Question

To construct a triangle similar to a given $△ABC$ with its sides in a ratio of $\frac{7}{3}$ with the corresponding sides of $△ABC$, draw a ray $BX$ making an 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 $C$.

• A. True
• B. False

Answer 1

The correct option is A True

Given ratio is $\frac{7}{3}>1$ so, the resulting triangle will be larger than given as $B_{7}C\parallel B_{3}C$ and $BX$ is equally divided into $7$ part as $(7>3).$

Steps of Construction:

• Draw the given triangle with the given specifications.
• Draw an acute angle $\angle CBX.$
• Divide $BX$ into $7$ equal parts and mark them $B_1,B_2,B_3,\dots ,B_7.$
• Produce $BC$ and $BA$ as shown in the figure.
• Join $B_{3}C.$
• Draw $B_7{C}^{\prime }\parallel B_{3}C.$ to intersect on $BC$ produced.
• Draw $C^{\prime }A\parallel AC.A^{\prime }$ on $BA$ produced $△A^{\prime }B{C}^{\prime }$ is required triangle.

Hence, the given statement is true.