Grade 11 Euclidean Geometry Notes

Euclidean Geometry: Chord Properties and Radius Calculations

Euclidean geometry involving circles often focuses on the relationship between chords, radii, and the center of the circle. In the provided problem 10.2, we examine a circle with center $O$ and a chord $AB$. The problem states that a line segment $OE$ bisects the chord $AB$. According to the fundamental theorem of circle geometry, a line drawn from the center of a circle to the midpoint of a chord is perpendicular to that chord. Therefore, we can established that $OE \perp AB$, which creates right-angled triangles within the circle, specifically $\triangle OEA$ and $\triangle OEB$.

To determine the radius $OB$ in terms of $x$, we first analyze the smaller right-angled triangle $\triangle ADE$. In this scenario, $A$ is a point on the circle, $E$ is the midpoint of the chord located on the line segment $OE$, and $D$ is a point on that same segment. Given the lengths $AD = 12\,cm$ and $ED = 8\,cm$, and knowing that $\angle AED = 90^{\circ}$, we use the Pythagorean theorem: $AE^2 + ED^2 = AD^2$. Substituting the known values, we have $AE^2 + 8^2 = 12^2$, which simplifies to $AE^2 + 64 = 144$, and thus $AE^2 = 80$. This means the length of half the chord, $AE$ (and by extension $EB$), is equivalent to $\sqrt{80}\,cm$.

In the larger right-angled triangle $\triangle OEB$, the hypotenuse is the radius $OB$. The length of the base $EB$ is $\sqrt{80}$. The length of the vertical segment $OE$ depends on the position of point $D$. Given $OD = x\,cm$ and $ED = 8\,cm$, if $E$ lies between $O$ and $D$, then $OE = x - 8$. Applying the Pythagorean theorem to $\triangle OEB$, we find $OB^2 = OE^2 + EB^2$. Substituting our values gives $OB = \sqrt{(x - 8)^2 + 80}$. Expanding this expression, we get $OB = \sqrt{x^2 - 16x + 64 + 80}$, which simplifies to the final expression for the radius: $OB = \sqrt{x^2 - 16x + 144}$.

To calculate the diameter of the circle as requested in question 10.2.2, we must first solve for $x$. In a circle, any point on the circumference is equidistant from the center. Since both $B$ and $D$ are points on the circumference (based on standard geometric problem structures where $D$ is labeled as a boundary point), we can equate their distances from the center: $OB = OD$. Thus, $x = \sqrt{x^2 - 16x + 144}$. Squaring both sides yields $x^2 = x^2 - 16x + 144$. Solving for $x$, we find $16x = 144$, which results in $x = 9\,cm$. Since the radius is $9\,cm$, the diameter is calculated as $2 \times 9\,cm = 18\,cm$.

Fundamental Circle Theorems: Angle at the Centre

Question 11.1 requires a formal proof of a cornerstone theorem in Euclidean geometry: the angle subtended by an arc at the center of a circle is double the size of the angle subtended by the same arc at any point on the circumference. In the provided diagram, $O$ is the center, and arc $AB$ subtends the angle $A\hat{O}B$ at the center and angle $A\hat{D}B$ at point $D$ on the circumference.

To prove that $A\hat{O}B = 2 A\hat{D}B$, a construction line must be drawn: join point $D$ to the center $O$ and extend this line to a point $P$ outside the circle. This construction creates two isosceles triangles, $\triangle OAD$ and $\triangle ODB$, because $OA = OD = OB$ (all being radii of the same circle). In $\triangle OAD$, because the base angles of an isosceles triangle are equal, $\hat{OAD} = \hat{ODA}$. Let these angles be represented by $x$. According to the exterior angle theorem of a triangle, the exterior angle $\hat{AOP}$ is equal to the sum of the two opposite interior angles, so $\hat{AOP} = x + x = 2x$.

Applying the same logic to $\triangle ODB$, the base angles $\hat{OBD}$ and $\hat{ODB}$ are equal. Let these be represented by $y$. The exterior angle $\hat{BOP}$ is therefore $y + y = 2y$. The total angle at the center, $A\hat{O}B$, is the sum of $\hat{AOP}$ and $\hat{BOP}$, which is $2x + 2y$. Factoring out the 2, we have $A\hat{O}B = 2(x + y)$. Since the total angle at the circumference $A\hat{D}B$ is the sum of the parts $\hat{ODA}$ ($x$) and $\hat{ODB}$ ($y$), we can conclude that $A\hat{O}B = 2 A\hat{D}B$.

Practical Application of Circle Theorems: Midpoints and Chords

Question 11.2 presents a circle designated as $RMPS$ with center $O$. A chord $RM$ is given, and $T$ is identified as the midpoint of this chord. The diagram also specifies that the angle $\hat{R}$ is $50^{\circ}$. This setup allows for the application of several geometric properties.

First, based on the theorem regarding lines from the center to a chord, since $T$ is the midpoint of chord $RM$, it is mandatory that the line $OT$ is perpendicular to the chord ($OT \perp RM$). This establishes that $\angle OTR = 90^{\circ}$. We are now dealing with a right-angled triangle $\triangle ORT$. In any triangle, the sum of the interior angles must equal $180^{\circ}$. Therefore, in $\triangle ORT$, we have $\hat{R} + \angle OTR + \angle ROT = 180^{\circ}$.

Substituting the known values into the equation, we get $50^{\circ} + 90^{\circ} + \angle ROT = 180^{\circ}$. Summing the known angles gives $140^{\circ} + \angle ROT = 180^{\circ}$, which allows us to find that $\angle ROT = 40^{\circ}$. Furthermore, because $OR$ and $OM$ are both radii, $\triangle ORM$ is an isosceles triangle. Since $OT$ is the altitude of this isosceles triangle, it also bisects the vertex angle $\hat{ROM}$. Consequently, the entire angle at the center subtended by chord $RM$ is $2 \times 40^{\circ} = 80^{\circ}$. This application demonstrates how midpoint properties and basic triangle theorems interact within circle geometry to solve for unknown angular values.