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Example 2

Construct a triangle similar to a given triangle ABC with its sides equal    its to 5/3 of the corresponding sides of the triangle ABC (scale factor 5/3).

Given scale factor = 5/3

Steps of Construction :

  1. Draw any ray BX making an acute angle with BC on the side opposite to the vertex A.
  2. Locate 5 (the greater of 5 and 3 in 5/4 )  points B_1,B_2,B_3,B_4 and B_5 on BX so that 〖BB〗_1=B_1 B_2=B_2 B_3=B_3 B_4=B_4 B_5.
  3. Join B_3 (3rd point as 3 is smaller in 5/3) to C and draw a line through  B_5 parallel to  B_3 C to intersecting extended BC at C′.
  4. Draw a line through C′ parallel to line CA to intersect extended BA at A′.

Thus, Δ A′BC′ is the required triangle.

Justification

Here,

 (BC^′)/BC=(BB_5)/(BB_3 ) = 5/3

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 =5/3 .

This justifies the construction.

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  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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