Question

Draw a triangle ABC with side BC = 7 cm, $\angle$ B = ${45}^{\circ }$, and $\angle$ A = ${105}^{\circ }$. Then, construct a triangle whose sides are $\frac{4}{3}$ times the corresponding sides of $\Delta$ ABC

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 $B_{3}C$.

5. From $B_4$, draw $B_{4}D\parallel B_{3}C$, meeting BC (produced) at D.

6. From D, draw $DE\parallel CA$, meeting ray BX at E.

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