Answer :

Construction of ΔABC

Steps:

a. Draw a base line BC = 6 cm

b. From point B make an arc with radius 8cm. From point C make an arc with radius 9cm.The two arcs meets at a point; this point is point A. Join AB and AC.

c. Hence, given ΔABC is constructed.

Now we will construct a triangle similar to ΔABC whose sides are of the corresponding sides of ΔABC.

Steps of construction:

a. Draw a ray BX making an acute angle with AB on the side opposite to vertex C. In the greater number is 5, so we will make 5 points, namely B_{1}, B_{2}, B_{3}, B_{4} and B_{5}, on ray BX at equal distance, i.e., BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}.

b. B_{3} is joined with C to form B_{3} C as 3 point is smaller. B_{5} C_{1} is drawn parallel to B_{3} C as 5 point is greater.

c. Now C_{1}A_{1}is drawn parallel to CA.

Thus, A_{1}BC_{1} is the required triangle.

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