Triangles $ABC$ and $PQR$ are similar. Angle $A$ matches angle $P$, and angle $B$ matches angle $Q$.

$AB=10$ cm, $AC=18$ cm, $PQ=15$ cm, and $QR=24$ cm. Find $PR$ and $BC$.

1. Use the matching angles to pair the sides: $AB\leftrightarrow PQ$, $AC\leftrightarrow PR$, and $BC\leftrightarrow QR$.

2. Use the known matching sides $AB$ and $PQ$ to find the scale factor from triangle $ABC$ to triangle $PQR$.

   $$\frac{PQ}{AB}=\frac{15}{10}=1.5$$

3. Since $AC$ corresponds to $PR$, multiply 18 by 1.5 to get 27 cm.

   $$PR=18\times 1.5=27$$

4. Since $BC$ corresponds to $QR$, divide 24 by 1.5 to get 16 cm.

   $$BC=24÷1.5=16$$

In triangle $ACD$, point $B$ lies on $AC$ and point $E$ lies on $AD$. The line $BE$ is parallel to $CD$.

$AB=10$ cm, $BC=5$ cm, $AE=8$ cm, and $CD=12$ cm. Find $ED$ and $BE$.

1. First find the full length $AC$ by adding the two parts.

   $$AC=10+5=15$$

2. Find the scale factor from the small triangle $ABE$ to the large triangle $ACD$.

   $$\frac{AC}{AB}=\frac{15}{10}=1.5$$

3. Use the same scale factor on $AE$ to find the full length $AD$, giving 12 cm.

   $$AD=8\times 1.5=12$$

4. Subtract to find the extra piece $ED$, giving 4 cm.

   $$ED=12-8=4$$

5. Since $CD$ is the large base, divide by 1.5 to find the small base $BE$, giving 8 cm.

   $$BE=12÷1.5=8$$

$AB$ is parallel to $XY$. The straight lines $AY$ and $BX$ meet at $P$.

$AB=7$ cm, $XY=21$ cm, and $XP=18$ cm. Find $BP$.

1. The similar triangles are triangle $ABP$ and triangle $XYP$.

2. Match the corresponding sides: $AB\leftrightarrow XY$ and $BP\leftrightarrow XP$.

3. Find the scale factor from triangle $ABP$ to triangle $XYP$.

   $$\frac{XY}{AB}=\frac{21}{7}=3$$

4. Since $XP$ is the larger matching side, divide by 3 to find $BP$, giving 6 cm.

   $$BP=18÷3=6$$

Two triangles meet at $C$. Points $A$, $C$, and $E$ are on a straight line, and points $B$, $C$, and $D$ are on a straight line. Also, $AB\parallel DE$.

$AB=8$ cm, $DE=12$ cm, $AC=10$ cm, and $CD=18$ cm. Find $CE$ and $BC$.

1. The similar triangles are triangle $ABC$ and triangle $EDC$.

2. Use the parallel sides to find the scale factor from triangle $ABC$ to triangle $EDC$.

   $$\frac{DE}{AB}=\frac{12}{8}=1.5$$

3. Since $AC$ corresponds to $CE$, multiply 10 by 1.5 to get 15 cm.

   $$CE=10\times 1.5=15$$

4. Since $BC$ corresponds to $CD$, divide 18 by 1.5 to get 12 cm.

   $$BC=18÷1.5=12$$

A small right-angled triangle is similar to a larger right-angled triangle. The side $AB$ corresponds to $AC$, and $BE$ corresponds to $CD$.

$AB=6$ cm, $BE=4$ cm, and $AB:AC=2:5$. Find $CD$ and $BC$.

1. Convert the ratio into a scale factor from the smaller triangle to the larger triangle.

   $$\frac{AC}{AB}=\frac{5}{2}=2.5$$

2. Since $BE$ corresponds to $CD$, multiply 4 by 2.5 to get 10 cm.

   $$CD=4\times 2.5=10$$

3. Find the full length $AC$ by multiplying 6 by 2.5 to get 15 cm.

   $$AC=6\times 2.5=15$$

4. If $B$ lies between $A$ and $C$, subtract to find $BC$, giving 9 cm.

   $$BC=15-6=9$$

A small rectangle measures 140 mm by 75 mm. A larger rectangle measures 180 mm by 90 mm. Show that the rectangles are not mathematically similar.

1. Compare the longer sides.

   $$\frac{180}{140}=\frac{9}{7}\approx 1.29$$

2. Compare the shorter sides.

   $$\frac{90}{75}=1.2$$

3. The two scale factors are different, so the rectangles are not mathematically similar.