Question

Draw a triangle ABC with $BC=6$cm, AB$=5$cm and $\angle$ABC $={60}^o$. Then construct a triangle whose sides are $\frac{3}{4}$ of the corresponding sides of the $\Delta$ ABC.

Answer 1

The $\Delta A^{\prime }B{C}^{\prime }$ whose sides are $\frac{3}{4}$ of the corresponding sides of $\Delta ABC$ can be drawn as follows:

Step 1: Draw a $\Delta ABC$ with side $BC=6cm,AB=5cm,\angle ABC={60}^{\circ }$

Step 2: Draw a ray $BX$ making an acute angle with $BC$ on the opposite side of vertex $A$.

Step 3: Locate $4$ points, $B_1,B_2,B_3,B_4$ on line segment $BX$.

Step 4: Join $B_{4}C$ and draw a line through $B_3$, parallel to $B_{4}C$ intersecting $BC$ at $C^{\prime }$.

Step 5: Draw a line through $C^{\prime }$ parallel to $AC$ intersecting $AB$ at $A^{\prime }$.

The triangle $A^{\prime }B{C}^{\prime }$ is the required triangle.