3. Triangles

1. If \(\triangle\)XYZ \(\sim\) \(\triangle\)LMN, write the corresponding angles of the two triangles and also write the ratios of corresponding sides.

Solution:

\(\triangle\)XYZ \(\sim\) \(\triangle\)LMN ... (Given)

∴ The corresponding angles are:
\(\angle\)X \(\cong\) \(\angle\)L
\(\angle\)Y \(\cong\) \(\angle\)M
\(\angle\)Z \(\cong\) \(\angle\)N

Also, ratios of corresponding sides are:

\(\displaystyle \frac{\text{XY}}{\text{LM}} = \frac{\text{YZ}}{\text{MN}} = \frac{\text{XZ}}{\text{LN}}\)

2. In \(\triangle\)XYZ, XY = 4 cm, YZ = 6 cm, XZ = 5 cm. If \(\triangle\)XYZ \(\sim\) \(\triangle\)PQR and PQ = 8 cm, then find the lengths of remaining sides of \(\triangle\)PQR.

Solution:

\(\triangle\)XYZ \(\sim\) \(\triangle\)PQR ... (Given)

∴ \(\displaystyle \frac{\text{XY}}{\text{PQ}} = \frac{\text{YZ}}{\text{QR}} = \frac{\text{XZ}}{\text{PR}}\) ... (c s s t)

∴ \(\displaystyle \frac{4}{8} = \frac{6}{\text{QR}} = \frac{5}{\text{PR}}\) ... (Given)

∴ \(\displaystyle \frac{1}{2} = \frac{6}{\text{QR}} = \frac{5}{\text{PR}}\)

∴ \(\displaystyle \frac{1}{2} = \frac{6}{\text{QR}}\)

∴ \(\text{QR} = 6 \times 2\)

∴ \(\text{QR} = 12\) cm ... (i)

Also, \(\displaystyle \frac{1}{2} = \frac{5}{\text{PR}}\)

∴ \(\text{PR} = 5 \times 2\)

∴ \(\text{PR} = 10\) cm ... (ii)

∴ \(\text{QR} = 12\) cm and \(\text{PR} = 10\) cm

3. Draw a sketch of a pair of similar triangles. Label them. Show their corresponding angles by the same sign. Show the lengths of corresponding sides by numbers in proportion.

Solution:

\(\triangle\)ABC \(\sim\) \(\triangle\)PQR
\(\angle\)A \(\cong\) \(\angle\)P
\(\angle\)B \(\cong\) \(\angle\)Q
\(\angle\)C \(\cong\) \(\angle\)R
... (Corresponding Angles of Similar Triangles)

And,

\(\displaystyle \frac{\text{AB}}{\text{PQ}} = \frac{\text{BC}}{\text{QR}} = \frac{\text{AC}}{\text{PR}} = \frac{3}{2}\)

... (Corresponding Sides of Similar Triangles)