Construct the following triangles,
A) $\Delta ABC,AB = 4cm,BC = 5cm,AC = 6cm$
B) $\Delta KLM,KL = LM = KM = 5cm$

Hint: For part (a), take AC= 6 cm side as the base. Take A as the centre and draw an arc of 4 cm. Similarly, take C as the centre and draw an arc of 5 cm intersecting the previous arc. Join the intersection of these arcs to A and C. This is the triangle.
For part (a), take LM= 5 cm side as the base. Take L as the centre and draw an arc of 5 cm. Similarly, take M as the centre and draw an arc of 5 cm intersecting the previous arc. Join the intersection of these arcs to L and M. This is the triangle.

Complete step-by-step answer:
A) We need to construct a triangle whose sides are 4 cm, 5 cm and 6 cm.
Follow the given steps to solve the question.
Let the base of the triangle be AC = 6 cm.

Taking A as the centre and radius 4 cm on the compass, draw an arc on the upper side of AC.

Taking C as the centre and radius 6cm on the compass, draw an arc on the upper side of AC, intersecting the previous arc. Mark this point of intersection as point B.

Join points B to A and B to C to complete the $\Delta ABC$.

B) We need to construct a triangle whose sides are 4 cm, 5 cm and 6 cm.
Follow the given steps to solve the question.
Let the base of the triangle be LM = 5 cm.

Taking L as the centre and radius 5 cm on the compass, draw an arc on the upper side of LM.

Taking M as the centre and radius 5 cm on the compass, draw an arc on the upper side of LM, intersecting the previous arc. Mark this point of intersection as point K.

Join points K to L and K to M to complete the $\Delta KLM$.


Note: First check whether it is possible to construct the given triangle using triangle inequality property. The triangle inequality property states that the sum of the lengths of any two sides of a triangle is greater than the length of the third side. We see that our triangle satisfies this and hence, it is possible to construct such a triangle.