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Ex 11.1, 3
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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.

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Justification
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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

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Δ
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AB’C’ ∼
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Δ
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ABC
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Since corresponding sides of

similar triangles are in the same ratio

(AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC

So,
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(AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC
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=7/5
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This justifies the construction.