Question

Draw a triangle $ABC$ with side $BC=6$ cm, $\angle C={30}^o$ and$\angle A={105}^o$. Then construct another triangle whose sides are $\frac{2}{3}$ times the corresponding sides of $\Delta ABC$.

Answer 1

Given:
$BC=6$ cm, $\angle C={30}^o$ and $\angle A={105}^o$
We know that in $\Delta ABC,\angle A+\angle B+\angle C={180}^o$
Therefore, $\angle B={180}^o-(\angle A+\angle C)$
$={180}^o-({105}^o+{30}^o)\Rightarrow {45}^o$
Steps of construction:
1. Draw a line $BC=6$ cm.
2. Draw a ray CN making an angle of ${30}^o$ at $C.$
3. Draw a ray BM making an angle of ${45}^o$ at $B.$
4. Locate the point of intersection of rays $CN$ and $BM$ and name it as $A.$
5. $ABC$ is the triangle whose similar triangle is to be drawn.
6. Draw any ray $BX$ making an acute angle with $BC$ on the side opposite to the vertex $A.$
7. Locate 3 (Greater of $2$ and $3$ in $\frac{2}{3}$) points $B_1,B_2$ and $B_3$ on $BX$ so that $B{B}_1=B_1{B}_2=B_2{B}_3$.
8. Join $B_{3}C$ and draw a line through $B_2$ (Smaller of $2$ and $3$ in $\frac{2}{3}$) parallel to $B_{3}C$ to intersect $BC$ at $C^{\prime }.$
9. Draw a line through $C^{\prime }$ parallel to the line $CA$ to intersect $BA$ at $A^{\prime }.$
10. $A^{\prime }B{C}^{\prime }$ is the required similar triangle whose sides are $\frac{2}{3}$ times the corresponding sides of $\Delta ABC$.