Answer :

Given Δ ABC,

- AB = 6 cm
- AC = 8 cm
- BC = 9 cm

__Steps of construction__:

- Draw a base AB of 6 cm.

2. With A as center and 8 cm as the radius, draw an arc.

- With B as center and 9 cm as the radius, draw an arc which cuts the previous arc at point C.

4. Join AC and BC. Thus ΔABC is the required triangle.

5. Now, from A, draw a ray AX which makes an acute angle on the opposite side of the vertex C.

6. With A as center, mark five points A

_{1}, A

_{2}, A

_{3}, A

_{4}and A

_{5}on AX such that they are equidistant. i.e. AA

_{1}= A

_{1}A

_{2}= A

_{2}A

_{3}= A

_{3}A

_{4}= A

_{4}A

_{5}.

7. Join A

_{5}B and then draw a line from A

_{3}parallel to A

_{5}B which meets the line AB at E.

8. From E, draw a line parallel to BC and meets the line AC at F.

Thus ΔAEF is the required triangle.

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