Question

Draw a right triangle in which the sides (other than hypotenuse) are of lengths $4\mathrm{cm}\mathrm{and}3\mathrm{cm}$. Then construct another triangle whose sides are $\frac{5}{3}$ times the corresponding sides of the given triangle. Give the justification for the construction

Answer 1

Draw the diagram with the help of a given statement.

The required triangle is the $AB\text{'}C\text{'}$ triangle as shown below:

1. let's create a line segment with the length $BC=3\mathrm{cm}$.
2. Now measure and draw a $90$-degree angle.
3. Draw an arc with a radius of $4\mathrm{cm}$ that intersects the ray at point $B$, using $B$ as the center.
4. Finally, combine the lines $AC$ to form the needed triangle $ABC$.
5. On the other side of vertex $A$, draw a ray $BX$ that forms an acute angle with $BC$.
6. On the ray $BX$, locate five such as $B_1,B_2,B_3,B_4,B_5$ so that $B{B}_1=B{B}_2=B{B}_3=B{B}_4=B{B}_5$.
7. Connect the $C{B}_3$ points.
8. Draw a line parallel to $C{B}_3$ through $B_5$ and intersecting the extended line $BC$ at $C\text{'}$.
9. Draw a line parallel to $AC$ through $C\text{'}$', intersecting the extended line $AB$ at $A\text{'}$.

The required triangle is the $A\text{'}BC\text{'}$ triangle as shown below:

The required diagram is shown below: