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  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC. Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Mark 4 points B_1, B_2, B_3, B_4 (as 4 is greater between 3 and 4) on line BX such that 〖BB〗_1=B_1 B_2=B_2 B_3=B_3 B_4. Join 〖CA〗_4 and draw a line through A_4 parallel to 〖CA〗_4 to interest BC at point C’. Draw a line through C’ parallel to the line AC to intersect AB at A’. ∴ ∆ A’BC’ is the required triangle. Justification Here, BC^′/BC=(BB_3)/(BB_4 )=3/4. Also, A’C’ is parallel to AC So, the will make the same angle with line BC ∴ ∠ A’C’B = ∠ ACB Now, In Δ A’BC’ and ABC ∠ B = ∠ B ∠ A’C’B = ∠ ACB Δ A’BC’ ∼ Δ ABC Since corresponding sides of similar triangles are in the same ratio (A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC So, (A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC =3/4. This justifies the construction.

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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.