Question

Construct an isosceles triangle whose base is 6 cm and altitude 4 cm. Then, construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the first triangle.

Answer 1

Steps of construction :

1. Construct a $\Delta$ ABC with given measurement.

2. Below BC, make an acute angle $\angle$ CBZ.

3. Along BZ, mark off four points $B_1,B_2,B_3$ and $B_4$ such that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$

4. Join $B_{4}C$

5. From $B_3$, draw $B_{3}E\parallel B_{4}C$, meeting BC at E

6. From E, draw $EF\parallel CA$, meeting BA at F.

7. $\Delta FBE$ is the required triangle, each of whose sides is $\frac{3}{4}$ times the corresponding side of $\Delta$ ABC