NCERT Solutions · Class 9 Mathematics I’m Up and Down, and Round and Round

Half the chord, the distance, and the radius form a right triangle.Let the circle have centre O and radius \(\displaystyle r=7\ \mathrm{cm}\), let AB be the chord, and let M be the foot of the perpendicular from O to AB, so \(\displaystyle \mathrm{OM}=6\ \mathrm{cm}\).First check such a chord can exist: the distance \(\displaystyle 6\ \mathrm{cm}\) is less than the radius \(\displaystyle 7\ \mathrm{cm}\), so the line does cut the circle.Step $\displaystyle 1$ — M is the midpoint. By Theorem $\displaystyle 5$, the perpendicular from the centre to a chord bisects it, so \(\displaystyle \mathrm{AM}=\tfrac{1}{2}\mathrm{AB}\).Step $\displaystyle 2$ — the right triangle. \(\displaystyle \triangle \mathrm{OMA}\) is right-angled at M. Its hypotenuse is \(\displaystyle \mathrm{OA}=r=7\ \mathrm{cm}\) (a radius, since A is on the circle), and its legs are \(\displaystyle \mathrm{OM}=6\ \mathrm{cm}\) and AM. By the Baudhāyana–Pythagoras theorem, \[\mathrm{OA}^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}, \] \[7^{2}=6^{2}+\mathrm{AM}^{2}, \] \[\mathrm{AM}^{2}=49-36=13, \qquad \mathrm{AM}=\sqrt{13}\ \mathrm{cm}. \]Step $\displaystyle 3$ — double it. \[\mathrm{AB}=2\,\mathrm{AM}=2\sqrt{13}\ \mathrm{cm}. \]Numerically \(\displaystyle \sqrt{13}\approx 3.606\), so \(\displaystyle \mathrm{AB}\approx 7.21\ \mathrm{cm}\).A quick plausibility check: the longest possible chord here is the diameter, \(\displaystyle 14\ \mathrm{cm}\), and this chord sits far out from the centre, so a value of about \(\displaystyle 7.2\ \mathrm{cm}\) — roughly half the diameter — is sensible.The chord is \(\displaystyle 2\sqrt{13}\ \mathrm{cm}\approx 7.2\ \mathrm{cm}\) long.

It is the previous question done with letters instead of numbers.Let the circle have centre O and radius \(\displaystyle r\), let AB be a chord, and let M be the foot of the perpendicular from O to AB, so the perpendicular distance of the chord from the centre is \(\displaystyle \mathrm{OM}=d\).Step $\displaystyle 1$ — M is the midpoint of AB. This is Theorem $\displaystyle 5$: the perpendicular from the centre of a circle to a chord bisects the chord. So \[\mathrm{AM}=\tfrac{1}{2}\mathrm{AB}, \qquad\text{that is,}\qquad \mathrm{AB}=2\,\mathrm{AM}. \] This step is the whole reason a single formula can work: it converts "the chord" into "twice one particular leg of a right triangle".Step $\displaystyle 2$ — build the right triangle. Join OA. Since A lies on the circle, \(\displaystyle \mathrm{OA}=r\). And \(\displaystyle \angle \mathrm{OMA}=90^\circ\), because OM is perpendicular to AB. So \(\displaystyle \triangle \mathrm{OMA}\) is right-angled at M, with hypotenuse OA.Step $\displaystyle 3$ — Baudhāyana–Pythagoras. \[\mathrm{OA}^{2}=\mathrm{OM}^{2}+\mathrm{AM}^{2}, \] \[r^{2}=d^{2}+\mathrm{AM}^{2}, \] \[\mathrm{AM}^{2}=r^{2}-d^{2}, \qquad \mathrm{AM}=\sqrt{r^{2}-d^{2}} \] (taking the positive square root, since AM is a length).Step $\displaystyle 4$ — double it. \[\mathrm{AB}=2\,\mathrm{AM}=2\sqrt{r^{2}-d^{2}}. \]Does the formula behave sensibly? Three checks: The formula also re-proves Theorem $\displaystyle 6$ and Theorem $\displaystyle 7$ in one line: for a fixed \(\displaystyle r\), the chord length depends on \(\displaystyle d\) alone, so equal \(\displaystyle d\) forces equal chord and equal chord forces equal \(\displaystyle d\). And since \(\displaystyle 2\sqrt{r^{2}-d^{2}}\) gets smaller as \(\displaystyle d\) gets bigger, it re-proves Theorem $\displaystyle 8$ as well.Hence the chord length is \(\displaystyle 2\sqrt{r^{2}-d^{2}}\).

