Answer :
Step 1: Draw a line segment AB = 4.2 cm.
Step 2: Taking A as center, draw an arc of radius 6.2 cm. Then, take B as center and draw an arc of radius 5 cm. These arcs will intersect at point C (say). Join AC and BC, such that AC = 6.2 cm and BC = 5 cm. Thus, ∆ABC is the required triangle.
Step 3: Draw a ray AX making acute angle with straight line AB to the opposite side of vertex C.
Step 4: Locate 3 points on ray AX, namely A 1 , A 2 and A 3 (since 3 is greater among 2 and 3) such that, AA 1 = A 1 A 2 = A 2 A 3 .
Step 5: Join BA 3 and then draw a line through A 2 such that it is parallel to BA 3 and intersects on AB at B’.
Step 6: Now draw a line through B’ such that it is parallel to BC and intersects on AC at C’.
∆AB’C’ is the required triangle with sides 2/3 of the corresponding sides of ∆ABC.
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NCERT - Mathematics