Class 9 Mathematics · Ganita Manjari Part 1

Class 9 Maths Chapter 5 – Exercise Set 5.3 Solutions

Question 1 — Converse of the perpendicular-to-chord theorem

Let $AB$ be a chord with centre $O$. Suppose $OM\perp AB$.

In $\triangle OMA$ and $\triangle OMB$:

• $OA=OB$ — radii.
• $OM=OM$ — common side.
• $\angle OMA=\angle OMB=90^\circ$.

Hence the triangles are congruent by RHS. Therefore, $AM=BM$ by CPCT.

Answer: The perpendicular from the centre to a chord bisects the chord.

Question 2 — Altitude of an inscribed isosceles triangle

Let $AB=AC$ and let $AD$ be the altitude to $BC$. Then $AD\perp BC$.

In $\triangle ABD$ and $\triangle ACD$, $AB=AC$, $AD$ is common and both angles at $D$ are right angles. Hence the triangles are congruent by RHS.

So $BD=DC$. Thus $AD$ is the perpendicular bisector of chord $BC$. The perpendicular bisector of a chord passes through the centre.

Answer: The altitude $AD$ passes through the centre of the circle.

Question 3 — Distance between the midpoints of two parallel chords

Given: radius $=5$ cm; chord lengths $6$ cm and $8$ cm, on opposite sides of the centre.

The perpendicular from the centre bisects each chord. Hence the half-chords are $3$ cm and $4$ cm.

For the $6$ cm chord:

For the $8$ cm chord:

Because the chords lie on opposite sides, the distance between their midpoints is

Answer: $\boxed{7\text{ cm}}$.