Construct a triangle ABC in which BC$\displaystyle =8$ cm, $\displaystyle \angle $B$\displaystyle =45^{o}$,$\displaystyle \angle $C$\displaystyle =30^{o}$.Construct another triangle similar to $\displaystyle \vartriangle $ABC such that its sides are $\frac{3}{4} $of the corresponding sides of $\displaystyle \vartriangle $ABC.
Given: Side $BC=8$ cm,$\angle B=45^{o}$, $\angle C=30^{o}$
To do: To construct$\vartriangle ABC$. And to construct another triangle similar to $\vartriangle ABC$ such that its sides are $\frac{3}{4}$ of the corresponding sides of $\vartriangle ABC$.
Solution:
Follow the steps:
1.draw a line $\displaystyle BC=8cm$.
2.draw $\displaystyle \angle PBC=45^{o} $ .
3.draw $\angle QCB=30^{o}$.
4.Line PB and QC intersects on point A.
$\vartriangle ABC$ is constructed.
Now we would construct another triangle similar to $\vartriangle ABC$ and its each side should be $\frac{3}{4}$ of the corresponding sides of $\vartriangle ABC$.
Follow these steps
1.Draw a ray BX making acute angle on the opposite side of vertex A with BC.
2. Now divide the ray BX into four equal parts $BB_{1} ,\ B_{1} B_{2} ,\ B_{2} B_{3} ,\ B_{3} B_{4}$.
3.draw a line from $B_{4} to\ C$.
4.from point $B_{3}$, draw a parallel line to $B_{4} C$ ,its end point on $BC$ is $C'$.
5.from point $C'$ draw parallel line to $AC$, its end point on line $AB$ is$\\ A'$.
$\vartriangle A'BC'$ is similar to $\vartriangle ABC$. And its each side is $\frac{3}{4}$ of $\vartriangle ABC$.
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