question_answer

Draw a triangle with sides 5 cm, 6 cm. and 7 cm. Then draw another triangle whose sides are \[\frac{4}{5}\] of the corresponding sides of first triangle.

Answer:

Steps of construction

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

(ii) With B as centre and radius equal to 5 cm, draw an arc.

(iii) With C as centre and radius equal to 7 cm, draw an arc.

(iv) Mark the point where the two arcs intersect as A. Join AB and AC

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

(v) Below BC, make an acute \[\angle CBX\].

(vi) Along BX, mark off five points \[{{B}_{1}},{{B}_{2}},{{B}_{3}},{{B}_{4}},{{B}_{5}}\] such that \[B\,{{B}_{1}}={{B}_{1}}{{B}_{2}}={{B}_{2}}{{B}_{3}}={{B}_{3}}{{B}_{4}}={{B}_{4}}{{B}_{5}}.\]

(vii) Join \[{{B}_{5}}C\].

(viii) From \[{{B}_{4}}\] draw \[{{B}_{4}}D\parallel {{B}_{5}}C\], meeting BC at D.

(ix) From D, draw \[DE\parallel CA\], meeting AB at E.

Then, \[\Delta \,EBD\] is the required triangle each of whose sides is \[\frac{4}{5}\] of the corresponding side of \[\Delta \,ABC\].