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  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle. Draw a line segment AB of 5 cm. Taking A and B as centre, draw arcs of 6 cm and 7 cm radius respectively. Let these arcs intersect each other at point C. ABC is the required triangle having length of sides as 5 cm, 6 cm, and 7 cm respectively. Draw a ray AX making acute angle with line AB on the opposite side of vertex C. Mark 7 points, A_1, A_2, A_3, A_4 A_5, A_6, A_7 (as 7 is greater between 5and 7), on line AX such that AA _1 = A_1 A_2 = A_2 A_3= A_3 A_4= A_4 A_5= A_5 A_6= A_6 A_7. Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B'. Draw a line through B parallel to BC intersecting the extended line segment AC at C . AB C is the required triangle. Justification Here, (AB^ )/AB=(AA_7)/(AA_5 ) = 7/5 Also, B C is parallel to BC So, the will make the same angle with line AB AB C = ABC Now, In AB C and ABC A = A AB C = ABC 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=7/5 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.