Q4:

Represent $\sqrt{9.3}$ on the number line.

Step 1: Initial Geometric Setup & Line Segment Definition

To represent the irrational number $\sqrt{9.3}$ on a number line, we utilize a geometric construction based on the properties of right-angled triangles and semicircles. We begin by translating the numerical value into a physical line segment.

• Draw a horizontal line and mark a starting point $A$.
• From point $A$, measure a distance of exactly $9.3$ units and mark this point as $B$. Thus, the length of the line segment $\overline{AB} = 9.3$ units.
• From point $B$, extend the line segment further by exactly $1$ unit and mark the new point as $C$. Thus, the length of $\overline{BC} = 1$ unit.

[By the Segment Addition Postulate], the total length of the line segment $\overline{AC}$ is:

$AC = AB + BC = 9.3 + 1 = 10.3 \text{ units}$

Step 2: Locating the Midpoint and Constructing the Semicircle

Next, we must find the geometric center of the segment $\overline{AC}$ to serve as the origin for a semicircle.

• Bisect the line segment $\overline{AC}$ to locate its midpoint, $O$.
• The distance from the midpoint $O$ to either endpoint ($A$ or $C$) represents the radius $r$ of the semicircle.

$r = OA = OC = \frac{AC}{2} = \frac{10.3}{2} = 5.15 \text{ units}$

With $O$ as the center and $r = 5.15$ units as the radius, construct a semicircle passing through points $A$ and $C$.

Step 3: Erecting the Perpendicular to Determine $\sqrt{9.3}$

At point $B$ (the boundary between the $9.3$ unit segment and the $1$ unit segment), construct a perpendicular line segment intersecting the semicircle.

• Draw a line perpendicular to $\overline{AC}$ at point $B$ ($\angle OBD = 90^\circ$).
• Let the point where this perpendicular intersects the semicircle be $D$.

The length of the segment $\overline{BD}$ is exactly $\sqrt{9.3}$ units.

Step 4: Rigorous Mathematical Proof of the Construction

To rigorously prove that $BD = \sqrt{9.3}$, we analyze the right-angled triangle $\triangle OBD$.

• The hypotenuse $\overline{OD}$ is a radius of the semicircle. Therefore, $OD = 5.15$ units.
• The base $\overline{OB}$ can be found by subtracting $\overline{BC}$ from the radius $\overline{OC}$: $OB = OC - BC = 5.15 - 1 = 4.15 \text{ units}$

[Per the Pythagorean Theorem], in right $\triangle OBD$:

$OD^2 = OB^2 + BD^2$

$BD^2 = OD^2 - OB^2$

Substitute the known values into the equation:

$BD^2 = (5.15)^2 - (4.15)^2$

Using the difference of squares identity, $a^2 - b^2 = (a - b)(a + b)$:

$BD^2 = (5.15 - 4.15)(5.15 + 4.15)$

$BD^2 = (1.00)(9.30) = 9.3$

$BD = \sqrt{9.3} \text{ units}$

Step 5: Mapping the Magnitude onto the Number Line

Now that we have a physical segment $\overline{BD}$ of length $\sqrt{9.3}$, we must map it onto a standard number line.

• Define point $B$ as the origin ($0$) of the number line.
• Consequently, point $C$ represents the number $1$ (since $BC = 1$ unit).
• Place the compass point at $B$ and the pencil point at $D$. Draw an arc downwards to intersect the extended number line at a new point, $E$.

Since the arc represents a circle with center $B$ and radius $BD$, the distance $BE$ is equal to $BD$. Therefore, point $E$ represents the exact location of $\sqrt{9.3}$ on the number line.

Final Solution: By treating point $B$ as the origin ($0$) and drawing an arc with radius $BD = \sqrt{9.3}$ units, the intersection of the arc with the number line at point $E$ accurately represents the irrational number $\sqrt{9.3}$.