Answer :

Step 1: Draw a line segment AB = 13 cm.

Step 2: Taking A as center, draw an arc of radius 5 cm. Then, take B as center and draw an arc of radius 12 cm. These arcs will intersect at point C (say). Join AC and BC, such that AC = 5 cm and BC = 12 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 5 points on ray AX, namely A _{ 1 } , A _{ 2 } , A _{ 3 } , A _{ 4 } and A _{ 5 } (since 5 is greater among 3 and 5) such that, AA _{ 1 } = A _{ 1 } A _{ 2 } = A _{ 2 } A _{ 3 } = A _{ 3 } A _{ 4 } = A _{ 4 } A _{ 5 } .

Step 5: Join BA _{ 5 } and then draw a line through A _{ 3 } such that it is parallel to BA _{ 5 } 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 3/5 of the corresponding sides of ∆ABC.

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