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Ex 11.1, 5
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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.

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Justification
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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

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Δ
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A’BC’ ∼
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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

(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC

So,
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(A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC
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=3/4
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.
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This justifies the construction.