Question
Construct a triangle with sides $5 cm, 4 cm$ and 6 cm . Then construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of first triangle.

Answer

Following are the steps of constructing a required triangle.
Step 1: Draw a line segment $A B=4 cm$ and then draw an are of radius 5 cm considering $A$ as a center.
Step 2: Draw an arc of radius 6 cm considering $B$ as center.
Step 3: Name the point where both the ares from step 1 and step 2 intersect as $C$.
Step 4: Join $B C$ and $A C . \triangle A B C$ is formed.
Step 5: Draw a ray $A Y$ which forms an acute angle with $A B$ on the opposite side of the vertex $C[1 / 2]$ Step 6: Locate three points $A_1, A_2$, and $A_3$ on lines $A Y$ such that $A A 1=A_1 A_2=A_2 A_3$.
Step 7: Join $B$ and $A_3$.
Step 8: Draw a line segment from point $A_2$ and parallel to $B A_3$ which intersects $A B$ at point $B$ '.
Step 9: Draw a line segment from $B^{\prime}$ and parallel to $B C$ which intersects $A C$ at the point $C^{\prime}$. [1/2] Thus, required triangle $\triangle A B^{\prime} C^{\prime}$ whose sides are $\frac{2}{3}$ times the corresponding sides of first triangle ( $\triangle A B C$ ) has been constructed.
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