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Ex 11.1, 3

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.

  1. 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.
  2. Draw a ray AX making acute angle with line AB on the opposite side of vertex C.
  3. 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.
  4. Join BA5 and draw a line through A7 parallel to BA5 to intersect extended line segment AB at point B'.
  5. 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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Transcript

Ex 11.1, 3 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, A’C’ is parallel to AC So, the will make the same angle with line BC ∴ ∠ 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′=𝐵𝐶/(𝐵^′ 𝐶^′ )=𝐴𝐶/(𝐴𝐶^′ ) So, 𝑨𝑩/𝑨𝑩′=𝑩𝑪/(𝑩^′ 𝑪^′ )=𝑨𝑪/(𝑨𝑪^′ )=𝟕/𝟓 This justifies the construction.

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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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