No. Test the claim with the formula \(\displaystyle 2\sqrt{r^{2}-d^{2}}\) — the relationship is not a proportion.Write \(\displaystyle r\) for the radius and let CD be at distance \(\displaystyle d\) from the centre, so AB is at distance \(\displaystyle 2d\). By the formula from Question $\displaystyle 2$, \[\mathrm{CD}=2\sqrt{r^{2}-d^{2}}, \qquad \mathrm{AB}=2\sqrt{r^{2}-4d^{2}}. \] (For AB to exist at all we need \(\displaystyle 2d<r\).)A counterexample settles it. Take \(\displaystyle r=5\ \mathrm{cm}\) and \(\displaystyle d=1\ \mathrm{cm}\), so CD is \(\displaystyle 1\ \mathrm{cm}\) from the centre and AB is \(\displaystyle 2\ \mathrm{cm}\) from the centre: \[\mathrm{CD}=2\sqrt{25-1}=2\sqrt{24}\approx 9.80\ \mathrm{cm}, \qquad \mathrm{AB}=2\sqrt{25-4}=2\sqrt{21}\approx 9.17\ \mathrm{cm}. \] CD is longer than AB, but only by about \(\displaystyle 7\%\) — nowhere near twice. So \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\) does not follow.Why the intuition fails. Doubling \(\displaystyle d\) does not double or halve anything inside the formula, because \(\displaystyle d\) enters as \(\displaystyle d^{2}\) and then under a square root, subtracted from \(\displaystyle r^{2}\). The chord length is not proportional to the distance, nor inversely proportional to it — the two are not in any fixed ratio at all.What we can say. AB is farther from the centre than CD, so by Theorem $\displaystyle 8$ the farther chord is the shorter one: \[\mathrm{CD}>\mathrm{AB}. \] That inequality is all that follows from the given information.When does \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\) actually happen? Solve for it: \[2\sqrt{r^{2}-d^{2}}=2\cdot 2\sqrt{r^{2}-4d^{2}} \] \[\sqrt{r^{2}-d^{2}}=2\sqrt{r^{2}-4d^{2}} \] Squaring both sides, \[r^{2}-d^{2}=4\left(r^{2}-4d^{2}\right)=4r^{2}-16d^{2} \] \[15d^{2}=3r^{2}, \qquad r^{2}=5d^{2}, \qquad r=d\sqrt{5}. \] So it happens for exactly one shape of picture: when the radius is \(\displaystyle \sqrt{5}\) times the distance of CD from the centre. For instance \(\displaystyle r=2\sqrt{5}\approx 4.47\ \mathrm{cm}\) with \(\displaystyle d=2\ \mathrm{cm}\) gives \[\mathrm{CD}=2\sqrt{20-4}=8\ \mathrm{cm}, \qquad \mathrm{AB}=2\sqrt{20-16}=4\ \mathrm{cm}, \] and indeed \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\). (Check \(\displaystyle 2d=4<4.47=r\), so AB really is a chord.) Every other choice of \(\displaystyle r\) and \(\displaystyle d\) gives a different ratio — and in fact, by pushing \(\displaystyle 2d\) close to \(\displaystyle r\), the ratio \(\displaystyle \mathrm{CD}/\mathrm{AB}\) can be made as large as you like.Answer: No, we cannot conclude \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\). Doubling the distance from the centre does not halve the chord. All that follows in general is \(\displaystyle \mathrm{CD}>\mathrm{AB}\); the equality \(\displaystyle \mathrm{CD}=2\,\mathrm{AB}\) holds only in the special case \(\displaystyle r=d\sqrt{5}\).