Question

To construct a triangle similar to given $\Delta ABC$ with its sides $\frac{4}{5}$ of that of $\Delta ABC$. Locate points $X_1,X_2,X_3,....$ on ray $BX$ at equal distances such that $\angle ABX$ is acute. The minimum number of points to be located on $BX$ is:

• A. $14$
• B. $5$
• C. $9$
• D. $20$

Answer 1

The correct option is B $5$

$△PQB$ is the required triangle.

Since side $BQ$ is $\frac{4}{5}$ times side $BA$.

$BQ=\frac{4}{5}\times (BQ+AQ)\Rightarrow 5BQ=4BQ+4AQ\Rightarrow BQ=4AQ\Rightarrow \frac{AQ}{BQ}=\frac{1}{4}$

Therefore, $Q$ divides $BA$ in ratio $4:1$.

So, the minimum number of points required on ray $BX$ is $1+4=5$.