Question
Construct a triangle similar to $\triangle\text{ABC}$ in which AB = 4.6cm, BC = 5.1cm, $\angle\text{A}=60^\circ$ with scale factor 4 : 5.
✓
Answer
Given that Construct a $\triangle\text{ABC}$ of given data, AB = 4.6cm, BC = 5.1cm and $\angle\text{A}=60^\circ$ and then a triangle similar to it whose sides are $\Big(4:5=\frac{4}{5}\Big)^\text{th}$ of the corresponding sides of $\triangle\text{ABC}.$ We follow the following steps to construct the given
Step of construction:
Step I: First of all we draw a line segment $A B=4.6 cm$.
Step II: With $A$ as centre draw an angle $\angle A=60^{\circ}$.
Step III: With $B$ as centre and radius $=B C=5.1 cm$, draw an arc, intersecting the arc drawn in step II at $C$.
Step IV: Joins $B C$ to obtain $\triangle A B C$.
Step V: Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.
Step VI: Along $A X$, mark off five points $A_1, A_2, A_3, A_4$ and $A_5$ such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5$.
Step VII: Join $A _5 B$.
Step VIII: Since we have to construct a triangle each of whose sides is $\left(\frac{4}{5}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.
So, we take four parts out of five equal parts on $A X$ from point $A_4$ draw $A_4 B \| A_5 B$, and meeting $A B$ at $B^{\prime}$.
Step IX: From $B^{\prime}$ draw $B^{\prime} C^{\prime} \| B C$, and meeting $A C$ at $C^{\prime}$. Thus, $\triangle A B^{\prime} C^{\prime}$ is the required triangle, each of whose sides is $\left(\frac{4}{5}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.
Step of construction:
Step I: First of all we draw a line segment $A B=4.6 cm$.
Step II: With $A$ as centre draw an angle $\angle A=60^{\circ}$.
Step III: With $B$ as centre and radius $=B C=5.1 cm$, draw an arc, intersecting the arc drawn in step II at $C$.
Step IV: Joins $B C$ to obtain $\triangle A B C$.
Step V: Below $A B$, makes an acute angle $\angle B A X=60^{\circ}$.
Step VI: Along $A X$, mark off five points $A_1, A_2, A_3, A_4$ and $A_5$ such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5$.
Step VII: Join $A _5 B$.
Step VIII: Since we have to construct a triangle each of whose sides is $\left(\frac{4}{5}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.
So, we take four parts out of five equal parts on $A X$ from point $A_4$ draw $A_4 B \| A_5 B$, and meeting $A B$ at $B^{\prime}$.
Step IX: From $B^{\prime}$ draw $B^{\prime} C^{\prime} \| B C$, and meeting $A C$ at $C^{\prime}$. Thus, $\triangle A B^{\prime} C^{\prime}$ is the required triangle, each of whose sides is $\left(\frac{4}{5}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.
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