construct an equilaterak triangle of each side 5cm
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Answer:
1.First Draw a line AB 0f 5 cm
2. From A and B draw 60°and cut 5 cm
from both
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Let ABC be the ∆ ...... (1)Draw AB as base = 5 cm......... (2) Draw perpendicular bisector of AB....... (3) Open the compass for 5 cm, keep the pointer at A and draw an arc such that it cuts the bisector........ (4) Do the same with point B and wherever those arcs meet name that point C......... (5) Thus ABC is the required ∆.
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