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.
No, we cannot conclude that CD = 2 AB. The longer the chord, the closer it is to the centre. Since the distance of chord AB from the centre is twice that of chord CD, AB is farther from the centre. This only implies that AB < CD, not that CD is exactly twice AB.
Using the Baudhāyana–Pythagoras Theorem, let the radius be r, distance of AB be 2d, and distance of CD be d. For AB: r² = (2d)² + (AB/2)². For CD: r² = d² + (CD/2)². Equating them gives 4d² + (AB/2)² = d² + (CD/2)², or 3d² + (AB/2)² = (CD/2)². This relation does not yield CD = 2 AB. Thus, the given conclusion is incorrect.
Explanation
The textbook establishes through Theorem 8 that a longer chord is closer to the centre. While the inverse relationship between chord length and distance holds, it is not linearly proportional. By applying the Baudhāyana–Pythagoras Theorem as demonstrated in the text, the square of the radius equals the sum of the squares of the perpendicular distance and half the chord length. Substituting the given distance relationship into this equation yields a non-linear equation, proving that halving the distance does not double the chord length.
Solution Steps
Step 1: Identify the inverse relationship: The longer the chord, the closer it is to the centre. Since AB is twice as far from the centre as CD, AB < CD.
Step 2: Apply the Baudhāyana–Pythagoras Theorem for chord AB (distance 2d): r² = (2d)² + (AB/2)².
Step 3: Apply the theorem for chord CD (distance d): r² = d² + (CD/2)².
Step 4: Equate the two expressions for r²: 4d² + (AB/2)² = d² + (CD/2)², which simplifies to 3d² + (AB/2)² = (CD/2)².
Step 5: Conclude that the derived equation does not simplify to CD = 2 AB, so the statement is false.