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 points to be joined in the next step are:

• A. $X_4,C$
• B. $X_5,C$
• C. $X_4,A$
• D. $X_5,A$

Answer 1

Answer: (D)

$X_5,A$

$△PQB$ 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, point $X_5$ should be connected to $A$, and $X_{4}Q$ should be drawn parallel to $X_{5}A$.