Answer :

__Steps of Construction:__

1. Draw a line segment BC = 3cm

2. With the help of compass, draw ∠B = 90°

3. Taking B as centre, 4cm as radius, draw an arc which intersects the ray at point A.

4. Join A to C

∴ ΔABC is the required triangle

Now, we need to make a triangle which is times its size

__Steps of construction:__

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

6. Mark 5 (the greater of 3 and 5 in ) points B_{1}, B_{2}, B_{3}, B_{4}, B_{5} on BX so that BB_{1} = B_{1}B_{2} = B_{2}B_{3} = B_{3}B_{4} = B_{4}B_{5}

7. Join B_{5}C and draw a line through B_{3} (the 3^{rd} point, 3 being smaller of 3 and 5 in ) parallel to B_{5}C, to intersect BC at C’

8. Draw a line through C’ parallel to the line AC to intersect AB at A’

Thus, A’BC’ is the required triangle.

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