**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.

- 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.

**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.