ΔPET, PE = 4.5 cm, ET = 5.4 cm, TP = 6.5 cm.
Step – 1 :
Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2 :
Draw a line segment PE of length 4.5 cm.
Step – 3 :
With centre P, draw an arc of radius 6.5 cm.
Step – 4 :
Draw another arc from E with radius – 5. 4 cm such that it intersects first arc at T.
Step – 5 :
Join P, T and E, T.
The required APET is constructed.
ΔABC, AB = 5.4 cm, BC = 4.5 cm and CA = 6.5
Step – 1 :
Draw a rough sketch of the triangle and label it with the given measurements.
Step – 2 :
Draw a line segment AB of length 5.4 cm.
Step – 3 :
With centre A, draw an arc of radius 6.5 cm.
Step – 4 :
Draw another arc from B with radius 4.5 cm such that it intersects first arc at C.
Step – 5 :
Join A, C and B,C.
The required AABC is constructed. If we place the ΔABC on ΔPET, the triangles are congruent. This is because, AB = TE AC = PT BC = PE
∴ ΔABC ≅ ΔTEP (SSS criteria)