question_answer

Draw a triangle ABC with \[BC=6\text{ }cm,\text{ }AB=5\text{ }cm\] and \[\angle ABC=60{}^\circ \]. Then construct a triangle whose sides are \[\frac{3}{4}\] of the corresponding sides of the \[\Delta \,ABC\].

Answer:

Steps of construction -

(i) Draw a line segment \[BC=6\text{ }cm\].

(ii) Construct \[\angle XBC=60{}^\circ \].

(iii) With B as centre and radius equal to 5 cm, draw an arc intersecting XB at A.

(iv) Join AC. Thus, \[\Delta \text{ }ABC\]is obtained.

(v) Draw an acute angle \[\angle CBY\] below of B.

(vi) Mark 4-equal parts on BY as \[{{B}_{1}},{{B}_{2}},{{B}_{3}}\] and \[{{B}_{4}}\].

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

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

(ix) Draw another line parallel to CA from C?, intersecting AB at A?.

(x) \[\Delta \,A'BC'\] is required triangle which is similar to \[\Delta \,ABC\] such that \[BC'=\frac{3}{4}BC\].