91Ó°ÊÓ

Given figure

Calculation.

Procedure:

1. Draw a line segment $\stackrel{→}{AB}$of any length.
2. Draw a line $l$.
3. Let $P$be a point on the line.
4. Measure the length of $AB$.
5. Use this measure as radius and $P$as center draw two arcs intersecting the line$l$.
   Let this point be Q and $S$.
6. Use a suitable radius and Qas center,draw an arc.
7. With same radius and S as center, draw an arc intersecting the arc drawn using Q as center.Label the point of intersection as X.
8. Draw $\stackrel{→}{PX}$.
9. Use the length of $AB$as radius and $P$as center draw an arc on $\stackrel{→}{PX}$. Label the point of intersection as $Y$.
10. Use the length of AB as radius and Xas center, draw an arc on $\stackrel{→}{PX}$. Label the point of intersection as $R$. Thus, length of $\stackrel{→}{PR}=2AB.$
11. Draw $\stackrel{→}{QR}$.

Thus, triangle $PQR$is obtained whose sides are in the ratio $1:2:\sqrt{5}$.

The figure drawn is shown below-