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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. ΔABC 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’. ΔAB’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 Δ AB’C’ and Δ ABC        ∠ A = ∠ A  ∠ AB’C’ = ∠ ABC Δ AB’C’ ∼ 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=7/5 This justifies the construction.

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CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .