Construct a triangle ABC with side $BC=7\ cm$, $\angle B = 45^{o}$, $\angle A = 105^{o}$. Then construct another triangle whose sides are $\frac{3}{4}$ times the corresponding sides of the ABC.

Given: Side $BC=7\ cm$,  $\angle B = 45^{o}$,  $\angle A = 105^{o}$.

To do: To Construct a triangle ABC with side $BC=7\ cm,\ \angle B = 45^{o}$, $\angle A = 105^{o}$ and then construct another triangle whose sides are $\frac{3}{4}$times the corresponding sides of the ABC. 

Here $BC=7\ cm$, $\angle B =45^{o}$ , $\angle A=105^{o}$

Draw $BC=7cm$ With help of a ruler

Take a protractor measure $\angle45^{o}$

From point B and draw a ray BX.

From point mark $30^{o}$ with help of protractor & draw a ray CY.

 

Now both BX and BY intersect at a point and this point is A Now we have ABC

To construct similar triangle with corresponding sides of the sides.

Step 1: Draw a ray BZ making an acute angle with BC.

Step 2: Along the ray BZ mark four points $B_{1} ,\ B_{2} ,\ B_{3}$ and $B_{4}$ such that $BB_{1} =B_{1} B_{2} =B_{2} B_{3} =B_{3} B_{4}$.

Step 3: Now join $B_{4}$ to C and draw a line parallel to $B_{4} C from $B_{3}$ intersecting the line BC at C'.

Step 4: Draw a line through C' parallel to CA which intersects AB at A'.

$\vartriangle A'BC$ is the required triangle.