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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Updated on: 10-Oct-2022

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