Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true. Class 9

Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true. Class 9

Question 1.

Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true. Class 9

Answer:

Given : C is the centre of the circle with radius r. Chord AB = Chord FG.

E and H are the midpoints of AB and FG respectively.

To show : CE = CH

Proof:

CE ⊥ AB and CH ⊥ FG ... [The line joining the centre and midpoint of a chord of a circle is perpendicular to the chord]

For chord AB,

AE = $\frac{\mathrm{AB}}{2}$ ... [E is the midpoint of AB]

In right-angled ∆CEA,

CA² = CE² + AE²

...[By Baudhayana-Pythagoras theorem]

∴ r² = CE² + $\left(\frac{\mathrm{AB}}{2}\right)^2$ ...(i)

For chord FG,

FH = $\frac{\mathrm{FG}}{2}$ ... [H is the midpoint of GF]

In right-angled ∆CHF,

CF² = CH² + FH²

...[By Baudhayana-Pythagoras theorem]

∴ r² = CH² + $\left(\frac{\mathrm{FG}}{2}\right)^2$ ...(ii)

From (i) and (ii), we get

$\mathrm{CE}^2+\left(\frac{\mathrm{AB}}{2}\right)^2=\mathrm{CH}^2+\left(\frac{\mathrm{FG}}{2}\right)^2$

But, AB = FG ... [Given]

∴ CE² = CH²

∴ CE = CH

Hence, chords of equal length are equidistant from the centre.