Question

Find the pair of similar triangles among the given pairs. State the similarity criterion and write the similarity relation in symbolic form.

(i)

(ii)

(iii)

(iv)

(v)

Answer 1

(i)
We have:
$$\begin{gathered} \angle BAC=\angle PQR=50° \\ \angle ABC=\angle QPR=60° \\ \angle ACB=\angle PRQ=70° \end{gathered}$$

Therefore, by AAA similarity theorem, $△ABC~QPR$

(ii)
We have:

$\frac{AB}{DF}=\frac{3}{6}=\frac{1}{2}and\frac{BC}{DE}=\frac{4.5}{9}=\frac{1}{2}$

But, $\angle ABC\hspace{0.17em}\ne \angle EDF$ (Included angles are not equal)
Thus, this does not satisfy SAS similarity theorem.
Hence, the triangles are not similar.

(iii)
We have:
$$\begin{gathered} \frac{CA}{QR}=\frac{8}{6}=\frac{4}{3}and\frac{CB}{PQ}=\frac{6}{4.5}=\frac{4}{3} \\ \Rightarrow \frac{CA}{QR}=\frac{CB}{PQ} \end{gathered}$$
Also, $\angle ACB=\angle PQR=80°$
Therefore, by SAS similarity theorem, $△ACB~△RQP$.

(iv)
We have
$$\begin{gathered} \frac{DE}{QR}=\frac{2.5}{5}=\frac{1}{2} \\ \frac{EF}{PQ}=\frac{2}{4}=\frac{1}{2} \\ \frac{DF}{PR}=\frac{3}{6}=\frac{1}{2} \\ \Rightarrow \frac{DE}{QR}=\frac{EF}{PQ}=\frac{DF}{PR} \end{gathered}$$

Therefore, by SSS similarity theorem, $△FED~△PQR$

(v)
$$\begin{gathered} \mathrm{In}△\mathrm{ABC} \\ \angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180°\left(\mathrm{Angle}\mathrm{Sum}\mathrm{Property}\right) \\ \Rightarrow 80°+\angle \mathrm{B}+70°=180° \\ \Rightarrow \angle \mathrm{B}=30° \end{gathered}$$
$\angle \mathrm{A}=\angle \mathrm{M}\mathrm{and}\angle \mathrm{B}=\angle \mathrm{N}$
Therefore, by AA similarity theorem, $△\mathrm{ABC}~△\mathrm{MNR}$