Question 23 of 50intermediate💡 UnderstandShort Answer3 marks

Explain why the following statement is true: If the perpendicular distance of a chord from the centre is dd and the radius is rr, then the chord length is 2r2−d22\sqrt{r^2 - d^2}.

Correct Answer

Let AB be the chord of a circle with centre C. The perpendicular distance from C to AB is dd, and the radius is rr.

By Theorem 5, the perpendicular from the centre of a circle to a chord bisects the chord. Thus, if the perpendicular from C meets AB at M, M is the midpoint of AB, meaning AM = MB. Also, ∠CMA=90∘\angle CMA = 90^\circ.

In the right-angled Δ\Delta CMA, we apply the Baudhāyana–Pythagoras Theorem: CA2=CM2+AM2CA^2 = CM^2 + AM^2. Substituting CA=rCA = r and CM=dCM = d, we get r2=d2+AM2r^2 = d^2 + AM^2. Solving for AM, we find AM=r2−d2AM = \sqrt{r^2 - d^2}. Since AB = 2AM, the chord length is 2r2−d22\sqrt{r^2 - d^2}.

Exercise: Exercise Set 5.5 | Q: 2 | (Page 14)
For More Understanding

Explanation

The explanation relies on two key concepts provided in the text: Theorem 5 (which states that a perpendicular from the centre bisects the chord) and the Baudhāyana–Pythagoras Theorem (used in Theorem 8's explanation to relate radius, perpendicular distance, and half-chord). By combining these, we form a right-angled triangle with the radius as the hypotenuse, allowing us to algebraically solve for the half-chord length and double it for the full length.

Solution Steps

  1. Step 1: Draw chord AB with centre C. Drop a perpendicular CM to AB, so CM = d and CA = r. By Theorem 5, M is the midpoint of AB, so AM = MB.

  2. Step 2: In right-angled triangle CMA, apply the Baudhāyana–Pythagoras Theorem: CA² = CM² + AM².

  3. Step 3: Substitute the given values: r² = d² + AM². Rearrange to find AM: AM² = r² - d², so AM = r2−d2\sqrt{r² - d²}.

  4. Step 4: Calculate the full chord length: AB = 2 * AM = 2r2−d2\sqrt{r² - d²}.

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