1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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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. 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. Draw any ray AX making an acute angle with BC on the side opposite to the vertex A 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. Join 〖BA〗_3 and draw a line through A_2 parallel to 〖BA〗_3 to interest AB at point B’. 5Draw a line through B’ parallel to the line BC to intersect AC at C’. ∴ ∆ AB’C’ is the required triangle. Justification Here, (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 Now, In Δ AB’C’ and Δ ABC ∠ A = ∠ A ∠ AB’C’ = ∠ ABC Δ A’BC’ ∼ Δ 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.

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.