Question

To construct a triangle similar to a given $∆ABC$ with its sides $\frac{7}{3}$ of the corresponding sides of $∆ABC$, draw a ray $BX$ making acute angle with $BC$ and $X$ lies on the opposite side of $A$ with respect to $BC$. The points $B_1,B_2,....,B_7$ are located at equal distances on $BX$, $B_3$ is joined to $C$ and then a line segment $B_{6}C\text{'}$ is drawn parallel to $B_{3}C$ where $C\text{'}$ lies on $BC$ produced. Finally, line segment $A\text{'}C\text{'}$ is drawn parallel to $AC$. Write ‘True’ or ‘False’ and justify your answer

• A. True
• B. False

Answer 1

The correct option is B False

The explanation for the statement:

Steps of construction:

1. Construct a line segment $BC$.
2. Taking $B$ and $C$ as centres, construct two arcs of suitable radius which intersects each other at $A$.
3. Now joining $BA$and $CA$. $∆ABC$ is the required triangle.
4. From point $B$, construct a ray $BX$ downwards which makes an acute angle $\angle CBX$.
5. On $BX$, mark $7$ points, $B_1,B_2,....,B_7$
6. Let us join $B_{3}C$ and from $B_7$ construct a line $B_{7}C\text{'}\left|\right|B_{3}C$ which intersects the extended line segment $BC$ at $C\text{'}$.
7. Construct C’A’||CA which intersects the extended line segment BA at A’.

So $∆A\text{'}B\text{'}C\text{'}$ is the required triangle whose sides are $\frac{7}{3}$ of the corresponding sides of $∆ABC$.

It is given that segment $B_{6}C\text{'}\left|\right|B_{3}C$

As the construction is not possible that segment $B_{6}C\text{'}\left|\right|B_{3}C$ as the similar triangle $∆A\text{'}B\text{'}C\text{'}$has its sides $\frac{7}{3}$of the corresponding sides of $∆ABC$. Here $B_{7}C\text{'}\left|\right|B_{3}C$

Hence, the given statement is false.