NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.3#
Theorem 4: The line joining the centre of a circle and the midpoint of a chord is perpendicular to the chord.
1. 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 Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.) #
Solution
Given: a circle with centre $C$, a chord $AB$, and $CM \perp AB$, so $\angle CMA = \angle CMB = 90°$.
To show: $AM = BM$.
Why is this true? Consider the right triangles $\triangle CMA$ and $\triangle CMB$.
| Part | Reason |
|---|---|
| $\angle CMA = \angle CMB = 90°$ | given |
| $CA = CB$ | radii of the circle (the hypotenuses) |
| $CM = CM$ | common side |
∴ By RHS congruence, $\triangle CMA \cong \triangle CMB$.
So the corresponding sides are equal:
$$AM = BM$$∴ The perpendicular from the centre to a chord bisects the chord. This is Theorem 5.
2. 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. #
Solution
Let $AD$ be the altitude from $A$, so $AD \perp BC$ and $\angle ADB = \angle ADC = 90°$.
Step 1: $AD$ bisects $BC$. Consider $\triangle ADB$ and $\triangle ADC$:
| Part | Reason |
|---|---|
| $\angle ADB = \angle ADC = 90°$ | $AD$ is the altitude |
| $AB = AC$ | given |
| $AD = AD$ | common side |
By RHS congruence, $\triangle ADB \cong \triangle ADC$, so $BD = DC$. So $D$ is the midpoint of $BC$, and $AD$ is the perpendicular bisector of $BC$.
Step 2: the perpendicular bisector of a chord passes through the centre. $BC$ is a chord of the circle. By Theorem 4, the line joining the centre $O$ to the midpoint $D$ of $BC$ is perpendicular to $BC$. At the point $D$ there is only one line perpendicular to $BC$, so the line $OD$ is the same as the line $AD$.
∴ The altitude $AD$ passes through the centre of the circle.
3. 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. #
Solution
Let $AB = 6$ cm and $CD = 8$ cm be the chords, with midpoints $M$ and $N$, and let $O$ be the centre. By Theorem 4, $OM \perp AB$ and $ON \perp CD$. The chords are parallel, so $OM$ and $ON$ lie along the same line, on opposite sides of $O$.
By Theorem 5, $M$ and $N$ are midpoints, so:
$$MB = \dfrac{6}{2} = 3 \text{ cm}$$$$ND = \dfrac{8}{2} = 4 \text{ cm}$$
In right $\triangle OMB$, by the Baudhāyana–Pythagoras theorem:
$$OM^2 = OB^2 - MB^2$$$$OM^2 = 5^2 - 3^2$$
$$OM^2 = 16$$
$$\Rightarrow OM = 4 \text{ cm}$$
In right $\triangle OND$:
$$ON^2 = OD^2 - ND^2$$$$ON^2 = 5^2 - 4^2$$
$$ON^2 = 9$$
$$\Rightarrow ON = 3 \text{ cm}$$
The chords are on opposite sides of the centre, so:
$$MN = OM + ON$$$$MN = 4 + 3$$
$$MN = 7 \text{ cm}$$
∴ The distance between the midpoints of the chords is 7 cm.