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Constructions — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsConstructions4 Marks
Question
Construct a triangle similar to a given $\triangle\text{ABC}$ such that each of its sides is $\Big(\frac{5}{7}\Big)^{\text{th}}$ of the corresponding sides of $\triangle\text{ABC}.$ It is given that AB - 5cm, BC = 7cm and $\angle\text{ABC} = 50^\circ.$
✓

Answer

Given that AB = 5cm, BC = 7cm and $\angle\text{ABC}=50^\circ$ Construct a triangle similar to a triangle ABC such that each of sides is $\Big(\frac{5}{7}\Big)^{\text{th}}$ of the corresponding sides of triangle ABC. We follow the following steps to construct the given
Step of construction:
Step I: First of all we draw a line segment.
Step II: With B as centre and draw an angle $\angle ABY =50^{\circ}$.
Step III: With B as centre and radius $=B C=7 cm$, draw an arc, cut the line BY drawn in step II at $C$.
Step IV: Joins $A C$ 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 seven points $A_1, A_2, A_3, A_4, A_5, A_6$ and $A_7$ such that $A A_1=A_1 A_2=A_2 A_3=A_3 A_4=A_4 A_5=$ $A _5 A_6= A _6 A_7$.
Step VII: Join $A_7 B$.
Step VIII: Since we have to construct a triangle each of whose sides is $\left(\frac{5}{7}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.
So, we take five parts out of seven equal parts on $A X$ from point $A_5$ draw $A_5 B^{\prime} \| A_7 B$, and meeting $A B$ at $B^{\prime}$.
Step IX: From $B^{\prime}$ draw $B^{\prime} C \| B C$, and meeting $A C$ at $C^{\prime}$. Thus, $\triangle A B^{\prime} C^{\prime}$ is the required triangle, each of whose sides is $\left(\frac{5}{7}\right)^{\text {th }}$ of the corresponding sides of $\triangle ABC$.

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