State which pairs of triangles in figure,

Solution:

(i) Yes. $\angle \mathrm{A}=\angle \mathrm{P}=60^{\circ}, \angle \mathrm{B}=\angle \mathrm{Q}=80^{\circ}$,

$\angle \mathrm{C}=\angle \mathrm{R}=40^{\circ}$

Therefore, $\Delta \mathrm{ABC} \sim \Delta \mathrm{PQR}$.

By AAA similarity criterion

(ii) Yes.

$\frac{A B}{Q R}=\frac{2}{4}=\frac{1}{2}, \frac{B C}{R P}=\frac{2.5}{5}=\frac{1}{2}, \frac{C A}{P Q}=\frac{3}{6}=\frac{1}{2}$

Therefore, $\Delta \mathrm{ABC} \sim \Delta \mathrm{QRP}$.

By SSS similarity criterion.

(iii) No.

$\frac{\mathrm{MP}}{\mathrm{DE}}=\frac{2}{4}=\frac{1}{2}, \frac{\mathrm{LP}}{\mathrm{DF}}=\frac{3}{6}=\frac{1}{2}, \frac{\mathrm{LM}}{\mathrm{EF}}=\frac{2.7}{5} \neq \frac{1}{2}$

i.e., $\frac{M P}{D E}=\frac{L P}{D F} \neq \frac{L M}{E F}$

Thus, the two triangles are not similar.

(iv) Yes,

$\frac{M N}{O P}=\frac{M L}{O R}=\frac{1}{2}$

and $\angle \mathrm{NML}=\angle \mathrm{PQR}=70^{\circ}$

By SAS similarity criterion

$\Delta \mathrm{NML} \sim \Delta \mathrm{PQR}$

(v) No,

$\frac{A B}{F D} \neq \frac{A C}{F E}$

Thus, the two triangles are not similar

(vi) In triangle DEF $\angle \mathrm{D}+\angle \mathrm{E}+\angle \mathrm{F}=180^{\circ}$

$70^{\circ}+80^{\circ}+\angle \mathrm{F}=180^{\circ}$

$\angle \mathrm{F}=30^{\circ}$

In triangle PQR

$\angle \mathrm{P}+80^{\circ}+30^{\circ}=180^{\circ}$

$\angle \mathrm{P}=70^{\circ}$

$\angle \mathrm{E}=\angle \mathrm{Q}=80^{\circ}$

$\angle \mathrm{D}=\angle \mathrm{P}=70^{\circ}$

$\angle \mathrm{r}=\angle \mathrm{K}=30$

By AAA similarity criterion

$\Delta \mathrm{DEF} \sim \Delta \mathrm{PQR}$