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# Draw a triangle ABC with side

$BC = 6\ cm, AB = 5\ cm$ and $∠ABC = 60^o$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $ABC$.

**Given: **

$BC = 6\ cm, AB = 5\ cm$ and $\angle ABC=60^{o}$.

**To do: **

We have to draw $\vartriangle ABC$ and then to construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the ABC.

**Solution:**

Steps of construction:

1. Draw a line segment $BC = 6\ cm$, draw a ray $BX$ making $60^{o}$ with $BC$.

2. Draw an arc with radius $5\ cm$ from $B$ so that it cuts $BX$ at $A$.

3. Now join AC to form $\vartriangle ABC $.

4. Draw a ray by making an acute angle with NC opposite to vertex A.

5. Locate 4 points on by such that $BB_{1}=B_{1}B_{2}=B_{2}B_{3}=B_{3}B_{4}$

6. Join $B_{4}$ to $C$.

And now draw a line from $B_{3}$ parallel to $B_{4}C$. so that it cuts $BC$ at $C\'$.

7. From C draw a line parallel to AC and cuts AB at A\' .

8. $\vartriangle A\' BC \'$ is the required triangle.

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