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Ex 11.1, 5

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.

  1. Draw a ΔABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°.
  2. 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.
  3. Join 〖CA〗_4 and draw a line through A_4 parallel to 〖CA〗_4 to interest BC at point C’.
  4. Draw a line through C’ parallel to the line AC to intersect AB at A’.

∴ ∆ A’BC’ is the required triangle.

Justification

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

Δ A’BC’ ∼ Δ ABC

Since corresponding sides of
similar triangles are in the same ratio

 (A^′ B)/AB=(A^′ C^′)/AC=(BC^′)/BC

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

This justifies the construction.

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  1. Class 10
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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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