question_answer

Draw a triangle ABC with \[BC=7\text{ }cm,\angle B=45{}^\circ \] and \[\angle A=105{}^\circ \]. Then construct a triangle whose sides are \[\frac{4}{5}\] times the corresponding sides of \[\Delta \text{ }ABC\].

Answer:

\[\angle B=45{}^\circ \] and \[\angle A=105{}^\circ \]

\[\because \] Sum of angles of triangle is \[180{}^\circ \]

\[\therefore \angle A+\angle B+\angle C=180{}^\circ \]

\[105{}^\circ +45{}^\circ +\angle C=180{}^\circ \]

\[\Rightarrow \angle C=180{}^\circ -(105{}^\circ +45{}^\circ )\]

\[\Rightarrow \angle C=30{}^\circ \]

Steps of construction:

(i) Draw a line segment \[BC=7\text{ }cm\]

(ii) Construct \[\angle B=45{}^\circ \] and \[\angle C=30{}^\circ \]

(iii) A is the intersecting point of ray through B and C.

Thus, \[\Delta \text{ }ABC\] is obtained.

(iv) Draw D on BC such that \[BD=\frac{4}{5}BC=\left( \frac{4}{5}\times 7 \right)cm=\frac{28}{5}cm=5.6\,cm\]

Then, \[\Delta \text{ }BDE\] is the required triangle similar to \[\Delta \text{ }ABC\]such that each side of \[\Delta \text{ }BDE\] is \[\frac{4}{5}\] times the corresponding side of\[\Delta \text{ }ABC\].