Q. 26

# Draw a ΔABC with

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 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.

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