Q1. Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.

Answer:

Theorem 6 states: Equal chords of a circle are equidistant from the centre.

Let AB and FG be equal chords of a circle with centre C. Let CE be the perpendicular from C to AB, and CH be the perpendicular from C to FG.

In right triangles CAE and CFH:

• CA = CF (radii)
• AE = FH (since AB = FG, and E and H are midpoints by Theorem 5)

Using the Baudhayana-Pythagoras theorem:

\( CE^2 = CA^2 - AE^2 \) and \( CH^2 = CF^2 - FH^2 \)

Since \( CA^2 = CF^2 \) and \( AE^2 = FH^2 \), we get \( CE^2 = CH^2 \).

Therefore, CE = CH, meaning the chords are equidistant from the centre.

Q2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF. [IMAGE NEEDED: Figure 5.15 from the textbook showing a circle with chords AB and GH]

Answer:

Given: CE = CH, CE ⊥ AB, CH ⊥ GH. AB and GH are chords of a circle with centre C.

In right triangles CAE and CGH:

• CA = CG (radii)
• CE = CH (given)

Using the RHS congruence criterion, \( \triangle CAE \cong \triangle CGH \).

Therefore, AE = GH (corresponding parts of congruent triangles).

Since E is the midpoint of AB (Theorem 5), AE = AB/2. Similarly, H is the midpoint of GH, so GH = GF/2.

Therefore, AB/2 = GF/2, which implies AB = GF.

Q3. Solve the previous question using the Baudhayana-Pythagoras theorem.

Answer:

Given: CE = CH, CE ⊥ AB, CH ⊥ GH. AB and GH are chords of a circle with centre C.

In right triangles CAE and CGH, using the Baudhayana-Pythagoras theorem:

\( AE^2 = CA^2 - CE^2 \) and \( GH^2 = CG^2 - CH^2 \)

Since CA = CG (radii) and CE = CH (given), we get AE² = GH², so AE = GH.

Since E is the midpoint of AB and H is the midpoint of GH, AB = 2AE and GH = 2GH. Therefore, AB = GH.