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⭕ Chapter 5: I'm Up and Down, and Round and Round • Exercise Set 5.4

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Exercise Set 5.1 3Think, Draw and Infer 1Exercise Set 5.2 2Exercise Set 5.3 3Exercise Set 5.4 2Exercise Set 5.5 3Exercise Set 5.6 1End-of-Chapter Exercises 12All Exercises 27
Q2

Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and $CE = CH$, show that $AB = GF$.

C A B E G F H Fig. 5.15: Equal perpendiculars CE = CH ⇔ AB = GF

Answer: The perpendiculars $CE$ and $CH$ from the centre $C$ to the chords $AB$ and $GF$ are given as equal ($CE = CH$). Theorem 7 states that chords of a circle that are equidistant from the centre have equal length. Therefore, $AB = GF$.

Q3

Solve the previous question using the Baudhāyana-Pythagoras theorem.

Answer: Draw radii $CA$ and $CG$. Both are equal to $r$.
In right $\Delta CEA$: $AE^2 = CA^2 - CE^2 = r^2 - CE^2$.
In right $\Delta CHG$: $GH^2 = CG^2 - CH^2 = r^2 - CH^2$.
Since we are given $CE = CH$, it follows that $AE^2 = GH^2$, so $AE = GH$.
Because a perpendicular from the centre bisects a chord, $AB = 2AE$ and $GF = 2GH$.
Therefore, $AB = GF$.