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Ex 11.1, 2

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

  1. Draw the line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ∆ABC is the required triangle.
  2. Draw any ray AX making an acute  angle with BC on the side opposite to  the vertex A
  3. Mark 3 points A_1,  A_2,  A_3 (as 3 is greater between 2 and 3) on line AX such that 〖AA〗_1=A_1 A_2=A_2 A_3.
  4. Join 〖BA〗_3 and draw a line through A_2 parallel to 〖BA〗_3 to interest AB at point B’.
  5. 5Draw a line through B’ parallel to the line BC to intersect AC at C’.

∴ ∆ AB’C’ is the required triangle.



(AB^′)/AB=(AA_2)/(AA_3 )=2/3.

Also, B’C’ is parallel to BC

So, the will make the same angle with line AC

∴ ∠ AB’C’ = ∠ ABC


In Δ AB’C’ and Δ ABC

        ∠ A = ∠ A

 ∠ AB’C’ = ∠ ABC


Since corresponding sides of
similar triangles are in the same ratio

∴  (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC

So,  (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC=2/3

This justifies the construction.


  1. Chapter 11 Class 10 Constructions
  2. Concept wise
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CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .