On your paper draw figures roughly like those shown. Use them in constructing the figures described in Exercise $17-24$.

A square with diagonals of length $b\sqrt{2}$

A line segment of length $b\sqrt{2}$is given.

A square with diagonal of length $b\sqrt{2}$having sides of length $b$.

So, a square is to be constructed with each side of length $b$.

At first, a straight-line $l$is drawn, then a point is chosen on $l$and is labelled as $A$.

The compass is set for radius $b$.

Using $A$as the center, an arc is drawn intersecting the line $l$.

The point of intersection is labelled as $B$.

Thus, a straight line is constructed of length $b$.

Now, all the interior angles of a square are of$90°$.
Using $A$as the center and any radius, arcs are drawn intersecting $l$at $P$and $Q$.

Using $P$as the center and radius greater than $PA$, an arc is drawn.

Using $Q$as the center and with same radius, an arc is drawn intersecting the arc with center $P$at the point $X$.

$AX$is drawn and is extended upward.

Thus, $\angle A=90°$is drawn.

Similarly,

Using $B$as the center and of any radius, arcs are drawn intersecting $l$at $R$and $S$.

Using $R$as the center and radius greater than $RB$, an arc is drawn.

Using $S$as the center and of same radius, an arc is drawn intersecting the arc with center $R$at the point $Y$.

$BY$is drawn and is extended upward.

Thus, $\angle B=90°$is drawn.

Now, the compass is set for radius $b$.

Using $A$as the center, an arc is drawn intersecting the line $AX$.

The point of intersection is labelled as $D$.

Using $B$as the center, an arc is drawn intersecting the line $BY$.

The point of intersection is labelled as $C$.

The points $C$and $D$are joined.

Thus, $\overline{AB}=\overline{BC}=\overline{CD}=\overline{DA}=b$and $\angle A=\angle B=\angle C=\angle D=90°$.

The diagonal $\overline{BD}=b\sqrt{2}$.

Therefore, a square $ABCD$is constructed with diagonal of length $b\sqrt{2}$.