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Answer :

i) Given, in \( \triangle \) ABC and \( \triangle \) PQR,

\( \angle \)A = \( \angle \)P = 60°

\( \angle \)B = \( \angle \)Q = 80°

\( \angle \)C = \( \angle \)R = 40°

Therefore by AAA similarity criterion,

\( \triangle \) ABC is similar to \( \triangle \) PQR

(ii) Given, in \( \triangle \) ABC and \( \triangle \) PQR,

\(\frac{AB}{QR} = \frac{BC}{RP} = \frac{CA}{PQ} \)

By SSS similarity criterion,

\( \triangle \) ABC is similar to \( \triangle \) QRP

(iii) Given, in \( \triangle \) LMP and \( \triangle \) DEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

\(\frac{MP}{DE} = \frac{2}{4} = \frac{1}{2} \)

\(\frac{PL}{DF} = \frac{3}{6} = \frac{1}{2} \)

\(\frac{LM}{EF} = \frac{2.7}{5} = \frac{27}{50} \)

Here , \(\frac{MP}{DE} = \frac{PL}{DF} \) \(\ne\) \(\frac{LM}{EF} \)

Therefore, \( \triangle \) LMP and \( \triangle \) DEF are not similar.

(iv) In \( \triangle \) MNL and \( \triangle \) QPR, it is given,

\(\frac{MN}{QP} = \frac{ML}{QR} = \frac{1}{2} \)

\( \angle \)M = \( \angle \)Q = 70°

Therefore, by SAS similarity criterion

\( \triangle \) MNL is similar to \( \triangle \) QPR

(v) In \( \triangle \) ABC and \( \triangle \) DEF, given that,

AB = 2.5, BC = 3, \( \angle \)A = 80°, EF = 6, DF = 5, \( \angle \)F = 80°

Here , \(\frac{AB}{DF} = \frac{2.5}{5} = \frac{1}{2} \)

And, \(\frac{BC}{EF} = \frac{3}{6} = \frac{1}{2} \)

\( \angle \)B \(\ne\) \( \angle \)F

Hence, \( \triangle \) ABC and \( \triangle \) DEF are not similar.

(vi) In \( \triangle \) DEF, by sum of angles of triangles, we know that,

\( \angle \)D + \( \angle \)E + \( \angle \)F = 180°

70° + 80° + \( \angle \)F = 180°

\( \angle \)F = 180° – 70° – 80°

\( \angle \)F = 30°

Similarly, In \( \triangle \) PQR,

\( \angle \)P + \( \angle \)Q + \( \angle \)R = 180 (Sum of angles of \( \triangle \) )

\( \angle \)P + 80° + 30° = 180°

\( \angle \)P = 180° – 80° -30°

\( \angle \)P = 70°

Now, comparing both the triangles, \( \triangle \) DEF and \( \triangle \) PQR, we have

\( \angle \)D = \( \angle \)P = 70°

\( \angle \)E = \( \angle \)Q = 80°

\( \angle \)F = \( \angle \)R = 30°

Therefore, by AAA similarity criterion,

Hence, \( \triangle \) DEF is similar to \( \triangle \) PQR

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- Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of \(\triangle\) PQR (see Fig 6.41). Show that \(\triangle\) ABC ~ \(\triangle\) PQR.
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- If AD and PM are medians of triangles ABC and PQR, respectively where \(\triangle\) ABC ~ \(\triangle\) PQR prove that AB/PQ = AD/PM.

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