Draw a ΔABC with
Construction of ΔABC
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 B1, B2, B3, B4 and B5, on ray BX at equal distance, i.e., BB1 = B1B2 = B2B3 = B3B4 = B4B5.
b. B3 is joined with C to form B3 C as 3 point is smaller. B5 C1 is drawn parallel to B3 C as 5 point is greater.
c. Now C1A1is drawn parallel to CA.
Thus, A1BC1 is the required triangle.
Rate this question :