NCERT Solutions for Class 9 Maths Chapter 5 — Exercise Set 5.4#
Theorem 6: Chords of a circle having the same length are all at the same distance from the centre.
Theorem 7: Chords of a circle that are equidistant from the centre have equal length.
The distance of a chord from the centre is the length of the perpendicular from the centre to the chord. By Theorem 5, this perpendicular meets the chord at its midpoint.
1. Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true. #
Solution
Let the circle have centre $C$ and radius $r$. Let $AB$ be any chord of length $l$, and let $E$ be the foot of the perpendicular from $C$ to $AB$. By Theorem 5, $E$ is the midpoint of $AB$:
$$AE = \dfrac{l}{2}$$$\triangle CEA$ is right-angled at $E$, with hypotenuse $CA = r$. By the Baudhāyana–Pythagoras theorem:
$$CE^2 + AE^2 = CA^2$$$$\Rightarrow CE^2 = r^2 - \left(\dfrac{l}{2}\right)^2$$
$$\Rightarrow CE = \sqrt{r^2 - \dfrac{l^2}{4}}$$
The distance $CE$ depends only on the radius $r$ and the chord length $l$. So any other chord of the same length $l$ in the same circle has the same distance from the centre.
∴ Chords of the same length are at the same distance from the centre, which is Theorem 6.
2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. #
Solution
Given: $CE \perp AB$, $CH \perp GF$ and $CE = CH$, where $C$ is the centre.
To show: $AB = GF$.
Why is this true? Join $CA$ and $CF$. Consider $\triangle CEA$ and $\triangle CHF$.
| Part | Reason |
|---|---|
| $\angle CEA = \angle CHF = 90°$ | given perpendiculars |
| $CA = CF$ | radii of the circle (the hypotenuses) |
| $CE = CH$ | given |
By RHS congruence, $\triangle CEA \cong \triangle CHF$, so:
$$AE = HF$$By Theorem 5, the perpendicular from the centre bisects the chord, so $E$ is the midpoint of $AB$ and $H$ is the midpoint of $GF$:
$$AB = 2AE$$$$GF = 2HF$$$$\Rightarrow AB = 2AE = 2HF = GF$$
∴ $AB = GF$. Chords equidistant from the centre are equal, which is Theorem 7.
3. Solve the previous question using the Baudhāyana–Pythagoras theorem. #
Solution
Let $r$ be the radius. Join $CA$ and $CF$, so $CA = CF = r$.
$\triangle CEA$ is right-angled at $E$:
$$AE^2 = CA^2 - CE^2$$$$AE^2 = r^2 - CE^2$$
$\triangle CHF$ is right-angled at $H$:
$$HF^2 = CF^2 - CH^2$$$$HF^2 = r^2 - CH^2$$
We are given $CE = CH$, so the right-hand sides are equal:
$$\Rightarrow AE^2 = HF^2$$$$\Rightarrow AE = HF$$
By Theorem 5, $AB = 2AE$ and $GF = 2HF$.
∴ $AB = GF$.