Question

Construct a triangle of sides$4\mathrm{cm},5\mathrm{cm},\mathrm{and}6\mathrm{cm}$ and then a triangle similar to it whose sides are $\frac{2}{3}$ of the corresponding sides of the first triangle. Give the justification for the construction

Answer 1

Step 1. Draw a line segment $AB$ that is $4$ cm long, that is, $AB=4$ cm.

Step 2. Draw an arc with a radius of $5$ cm with point $A$ as the centre. Using point $B$ as the center, draw an arc with a radius of $6$ cm. At point $C$, the arcs you've drawn will connect.

Now we get $AC=5$ cm and $BC=6$ cm, and the required triangle is $ABC$.

Step 4. Make a similar triangle with the scale factor $\frac{2}{3}$. Now draw a ray $AX$ on the opposite side of vertex $C$ that forms an acute angle with the line segment $AB$. Draw a line $AX$ with three points $A_1,A_2,\mathrm{and}{A}_3$ , where $3$ is higher between $2$ and $3$ such that $A{A}_1=A{A}_2=A{A}_3$.

Step 5. Connect the points $B{A}_3$ and $A_2$ by drawing a line through $A_2$ that is parallel to $B{A}_3$ that intersects $AB$ at point $B\text{'}$.Draw a line parallel to the line $BC$ that intersects the line $AC$ at $C\text{'}$ through the point $B\text{'}$.

As a result, the needed triangle is $AB\text{'}C\text{'}$.

Hence, the required graph is shown below:

Step 6. Make a justification.

Since the scale factor is $\frac{2}{3}$. So, prove that $\frac{AB\text{'}}{AB}=\frac{B\text{'}C\text{'}}{BC}=\frac{AC\text{'}}{AC}=\frac{2}{3}$.

From the construction, $B\text{'}C\text{'}\parallel BC$.

So, $\angle AB\text{'}C\text{'}=\angle ABC\cdots \left(1\right)$ (Corresponding angles)

In $∆AB\text{'}C\text{'}$ and $∆ABC$ ,

$\angle AB\text{'}C\text{'}=\angle ABC\left[\mathrm{from}\left(1\right)\right]$

$\angle A=\angle A\left(\mathrm{Common}\right)$

$∆AB\text{'}C\text{'}=∆ABC$ (From AA similarity criterion)

Since corresponding sides are in the same ratio.

Therefore, $\frac{AB\text{'}}{AB}=\frac{B\text{'}C\text{'}}{BC}=\frac{AC\text{'}}{AC}\cdots \left(2\right)$

From the construction, $\frac{AB\text{'}}{AB}=\frac{A{A}_2}{A{A}_3}\cdots \left(3\right)$

So, from the equations $\left(2\right)$ and $\left(3\right)$, we get

$\frac{AB\text{'}}{AB}=\frac{B\text{'}C\text{'}}{BC}=\frac{AC\text{'}}{AC}=\frac{2}{3}$

Hence, the proof is justified.