Question

Construct a triangle with sides $5\mathrm{cm},6\mathrm{cm},\mathrm{and}7\mathrm{cm}$ and then another triangle whose sides are $\frac{7}{5}$ of the corresponding sides of the first triangle. Give the justification of the construction

Answer 1

Step 1. Draw a line segment with the length $AB=6cm$

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Step 2. Formed the triangle:

Using $A$ and $B$ as the centers, draw arcs with radiuses of $7cm$ cm and $5cm$ , respectively. These arcs will cross at point $C$ , $ABC$ is the requisite triangle, with sides of $5\mathrm{cm},6\mathrm{cm},\mathrm{and}7\mathrm{cm}$, respectively.

Step 3. Divide the side in the given ratio.

Draw a ray $AX$ that intersects the line segment $AB$ on the other side of vertex $C$ at an acute angle. On line $AX$, find the $7$ points

$A_1,A_2,A_3,A_4,A_5,A_6,A_7$, are mark such that $A{A}_1=A_1{A}_2=A_2{A}_3=A_3{A}_4=A_4{A}_5=A_5{A}_6=A_6{A}_7$ is formed.

Draw a line from point $A_7$ to point $B{A}_5$ that is parallel to the line $B{A}_5$ where it crosses the extended line segment $AB$ at point $B\text{'}$. And draw a line from $B\text{'}$ to $C\text{'}$ that is parallel to the line $BC$ and intersects to form a triangle.

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

Hence, the required graph is shown below:

Step 4. Make a justification.

Since the scale factor is $\frac{7}{3}$. So, prove that $\frac{AB\text{'}}{AB}=\frac{B\text{'}C\text{'}}{BC}=\frac{AC\text{'}}{AC}=\frac{7}{5}$.

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

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)

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

From the construction, $\frac{AB\text{'}}{AB}=\frac{A{A}_5}{A{A}_7}=\frac{7}{5}\cdots \left(3\right)$.

Therefore, form $\left(2\right)$ and $\left(3\right)$

$\frac{AB\text{'}}{AB}=\frac{B\text{'}C\text{'}}{BC}=\frac{AC\text{'}}{AC}=\frac{7}{5}$

Hence, the proof is justified.