Question

A triangle $ABC$ with $BC=6cm,AB=5cm$ and $\angle ABC={60}^{\circ }$. 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. 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'.

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

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

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

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

Answer 1

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

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'.