Question 1

We will use the following similarity criterion:

• AAA (Angle-Angle-Angle): If all three angles of one triangle are equal to the corresponding angles of another triangle, the triangles are similar.
• SSS (Side-Side-Side): If the ratios of all three corresponding sides of two triangles are equal, the triangles are similar.
• SAS (Side-Angle-Side): If two sides of one triangle are proportional to two corresponding sides of another triangle, and the included angles are equal, the triangles are similar.

We check for similarity using AAA, SSS, or SAS criteria.

Step 1 — Analyze pair (i)

Let's look at the angles of $\triangle ABC$. We have $\angle A = \textbf{60}^\circ$. We have $\angle B = \textbf{80}^\circ$. We have $\angle C = \textbf{40}^\circ$. Let's look at the angles of $\triangle PQR$. We have $\angle P = \textbf{60}^\circ$. We have $\angle Q = \textbf{80}^\circ$. We have $\angle R = \textbf{40}^\circ$. We see that all corresponding angles are equal. So, $\triangle ABC$ is similar to $\triangle PQR$.

Step 2 — Analyze pair (ii)

Let's look at the sides of $\triangle ABC$. We have AB = $\textbf{2}$, BC = $\textbf{2.5}$, AC = $\textbf{3}$. Let's look at the sides of $\triangle PQR$. We have PQ = $\textbf{6}$, QR = $\textbf{4}$, PR = $\textbf{5}$. We compare the ratios of corresponding sides. Let's try matching AB with QR, BC with RP, and AC with PQ. $\frac{AB}{QR} = \frac{2}{4}$ $= \frac{1}{2}$ $\frac{BC}{RP} = \frac{2.5}{5}$ $= \frac{1}{2}$ $\frac{AC}{PQ} = \frac{3}{6}$ $= \frac{1}{2}$ All corresponding sides are proportional. So, $\triangle ABC$ is similar to $\triangle QRP$.

Step 3 — Analyze pair (iii)

Let's look at the sides of $\triangle LMP$. We have LM = $\textbf{2.7}$, MP = $\textbf{2}$, LP = $\textbf{3}$. Let's look at the sides of $\triangle DEF$. We have DE = $\textbf{4}$, EF = $\textbf{5}$, DF = $\textbf{6}$. We compare the ratios of corresponding sides. $\frac{LM}{DE} = \frac{2.7}{4}$ $= 0.675$ $\frac{MP}{EF} = \frac{2}{5}$ $= 0.4$ $\frac{LP}{DF} = \frac{3}{6}$ $= 0.5$ The ratios of corresponding sides are not equal. So, the triangles are not similar.

Step 4 — Analyze pair (iv)

Let's look at $\triangle NML$. We have NM = $\textbf{2.5}$, $\angle N = \textbf{70}^\circ$, ML = $\textbf{5}$. Let's look at $\triangle PQR$. We have PQ = $\textbf{5}$, $\angle Q = \textbf{70}^\circ$, QR = $\textbf{10}$. We compare the ratios of two sides. $\frac{NM}{PQ} = \frac{2.5}{5}$ $= \frac{1}{2}$ $\frac{ML}{QR} = \frac{5}{10}$ $= \frac{1}{2}$ We have $\angle N = \angle Q = \textbf{70}^\circ$. However, the angle $\angle N$ is not included between sides NM and ML. The SAS similarity criterion requires the included angle. So, the triangles are not similar.

Step 5 — Analyze pair (v)

Let's look at $\triangle ABC$. We have AB = $\textbf{2.5}$, $\angle A = \textbf{80}^\circ$, AC = $\textbf{3}$. Let's look at $\triangle DEF$. We have DF = $\textbf{5}$, EF = $\textbf{6}$, $\angle F = \textbf{80}^\circ$. We check if the corresponding sides are proportional for any valid correspondence. We find that the triangles are not similar. The corresponding sides are not proportional.

Step 6 — Analyze pair (vi)

Let's find the third angle in $\triangle DEF$. $\angle D + \angle E + \angle F = 180^\circ$ $\textbf{70}^\circ + \textbf{80}^\circ + \angle F = 180^\circ$ $\textbf{150}^\circ + \angle F = 180^\circ$ $\angle F = 180^\circ - \textbf{150}^\circ$

$\boxed{\angle F = 30^\circ}$ Let's find the third angle in $\triangle PQR$. $\angle P + \angle Q + \angle R = 180^\circ$ $\angle P + \textbf{80}^\circ + \textbf{30}^\circ = 180^\circ$ $\angle P + \textbf{110}^\circ = 180^\circ$ $\angle P = 180^\circ - \textbf{110}^\circ$ $\boxed{\angle P = 70^\circ}$ Now we compare the angles of $\triangle DEF$ and $\triangle PQR$. We have $\angle D = \textbf{70}^\circ = \angle P$. We have $\angle E = \textbf{80}^\circ = \angle Q$. We have $\angle F = \textbf{30}^\circ = \angle R$. All corresponding angles are equal. So, $\triangle DEF$ is similar to $\triangle PQR$.

Answer

(i) $\triangle ABC \sim \triangle PQR$ [AAA similarity] (ii) $\triangle ABC \sim \triangle QRP$ [SSS similarity] (iii) Triangles are not similar because the corresponding sides are not proportional. (iv) Triangles are not similar because the corresponding sides are not proportional. (v) Triangles are not similar because the corresponding sides are not proportional. (vi) $\triangle DEF \sim \triangle PQR$ [AAA similarity]