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Ex 11.1, 6

Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of Δ ABC.

In Δ ABC

  ∠B = 45°, ∠A = 105°

Sum of all interior angles in a triangle is 180°.

  ∠A + ∠B + ∠C = 180°

105° + 45° + ∠C = 180°

∠C = 180° − 150°

∠C = 30°

Steps of construction

  1. Draw a ΔABC with side BC = 7 cm, ∠B = 45°, ∠C = 30°.
  2. Draw a ray BX making an acute angle with BC on the opposite side of vertex A.
  3. Locate 4 points (as 4 is greater in 4 and 3), B_1, B_2, B_3, B_4, on BX.
  4. Join B_3C. Draw a line through B_4 parallel to B_3C intersecting extended BC at C'.
  5. Through C', draw a line parallel to AC intersecting extended line segment at C'.

ΔA'BC' is the required triangle.

Justification

Here,

 (BC^′)/BC=(BB_4)/(BB_3 ) = 4/3

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 =4/3 .

This justifies the construction.

-ev-

  1. Chapter 11 Class 10 Constructions
  2. Concept wise
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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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