Question

A triangle ABC with sides BC =7cm, $\angle B={45}^{\circ },\angle A={105}^{\circ }$ is given. The image of constructing a similar triangle of $\Delta ABC$ whose sides are $\frac{3}{4}$ times the corresponding sides of the $\Delta ABC$ is given below. Arrange the steps of construction in the correct order.

1.Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

2.Join $B_{4}C$ and raw a line through $B_3$( the third point, 3 being smaller of 4 in$\frac{3}{4}$) parallel to $B_{4}C$ to intersect BC at C'.

3.Draw a line through C' parallel to the line CA to intersect BA at A'.

4.Locate 4(the greater of 3 and 4 in$\frac{3}{4}$) points $B_1,B_2,B_3\text{and}{B}_4$ on BX so that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$

• A. 1, 3, 2, 4
• B. 1, 4, 2, 3
• C. 2, 1, 3, 4
• D. 2, 3, 1, 4

Answer 1

The correct option is B 1, 4, 2, 3

Steps to be followed to for the construction of a similar triangle of $\Delta ABC$ is given below.

1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.

2.Locate 4(the greater of 3 and 4 in$\frac{3}{4}$) points $B_1,B_2,B_3\text{and}{B}_4$ on BX so that $B{B}_1=B_1{B}_2=B_2{B}_3=B_3{B}_4$

3.Join $B_{4}C$ and raw a line through $B_3$( the third point, 3 being smaller of 4 in$\frac{3}{4}$) parallel to $B_{4}C$ to intersect BC at C'.

4.Draw a line through C' parallel to the line CA to intersect BA at A'.