Question -
Answer -
Steps of construction :
(i) Draw a line segment AB = 5 cm.
(ii) At A, draw a perpendicular and cut off AE = 3 cm.
(iii) From E, draw EF || AB.
(iv) From B, draw a ray making an angle of 60 meeting EF at C.
(v) Join CA. Then ABC is the triangle.
(vi) From A, draw a ray AX making an acute angle with AB and cut off 3 equal parts making A A1= A1A2 = A2A3.
(vii) Join A2 and B.
(viii) From A , draw A^B’ parallel to A2B and B’C’ parallel toBC.
Then ΔC’AB’ is the required triangle.