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⭕ Chapter 5: I'm Up and Down, and Round and Round • Exercise Set 5.3

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Exercise Set 5.1 3Think, Draw and Infer 1Exercise Set 5.2 2Exercise Set 5.3 3Exercise Set 5.4 2Exercise Set 5.5 3Exercise Set 5.6 1End-of-Chapter Exercises 12All Exercises 27
Q1

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?

Answer: Let the circle have centre $C$, chord $AB$, and let $CM$ be perpendicular to $AB$ (so $\angle CMA = \angle CMB = 90^\circ$). We want to show $AM = BM$.
Consider $\Delta CMA$ and $\Delta CMB$:
1. $CA = CB$ (radii of the same circle).
2. $CM = CM$ (common side).
3. $\angle CMA = \angle CMB = 90^\circ$ (given perpendicular).
By the Right-Angle Hypotenuse Side (RHS) congruence criterion, $\Delta CMA \cong \Delta CMB$. Therefore, by CPCTC, $AM = BM$, meaning the perpendicular bisects the chord.

C A B M
By RHS Congruence: ΔCMA ≅ ΔCMB ⇒ AM = BM
Q2

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.

Answer: Let $AD$ be the altitude from $A$ to $BC$. In the isosceles triangle $\Delta ABC$ ($AB = AC$), the altitude to the base $BC$ also bisects the base. Thus, $AD$ is the perpendicular bisector of chord $BC$. Since the perpendicular bisector of any chord of a circle must pass through the centre of that circle, the altitude $AD$ must pass through the centre.

O (Centre) A B C D
Altitude AD is the perpendicular bisector of chord BC, so it passes through centre O
Q3

Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.

Answer: Let the chords be $AB = 6\text{ cm}$ and $CD = 8\text{ cm}$, and the radius be $r = 5\text{ cm}$. Let $O$ be the centre, and let $M$ and $N$ be the midpoints of $AB$ and $CD$ respectively. A line through the centre perpendicular to parallel chords will pass through both their midpoints.
For chord $AB$: $AM = 6/2 = 3\text{ cm}$. In right $\Delta OMA$, $OM = \sqrt{OA^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$.
For chord $CD$: $CN = 8/2 = 4\text{ cm}$. In right $\Delta ONC$, $ON = \sqrt{OC^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$.
Since the chords are on opposite sides of the centre, the total distance between their midpoints is $OM + ON = 4 + 3 = \mathbf{7\text{ cm}}$.

O A B M (AM=3) C D N (CN=4) OM = 4 cm ON = 3 cm 5 cm 5 cm
Total distance between midpoints MN = OM + ON = 4 cm + 3 cm = 7 cm