Question

To construct a triangle similar to given $△ABC$ with its sides $\frac{2}{3}$ of that of $\Delta ABC$, locate points on ray $BX$ at equal distances as $B_1,B_2,B_3,....$ such that $\angle CBX$ is acute. The points to be joined in the next step are:

• A. $B_4,C$
• B. $B_3,C$
• C. $B_1,C$
• D. $B_2,C$

Answer 1

The correct option is B $B_3,C$

$△PQB$ is the required triangle.

Since side $BQ$ is $\frac{2}{3}$ times side $BC$.

$BQ=\frac{2}{3}\times (BQ+CQ)\Rightarrow 3BQ=2BQ+2CQ\Rightarrow BQ=2CQ\Rightarrow \frac{CQ}{BQ}=\frac{1}{2}$

Therefore, $Q$ divides $BC$ in ratio $2:1$.

So, point $B_3$ should be connected to $C$, and $B_{2}Q$ should be drawn parallel to $B_{3}C$.