question_answer

Construct a triangle ABC with side\[BC=7\text{ }cm,\text{ }\angle B=45{}^\circ ,\text{ }\angle A=105{}^\circ \]. Then construct another triangle whose sides are \[\frac{3}{4}\] times the corresponding sides of the \[\Delta \text{ }ABC\].

Answer:

\[BC=7\text{ }cm,\,\angle B=45{}^\circ ,\text{ }\angle \text{ }A=105{}^\circ \]

\[\angle C=180{}^\circ -(\angle B+\angle A)\]

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

\[=180{}^\circ -150{}^\circ \]

\[=30{}^\circ \]

Steps of construction-

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

(ii) Draw an angle \[45{}^\circ \] at B and \[30{}^\circ \] at C. They intersect at A.

(iii) Draw an acute angle at B.

(iv) Divide angle ray in 4 equal parts as \[{{B}_{1}},{{B}_{2}},{{B}_{3}}\] and \[{{B}_{4}}\].

(v) Join \[{{B}_{4}}\] to C?.

(vi) From \[{{B}_{3}}\]draw a line parallel to \[{{B}_{4}}C\] intersecting BC at C?.

(vii) Draw another line parallel to CA from C' intersecting AB ray at A'.

(viii) \[\Delta \text{ }A'BC'\] is required triangle such that \[\Delta \text{ }A'BC'\sim \Delta \,ABC\] with \[A'B=\frac{3}{4}AB\].