Question

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 1

Steps of construction

(i) Draw a line segment $BC=6cm$

(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, $\Delta ABC$ 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 $B_{4}D||B_{5}C$, meeting BC at D.

(ix) From D, draw $DE||CA$, meeting AB at E.

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