Exercise 5.5 Practice
00:00
Overview
This page provides comprehensive Ch 5: Circles - Exercise Set 5.5 Practice. Solve problems on computing chord lengths from radius and perpendicular height, general algebraic chord formulas, and disproving linear scaling of chords relative to perpendicular distances with step-by-step solutions.
Calculations of Chord Lengths & Perpendicular Heights
Q1: Chord Length Calculation
Find the length of the chord of a circle where the radius is $7\text{ cm}$ and perpendicular distance is $6\text{ cm}$.
Let $AB$ be the chord, and let $M$ be its midpoint. Let $O$ be the centre of the circle.
• Perpendicular distance $OM = 6\text{ cm}$.
• Radius $OA = 7\text{ cm}$.
• Angle $\angle OMA = 90^\circ$.
• Perpendicular distance $OM = 6\text{ cm}$.
• Radius $OA = 7\text{ cm}$.
• Angle $\angle OMA = 90^\circ$.
Using the Baudhāyana-Pythagoras theorem in right-angled triangle $\Delta OMA$:
$$OA^2 = OM^2 + AM^2 \implies 7^2 = 6^2 + AM^2$$
$$49 = 36 + AM^2 \implies AM^2 = 49 - 36 = 13 \implies AM = \sqrt{13}\text{ cm} \approx 3.61\text{ cm}$$
$$OA^2 = OM^2 + AM^2 \implies 7^2 = 6^2 + AM^2$$
$$49 = 36 + AM^2 \implies AM^2 = 49 - 36 = 13 \implies AM = \sqrt{13}\text{ cm} \approx 3.61\text{ cm}$$
Since the perpendicular from the centre bisects the chord:
$$\text{Chord Length } AB = 2 \times AM = 2\sqrt{13}\text{ cm} \approx \mathbf{7.21\text{ cm}}$$
$$\text{Chord Length } AB = 2 \times AM = 2\sqrt{13}\text{ cm} \approx \mathbf{7.21\text{ cm}}$$
Length of chord = 2√13 cm (≈ 7.21 cm)
Q2: Chord Length Formula Derivation
Explain why the following statement is true: If the perpendicular distance of a chord from the centre is $d$ and the radius is $r$, then the chord length is $2\sqrt{r^2 - d^2}$.
Let $AB$ be a chord of a circle of radius $r$ with centre $O$.
Draw perpendicular $OM$ from $O$ to $AB$. Thus, the perpendicular distance $OM = d$.
Draw perpendicular $OM$ from $O$ to $AB$. Thus, the perpendicular distance $OM = d$.
By the chord bisector theorem, the perpendicular from the centre bisects the chord, so $M$ is the midpoint of $AB$:
$$\text{Chord length } AB = 2 \times AM$$
$$\text{Chord length } AB = 2 \times AM$$
In right-angled triangle $\Delta OMA$, using the Baudhāyana-Pythagoras theorem:
$$OA^2 = OM^2 + AM^2 \implies r^2 = d^2 + AM^2$$
$$AM^2 = r^2 - d^2 \implies AM = \sqrt{r^2 - d^2}$$
$$OA^2 = OM^2 + AM^2 \implies r^2 = d^2 + AM^2$$
$$AM^2 = r^2 - d^2 \implies AM = \sqrt{r^2 - d^2}$$
Multiply by 2 to get the full chord length:
$$AB = 2 \times AM = \mathbf{2\sqrt{r^2 - d^2}}$$
This proves the statement.
$$AB = 2 \times AM = \mathbf{2\sqrt{r^2 - d^2}}$$
This proves the statement.
Proved: Chord length is 2√(r² - d²).
Q3: Non-Linear Chord Scaling
In a circle, if the distance of chord $AB$ from the centre is twice the distance of another chord $CD$ from the centre, then can we conclude that $CD = 2 AB$? Give reasons for your answer.
Let the radius of the circle be $r$.
Let the perpendicular distance of chord $CD$ from the centre be $x$.
Then, the perpendicular distance of chord $AB$ from the centre is given as $2x$.
Let the perpendicular distance of chord $CD$ from the centre be $x$.
Then, the perpendicular distance of chord $AB$ from the centre is given as $2x$.
Using the chord length formula derived in Question 2:
$$CD = 2\sqrt{r^2 - x^2}$$
$$AB = 2\sqrt{r^2 - (2x)^2} = 2\sqrt{r^2 - 4x^2}$$
$$CD = 2\sqrt{r^2 - x^2}$$
$$AB = 2\sqrt{r^2 - (2x)^2} = 2\sqrt{r^2 - 4x^2}$$
For $CD = 2 AB$ to be true:
$$2\sqrt{r^2 - x^2} = 2 \times \left(2\sqrt{r^2 - 4x^2}\right) = 4\sqrt{r^2 - 4x^2}$$
$$\sqrt{r^2 - x^2} = 2\sqrt{r^2 - 4x^2}$$
Squaring both sides:
$$r^2 - x^2 = 4(r^2 - 4x^2) = 4r^2 - 16x^2$$
$$15x^2 = 3r^2 \implies 5x^2 = r^2 \implies x = \frac{r}{\sqrt{5}}$$
$$2\sqrt{r^2 - x^2} = 2 \times \left(2\sqrt{r^2 - 4x^2}\right) = 4\sqrt{r^2 - 4x^2}$$
$$\sqrt{r^2 - x^2} = 2\sqrt{r^2 - 4x^2}$$
Squaring both sides:
$$r^2 - x^2 = 4(r^2 - 4x^2) = 4r^2 - 16x^2$$
$$15x^2 = 3r^2 \implies 5x^2 = r^2 \implies x = \frac{r}{\sqrt{5}}$$
• The relationship $CD = 2 AB$ is **only** true when the distance of chord $CD$ is exactly $x = \frac{r}{\sqrt{5}}$.
• For any other distance, the relationship is false. Therefore, we **cannot** conclude that $CD = 2 AB$ in general.
• The relationship between chord length and perpendicular distance is non-linear due to the square root and square operations.
• For any other distance, the relationship is false. Therefore, we **cannot** conclude that $CD = 2 AB$ in general.
• The relationship between chord length and perpendicular distance is non-linear due to the square root and square operations.
No, we cannot conclude CD = 2 AB because chord length scales non-linearly.