Q. 23 - Geometry

Question

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 $b\sqrt{2}$

Step-by-Step Solution

Answer

The figure is,

Step 1. Given information.

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

Step 2. Concept used.

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 $b$.

Step 3. Consider the following ways.

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 $b\sqrt{2}$.