Question

Construct a $\Delta ABC$ in which $CA=6cm,AB=5cm$ and $\angle BAC={45}^o$. Then construct a triangle whose sides are $\frac{3}{5}$ of the corresponding sides of $\Delta ABC$.

Answer 1

Steps Of construction

Step $1:$ Draw a line $AB=5\text{cm}$

Step $2:$ With $A$ as center Measure ${45}^{\circ }$ using Protractor and draw a ray $\overrightarrow{AX}$

Step $3:$ With $B$ as center Cut an arc of Radius $6\text{cm}$ on the $\overrightarrow{AX}$

Step $4:$ Mark the intersection as $C$ Join $B,C$

Step $5:$ Now divide the line $AB$ in the ratio $\frac{3}{5}$

Step $6:$ Mark a point $B^{\prime }$ $3\text{cm}$ apart from $A$ on the line $AB$

Step $7:$ Draw a line parallel to $BC$ passing through $B^{\prime }$

Step $8:$ Mark the intersection as $C^{\prime }$