Exercise 5.3 Practice
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Overview
This page provides comprehensive Ch 5: Circles - Exercise Set 5.3 Practice. Solve proofs on why perpendiculars from center bisect chords, altitude path proofs of inscribed isosceles triangles, and distance calculations between parallel chords of different lengths on opposite sides of the center with step-by-step solutions.
Perpendicular Bisectors & Parallel Chord Distances
Q1: Converse of Perpendicular Chord Theorem
Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
(Hint: Use a right triangle congruence argument. Let the centre be $C$ and perpendicular meet chord $AB$ at $M$ such that $\angle CMA = \angle CMB = 90^\circ$. Show $AM = BM$.)
(Hint: Use a right triangle congruence argument. Let the centre be $C$ and perpendicular meet chord $AB$ at $M$ such that $\angle CMA = \angle CMB = 90^\circ$. Show $AM = BM$.)
Consider a circle with centre $C$ and chord $AB$. A line $CM$ is drawn perpendicular to $AB$ meeting it at $M$, so:
$$\angle CMA = \angle CMB = 90^\circ$$
We want to prove that $M$ is the midpoint of $AB$ (i.e. $AM = BM$).
$$\angle CMA = \angle CMB = 90^\circ$$
We want to prove that $M$ is the midpoint of $AB$ (i.e. $AM = BM$).
Compare right-angled triangles $\Delta CMA$ and $\Delta CMB$:
1. $\angle CMA = \angle CMB = 90^\circ$ (perpendicularity)
2. $CA = CB$ (hypotenuses are equal as they are radii $= r$)
3. $CM = CM$ (common side)
1. $\angle CMA = \angle CMB = 90^\circ$ (perpendicularity)
2. $CA = CB$ (hypotenuses are equal as they are radii $= r$)
3. $CM = CM$ (common side)
By the **RHS (Right angle-Hypotenuse-Side) Congruence Criterion**:
$$\Delta CMA \cong \Delta CMB$$
$$\Delta CMA \cong \Delta CMB$$
Since the triangles are congruent, their corresponding parts are equal (CPCT):
$$\mathbf{AM = BM}$$
Thus, the perpendicular from the centre bisects the chord.
$$\mathbf{AM = BM}$$
Thus, the perpendicular from the centre bisects the chord.
Proved: AM = BM by RHS congruence.
Q2: Inscribed Isosceles Altitude
An isosceles triangle $ABC$ is inscribed in a circle, with $AB = AC$. Show that the altitude from $A$ to $BC$ passes through the centre of the circle.
Let $AD$ be the altitude from $A$ to the side $BC$ in $\Delta ABC$, where $D$ lies on $BC$.
• Since $AB = AC$, the altitude $AD$ in an isosceles triangle also bisects the base $BC$.
• Thus, $AD \perp BC$ and $BD = CD$. This makes $AD$ the **perpendicular bisector** of the segment $BC$.
• Since $AB = AC$, the altitude $AD$ in an isosceles triangle also bisects the base $BC$.
• Thus, $AD \perp BC$ and $BD = CD$. This makes $AD$ the **perpendicular bisector** of the segment $BC$.
• In circle geometry, $BC$ forms a chord of the circumcircle.
• By theorem, the perpendicular bisector of any chord of a circle must pass through the centre of that circle.
• By theorem, the perpendicular bisector of any chord of a circle must pass through the centre of that circle.
Since $AD$ is the perpendicular bisector of the chord $BC$, the line containing altitude $AD$ must pass through the circle's centre $O$.
Proved: Altitude AD is the perpendicular bisector of chord BC, so it must pass through the centre.
Q3: Distance Between Parallel Chords
Two parallel chords of lengths $6\text{ cm}$ and $8\text{ cm}$ are on opposite sides of the centre of a circle. If the radius of the circle is $5\text{ cm}$, find the distance between the midpoints of the chords.
Let the centre of the circle be $O$ and radius $r = 5\text{ cm}$.
Let $AB = 6\text{ cm}$ and $CD = 8\text{ cm}$ be the parallel chords.
Draw perpendiculars from $O$ to both chords, meeting $AB$ at midpoint $M$ and $CD$ at midpoint $N$.
Since chords are parallel and on opposite sides, $M, O, N$ are collinear, and the distance between midpoints is:
$$MN = OM + ON$$
Let $AB = 6\text{ cm}$ and $CD = 8\text{ cm}$ be the parallel chords.
Draw perpendiculars from $O$ to both chords, meeting $AB$ at midpoint $M$ and $CD$ at midpoint $N$.
Since chords are parallel and on opposite sides, $M, O, N$ are collinear, and the distance between midpoints is:
$$MN = OM + ON$$
Find $OM$ using right triangle $\Delta OMA$:
• $AM = \frac{1}{2} AB = 3\text{ cm}$.
• Hypotenuse $OA = 5\text{ cm}$.
$$OM = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$
• $AM = \frac{1}{2} AB = 3\text{ cm}$.
• Hypotenuse $OA = 5\text{ cm}$.
$$OM = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$
Find $ON$ using right triangle $\Delta ONC$:
• $CN = \frac{1}{2} CD = 4\text{ cm}$.
• Hypotenuse $OC = 5\text{ cm}$.
$$ON = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$$
• $CN = \frac{1}{2} CD = 4\text{ cm}$.
• Hypotenuse $OC = 5\text{ cm}$.
$$ON = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$$
Calculate total distance $MN$:
$$MN = OM + ON = 4\text{ cm} + 3\text{ cm} = \mathbf{7\text{ cm}}$$
$$MN = OM + ON = 4\text{ cm} + 3\text{ cm} = \mathbf{7\text{ cm}}$$
Distance = 7 cm