Question

Draw a triangle $\mathrm{ABC}$ with side $\mathrm{BC}=6\mathrm{cm},\mathrm{AB}=5\mathrm{cm}$ and $\angle \mathrm{ABC}=60°$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $\mathrm{ABC}$. Give the justification of the construction

Answer 1

Step 1. Draw a triangle $\mathrm{ABC}$ with side $\mathrm{BC}=6\mathrm{cm},\mathrm{AB}=5\mathrm{cm}$ and $\angle \mathrm{ABC}=60°$.

Draw a line $\mathrm{BC}$ of side $6\mathrm{cm}$:

Draw an $\angle \mathrm{B}=60°$:

Take $\mathrm{B}$ as a center, $5\mathrm{cm}$ as radius, cut an arc at $\mathrm{A}$ on the same line so that $\mathrm{AB}=5\mathrm{cm}$ and then join $AC$:

$△\mathrm{ABC}$ is formed.

Step 2. Draw a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the triangle $\mathrm{ABC}$.

Draw a ray $\mathrm{BX}$ making an acute angle with $\mathrm{BC}$ on the opposite side of a vertex $\mathrm{A}$:

Locate $4$ points ${\mathrm{B}}_1,{\mathrm{B}}_2,{\mathrm{B}}_3$ and ${\mathrm{B}}_4$ on the line segment $\mathrm{BX}$.so that ${\mathrm{BB}}_1={\mathrm{B}}_1{\mathrm{B}}_2={\mathrm{B}}_2{\mathrm{B}}_3={\mathrm{B}}_3{\mathrm{B}}_4$:

Join ${\mathrm{B}}_4\mathrm{C}$ and draw a line through ${\mathrm{B}}_3$ parallel to ${\mathrm{B}}_4\mathrm{C}$ and then intersect $\mathrm{BC}$ at $C\text{'}$:

Draw a line through $C\text{'}$parallel to $AC$ intersecting $\mathrm{AB}$ at $A\text{'}$:

$△A\text{'}\mathrm{BC}\text{'}$ is formed.

Step 3. Justification of construction.

This construction is justified by proving $\mathrm{A}\text{'}\mathrm{B}=\frac{3}{4}\mathrm{AB},BC\text{'}=\frac{3}{4}BC$ and $\mathrm{A}\text{'}\mathrm{C}\text{'}=\frac{3}{4}\mathrm{A}C$.

In $△A\text{'}\mathrm{BC}\text{'}$ and $△\mathrm{ABC}$,

$\begin{array}{rcl}\angle \mathrm{A}\text{'}\mathrm{C}\text{'}\mathrm{B}& =& \angle \mathrm{ACB}\left[\mathrm{Corresponding}\mathrm{angles}\right]\\ \angle \mathrm{A}\text{'}\mathrm{BC}\text{'}& =& \angle \mathrm{ABC}\left[\mathrm{Common}\right]\\ & & \therefore △\mathrm{A}\text{'}\mathrm{BC}\text{'}~△\mathrm{ABC}\left[\mathrm{AA}\mathrm{Similarity}\mathrm{criterion}\right]\\ & & \because △\mathrm{A}\text{'}\mathrm{BC}\text{'}~△\mathrm{ABC}\\ \frac{\mathrm{A}\text{'}\mathrm{B}}{\mathrm{AB}}& =& \frac{\mathrm{BC}\text{'}}{\mathrm{BC}}=\frac{\mathrm{A}\text{'}\mathrm{C}\text{'}}{\mathrm{AC}}\dots \left(1\right)\end{array}$

Now,

In $△{\mathrm{BB}}_3\mathrm{C}\text{'}$ and $△{\mathrm{BB}}_4\mathrm{C}$,

$\begin{array}{rcl}\angle {\mathrm{B}}_3\mathrm{BC}\text{'}& =& \angle B_{4}BC\left[\mathrm{Corresponding}\mathrm{angles}\right]\\ \angle {\mathrm{BB}}_3\mathrm{C}\text{'}& =& \angle {\mathrm{BB}}_4\mathrm{C}\left[\mathrm{Common}\right]\\ & & \therefore △{\mathrm{BB}}_3\mathrm{C}\text{'}~△{\mathrm{BB}}_4\mathrm{C}\left[\mathrm{AA}\mathrm{Similarity}\mathrm{criterion}\right]\\ & & \because △{\mathrm{BB}}_3\mathrm{C}\text{'}~△{\mathrm{BB}}_4\mathrm{C}\\ \frac{BC\text{'}}{BC}& =& \frac{{\mathrm{BB}}_3}{{\mathrm{BB}}_4}\\ \frac{BC\text{'}}{BC}& =& \frac{3}{4}\dots \left(2\right)\end{array}$

From equations $\left(1\right)$ and $\left(2\right)$:

$\begin{array}{rcl}\frac{\mathrm{A}\text{'}\mathrm{B}}{\mathrm{AB}}& =& \frac{\mathrm{BC}\text{'}}{\mathrm{BC}}=\frac{\mathrm{A}\text{'}\mathrm{C}\text{'}}{\mathrm{AC}}=\frac{3}{4}\end{array}$

$\mathrm{A}\text{'}\mathrm{B}=\frac{3}{4}\mathrm{AB},BC\text{'}=\frac{3}{4}BC$ and $\mathrm{A}\text{'}\mathrm{C}\text{'}=\frac{3}{4}\mathrm{A}C$

Hence, the sides of the constructed triangle $△A\text{'}\mathrm{BC}\text{'}$ are $\frac{3}{4}$ of the corresponding sides of the triangle $∆\mathrm{ABC}$.