Verified step-by-step solutions for your class notes.
Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
Answer: Let the radius be $r = 7\text{ cm}$ and the perpendicular distance to the chord be $d = 6\text{ cm}$. Let half the chord length be $x$.
Using Pythagoras' theorem: $x^2 + d^2 = r^2$
$x^2 + 6^2 = 7^2 \implies x^2 + 36 = 49 \implies x^2 = 13 \implies x = \sqrt{13}$.
The total chord length is $2x = \mathbf{2\sqrt{13}\text{ cm}}$ (or approx. $7.21\text{ cm}$).
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 $2\sqrt{r^2 - d^2}$.
Answer: The radius $r$, the perpendicular distance $d$ from the centre to the chord, and half the chord length ($x$) form a right-angled triangle. By the Baudhāyana-Pythagoras theorem, $x^2 + d^2 = r^2$, which gives $x^2 = r^2 - d^2$, so $x = \sqrt{r^2 - d^2}$. Since the perpendicular from the centre bisects the chord (Theorem 5), the total length of the chord is twice $x$, which is $2\sqrt{r^2 - d^2}$.
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 $CD = 2AB$? Give reasons for your answer.
Answer: No, we cannot conclude that $CD = 2AB$.
The length of a chord based on its distance from the centre is given by $2\sqrt{r^2 - d^2}$. The relationship is non-linear due to the square and square root. While Theorem 8 states that a chord closer to the centre is longer, the length does not scale proportionally with the distance.
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