Class 9 Mathematics · Ganita Manjari Part 1

Class 9 Maths Chapter 5 – Exercise Set 5.4 Solutions

Question 1 — Prove Theorem 6 using Baudhāyana–Pythagoras

Let $AB$ and $CD$ be equal chords of a circle with centre $O$. Let $OM\perp AB$ and $ON\perp CD$.

The perpendicular from the centre bisects a chord, so $AM=CN$ because $AB=CD$. Also $OA=OC$ as both are radii.

By Baudhāyana–Pythagoras:

Since $OA=OC$ and $AM=CN$, we get $OM^2=ON^2$, hence $OM=ON$.

Answer: Equal chords are at equal distances from the centre.

Question 2 — Equal perpendicular distances imply equal chords

Suppose $CE\perp AB$, $CH\perp GF$ and $CE=CH$ in the given circle.

The perpendiculars from the centre bisect the chords, so $AE=EB$ and $GH=HF$.

In the right triangles $\triangle CEA$ and $\triangle CHG$, the hypotenuses are equal radii, the perpendicular sides are equal by hypothesis, and the included angles are right angles. Hence the triangles are congruent by RHS.

Thus $AE=GH$, so $AB=2AE=2GH=GF$.

Answer: $\boxed{AB=GF}$.

Question 3 — Solve Question 2 using Baudhāyana–Pythagoras

From the perpendicular-bisector property, $AE=AB/2$ and $GH=GF/2$.

Applying Baudhāyana–Pythagoras:

Here $CA=CG$, because both are radii, and $CE=CH$. Therefore $AE^2=GH^2$, so $AE=GH$.

Answer: $\boxed{AB=GF}$.