Question

Draw an isosceles $\Delta ABC$ in which $BC=5.5$cm and altitude $AL=3$ cm. Then construct another triangle whose sides are $\frac{3}{4}$ of the corresponding sides of $\Delta ABC$.

Answer 1

Steps of construction:

(i) Draw a line segment $BC=5.5$ cm.

(ii) Draw a perpendicular bisector of BC intersecting



BC at L such that $AL=3$ cm.

(iii) Join AB and AC

Thus, $\Delta ABC$ is obtained.

(iv) Below BC, make an acute angle $\angle CBX$.

(v) Along BX, mark off four points $B_1,B_2,B_3,B_4$

such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$.

(vi) Join $B_{4}C$.

(vii) From $B_3$ draw $B_{3}D||B_{4}C$, meeting BC at D.

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

Then $\Delta EBD$ is the required triangle each of whose

sides is $\frac{3}{4}$ of the corresponding side of $\Delta ABC$.