similarity and Dilations: practice solutions, Grade 10 – download the PDF

1 Dilating points from the origin ★★★

A dilation from the origin sends \((x,y)\) to \((3x,3y)\).

1. \(A'=(6,-3)\)
2. \(B'=(-12,15)\)
3. \(C'=(0,18)\)

2 Finding the scale factor ★★★

1. \(k=\dfrac{10}{4}=\dfrac{15}{6}=\dfrac{20}{8}=2.5\). The scale factor is 2.5.
2. Going back, \(k=\dfrac{4}{10}=0.4\), the reciprocal of 2.5.

3 True or false? ★★★

1. True. All angles are \(90^\circ\) and the sides have one common ratio (side over side).
2. False. A 2 by 6 rectangle and a 4 by 8 rectangle have equal angles, but \(\tfrac42\neq\tfrac86\).
3. True. All angles are \(60^\circ\), so AA applies.
4. True. Since \(0

4 Corresponding sides ★★★

The scale factor is \(k=\dfrac{DE}{AB}=\dfrac86=\dfrac43\).

\(EF=\dfrac43\cdot 9=12\) and \(DF=\dfrac43\cdot12=16\).

5 Enlarging a photo ★★★

Scale factor: \(k=\dfrac{10}{4}=2.5\).

New height: \(6\times2.5=15\) in. The enlarged photo is 10 in. by 15 in.

6 Reading a map scale ★★★

1. \(3.5\times15=52.5\) miles.
2. \(\dfrac{120}{15}=8\) inches on the map.

7 Angle sum and AA ★★★

\(\angle C=180^\circ-52^\circ-71^\circ=57^\circ\).

So \(\angle B=\angle D=71^\circ\) and \(\angle C=\angle F=57^\circ\). By AA, \(\triangle BCA\sim\triangle DFE\). The triangles are similar (with \(A\) matching \(E\)).

8 Shadows and height ★★★

The sun’s rays are parallel, so the angles with the ground match, and both the person and the tree make a right angle with the ground. By AA the right triangles are similar.

\(\dfrac{h}{1.8}=\dfrac{14.4}{2.4}=6\), so \(h=1.8\times6=10.8\) m. The tree is 10.8 m tall.

9 SSS or not? ★★★

1. \(\tfrac96=\tfrac{12}{8}=\tfrac{15}{10}=1.5\). Similar by SSS, \(k=1.5\).
2. \(\tfrac{10}{5}=2\), \(\tfrac{14}{7}=2\), but \(\tfrac{19}{9}\approx2.11\). The ratios differ, so the triangles are not similar.

10 Using SAS ★★★

1. \(\dfrac{DE}{AB}=\dfrac{10}{6}=\dfrac53\) and \(\dfrac{DF}{AC}=\dfrac{15}{9}=\dfrac53\). The included angles \(A\) and \(D\) are congruent, so \(\triangle ABC\sim\triangle DEF\) by SAS.
2. \(EF=\dfrac53\cdot7.2=12\).

11 A line parallel to a side ★★★

By the triangle proportionality theorem, \(\dfrac{AD}{DB}=\dfrac{AE}{EC}\), so \(\dfrac64=\dfrac{9}{EC}\).

\(EC=\dfrac{9\cdot4}{6}=6\), and \(AC=9+6=15\).

12 Is it parallel? ★★★

Use the converse of the triangle proportionality theorem.

1. \(\tfrac{5}{10}=\tfrac12\) and \(\tfrac48=\tfrac12\). The ratios are equal, so \(DE\parallel BC\).
2. \(\tfrac35=0.6\) but \(\tfrac46\approx0.67\). Not equal, so \(DE\) is not parallel to \(BC\).

13 Bisecting an angle ★★★

By the angle bisector theorem, \(\dfrac{BD}{DC}=\dfrac{12}{20}=\dfrac35\).

So \(BD=\dfrac38\cdot24=9\) and \(DC=\dfrac58\cdot24=15\). Check: \(9+15=24\).

14 Perimeter and area of a model ★★★

Perimeter: \(14\times3=42\) cm.

Area: \(5\times3^2=45\) cm².

15 Altitude on the hypotenuse ★★★

1. Angle \(A\) is shared by both triangles. Also \(\angle ADC=90^\circ=\angle ACB\). By AA, \(\triangle ACD\sim\triangle ABC\).
2. Matching sides: \(\dfrac{AD}{AC}=\dfrac{AC}{AB}\), so \(AD=\dfrac{15^2}{25}=9\). Also \(\dfrac{CD}{BC}=\dfrac{AC}{AB}\), so \(CD=\dfrac{20\cdot15}{25}=12\). Check: \(9^2+12^2=225=15^2\).

16 Diagonals of a trapezoid ★★★

1. \(\angle BAE\cong\angle DCE\) (alternate interior angles, \(AB\parallel CD\)) and \(\angle AEB\cong\angle CED\) (vertical angles). By AA, \(\triangle ABE\sim\triangle CDE\).
2. \(\dfrac{AE}{CE}=\dfrac{AB}{CD}=\dfrac{12}{8}\), so \(CE=\dfrac{9\cdot8}{12}=6\) and \(AC=9+6=15\).

17 Areas of similar triangles ★★★

1. The ratio of areas is \(\dfrac{50}{18}=\dfrac{25}{9}=k^2\), so \(k=\dfrac53\).
2. \(21\times\dfrac53=35\) cm.

18 Scale model of a building ★★★

1. \(9.5\times4=38\) ft (that is, \(9.5\times48=456\) in. = 38 ft, about 11.6 m).
2. The scale factor is 48, so the area ratio is \(48^2=2304\). \(150\times2304=345{,}600\) in², and \(\dfrac{345{,}600}{144}=2400\) ft².

19 Dilation with another center ★★★

1. \(A'=(1+2\cdot2,\;2+2\cdot2)=(5,6)\).
2. \(B'=(1+2(-1),\;2+2(-2))=(-1,-2)\).
3. \(C'=(1+0.5\cdot4,\;2+0.5\cdot(-4))=(3,0)\).

20 Solving with an unknown ★★★

\(\dfrac{AD}{DB}=\dfrac{AE}{EC}\) gives \(\dfrac{x}{6}=\dfrac{x+1}{9}\).

Cross-multiplying: \(9x=6x+6\), so \(3x=6\) and \(x=2\). Then \(AD=2\) and \(AE=3\). Check: \(\tfrac26=\tfrac39=\tfrac13\).

21 Proving the theorem ★★★

1. \(\angle A\cong\angle A\) (shared angle).
2. \(\angle ADE\cong\angle ABC\): corresponding angles, since \(DE\parallel BC\) and \(AB\) is a transversal.
3. By AA, \(\triangle ADE\sim\triangle ABC\).
4. Corresponding sides of similar triangles are proportional: \(\dfrac{AD}{AB}=\dfrac{AE}{AC}=\dfrac{DE}{BC}\).