construct ABC an equliaterql triangle with side 5 cm
Answers
Answer:
Steps of construction are:-
(i) Firstly draw a line segment BC of length 5 cm.
(ii) Now cut an arc of radius 5 cm from point B and an arc of 5 cm from point C.
(iii) Name the point of intersection of arcs to be point A.
(iv) Now join point AC and BC. Thus \Delta ABC is the required triangle.
(v) Draw a line AX which makes an acute angle with BC and is opposite of vertex A.
(vi) Cut three equal parts of line AD namely BB1, BB2, BB3.
(vii) Now join B3 to C. Draw a line B2C' parallel to B3C
(viii) And then draw a line A'C' parallel to AC.
Hence \Delta BA'C' is the required triangle.
Answer:
Steps of construction are:-
(i) Firstly draw a line segment BC of length 5 cm.
(ii) Now cut an arc of radius 5 cm from point B and an arc of 5 cm from point C.
(iii) Name the point of intersection of arcs to be point A.
(iv) Now join point AC and BC. Thus \Delta ABC is the required triangle.
(v) Draw a line AX which makes an acute angle with BC and is opposite of vertex A.
(vi) Cut three equal parts of line AD namely BB1, BB2, BB3.
(vii) Now join B3 to C. Draw a line B2C' parallel to B3C
(viii) And then draw a line A'C' parallel to AC.
Hence \Delta BA'C' is the required triangle.
