Construct a triangle ABC similar to a given triangle AB'C' with its sides equal to $ \frac{5}{3} $ of the corresponding sides of the triangle AB'C' (i.e., of scale factor $ \frac{5}{3} $ ).

Academic Mathematics NCERT Class 10

To do:

Construct a triangle ABC similar to a given triangle AB'C' with its sides equal to \( \frac{5}{3} \) of the corresponding sides of the triangle AB'C'.

Solution:

Let us construct a $\vartriangle AB'C'$ in which $C'A= 6\ cm$, $AB'= 3\ cm$ and $\angle B'AC'= 45^{o}$.

We have to construct a triangle ABC with its sides equal to \( \frac{5}{3} \) of the corresponding sides of the triangle AB'C'.

Steps of construction:

1. Draw $AB'=3\ cm$. With A as centre, draw $\angle B'AC'= 45^{o}$. Join B'C'. AB'C' is thus formed.

2. Draw AX such that $\angle B'AX$ is an acute angle.

3. Cut 5 equal arcs such that $AA_{1}=A_{1}A_{2}=A_{2}A_{3}=A_{3}A_{4}=A_{4}A_{5}$

4. Join $A_{5}$ to $B'$ and draw a line through $A_{5}$ parallel to $A_{3}B'$ which meets AB' at B.

Here, $AB=\frac{5}{3}AB'$.

Now draw a line through $B$ parallel to $B'C'$ which joins $AC'$ extended at $C$.

Here,

$BC=\frac{5}{3}B'C'$ and $AC=\frac{5}{3}AC'$

Thus, $\vartriangle ABC$ is the required triangle.

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Updated on: 10-Oct-2022

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