Question

Construct a TABC in which AB = 4 cm, BC = 5cm and AC = 6cm. Then construct another triangle whose sides are 3

2

times the corresponding sides of TABC​

Answer 1

$\triangle A^{'}BC^{'}$ is the triangle whose sides are $\frac{3}{2}$ to the $\triangle ABC$

Step-by-step explanation:

Given,

Construct a $\triangle ABC$ in which $AB=4\ cm,BC=5\ cm\ and\ AC=6\ cm$. Then construct another triangle whose sides are $\frac{3}{2}$

Following steps:

• Draw a line segment $AC=6\ cm$ using scale.
• $A$ as center $4\ cm$ as radius draw an arc and $C$ as center $5\ cm$ as radius draw an arc which cuts before arc known as $B$

• Join $AB$ and $BC$

• $\triangle ABC$ is the required triangle.

• Below the line $AC$ draw an angle like $\angle CAX=\theta$

• Divide line $AX$ into $5$ equal parts like $A_{1},A_{2},........,X$

• Join $A_{2}C$

• Draw a parallel line $A_{2}C$ which intersect the extended line $AC$ at $C^{'}$

• Draw a line through $C^{'}$ parallel to the line $BC$ to intersect $AB$ extended at $A^{'}$

Hence, $\triangle A^{'}BC^{'}$ is the triangle whose sides are $\frac{3}{2}$ to the $\triangle ABC$