Exercise 5.3 Practice
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Overview
Step-by-step NCERT solutions and practice problems for CBSE Class 9 Maths Circles (Exercise 5.3 Practice), providing clear algebraic methods and exam-oriented 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