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**Ex 11.1, 2**

Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of the corresponding sides of the first triangle.

- Draw the line segment AB = 4 cm. Taking point A as centre, draw an arc of 5 cm radius. Similarly, taking point B as its centre, draw an arc of 6 cm radius. These arcs will intersect each other at point C. Now, AC = 5 cm and BC = 6 cm and ∆ABC is the required triangle.
- Draw any ray AX making an acute angle with BC on the side opposite to the vertex A
- Mark 3 points A_1, A_2, A_3 (as 3 is greater between 2 and 3) on line AX such that 〖AA〗_1=A_1 A_2=A_2 A_3.
- Join 〖BA〗_3 and draw a line through A_2 parallel to 〖BA〗_3 to interest AB at point B’.
- 5Draw a line through B’ parallel to the line BC to intersect AC at C’.

∴ ∆ AB’C’ is the required triangle.

**Justification**

Here,

(AB^′)/AB=(AA_2)/(AA_3 )=2/3.

Also, B’C’ is parallel to BC

So, the will make the same angle with line AC

∴ ∠ AB’C’ = ∠ ABC

Now,

In Δ AB’C’ and Δ ABC

∠ A = ∠ A

∠ AB’C’ = ∠ ABC

**Δ**** A’BC’ ∼ ****Δ**** ABC**

Since corresponding sides of

similar triangles are in the same ratio

∴ (AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC

So, **(AB^′)/AB=(B^′ C^′)/BC=(AC^′)/AC=2/3**

This justifies the construction.