Practical Geometry - Constructing a Perpendicular Bisector of a Line Segment

Draw a line segment $\overline{AB}$ of length $7 \text{ cm}$ and construct its perpendicular bisector using a ruler and compass.

1. Draw a line segment $\overline{AB} = 7 \text{ cm}$ using a ruler.
2. With $A$ as the center and a radius more than half of $AB$ (e.g., $4 \text{ cm}$), draw two arcs—one above $\overline{AB}$ and one below it.
3. Keeping the same radius and with $B$ as the center, draw two more arcs intersecting the previous arcs at points $P$ and $Q$.
4. Join point $P$ to point $Q$ using a ruler. Let $PQ$ intersect $AB$ at point $M$.
5. The line $PQ$ is the required perpendicular bisector, and $M$ is the midpoint.
6. Verification: Measure $AM$ and $MB$ with a ruler. Both should be $3.5 \text{ cm}$, since $\frac{7 \text{ cm}}{2} = 3.5 \text{ cm}$.

The compass radius must be greater than $3.5 \text{ cm}$ to ensure the arcs from $A$ and $B$ cross each other. The points of intersection $P$ and $Q$ provide the vertical path that cuts $\overline{AB}$ exactly in half at a $90^{\circ}$ angle.

If a line $XY$ is the perpendicular bisector of a segment $\overline{PQ}$ and they intersect at point $O$, find the length of $PQ$ if $PO = 4.2 \text{ cm}$.

1. Since $XY$ is the perpendicular bisector of $\overline{PQ}$, point $O$ must be the midpoint of $\overline{PQ}$.
2. By the property of a midpoint, $PO = OQ$.
3. Given $PO = 4.2 \text{ cm}$, it follows that $OQ = 4.2 \text{ cm}$.
4. The total length $PQ = PO + OQ$.
5. $PQ = 4.2 \text{ cm} + 4.2 \text{ cm} = 8.4 \text{ cm}$.

This problem uses the definition of a bisector. A perpendicular bisector always passes through the midpoint, meaning it splits the segment into two equal halves. Multiplying the length of one half by $2$ gives the total length.

1. Draw a line segment $MN = 6\text{ cm}$ using a ruler.
2. With $M$ as center and radius $> 3\text{ cm}$, draw arcs above and below $MN$.
3. With $N$ as center and the same radius, draw arcs intersecting the previous arcs at $X$ and $Y$.
4. Join $XY$. The point where $XY$ intersects $MN$ is $O$.
5. Length $MO = \frac{1}{2} \times MN = \frac{1}{2} \times 6 = 3\text{ cm}$.

Since the line $XY$ is the perpendicular bisector, it divides $MN$ into two equal parts. Thus, $MO = ON = 3\text{ cm}$.

1. In a perpendicular bisector, any point on the bisector is equidistant from the endpoints of the segment.
2. Since $P$ lies on the perpendicular bisector $CD$ of segment $EF$, then $PE = PF$.
3. Given $PE = 5\text{ cm}$, therefore $PF = 5\text{ cm}$.

This property of the perpendicular bisector ensures that the triangle $PEF$ is an isosceles triangle with $PE = PF$.