Radius OA is perpendicular to tangent BC, bisecting it into segments of length l/2. In right-angled triangle OAB, R² – r² = (l/2)² = l²/4. The ring area is pi(R² – r²) = (1/4)pi l².
In Fig. 6.51 we see two concentric circles with a common centre O. A chord BC of the larger circle is drawn, touching the smaller circle at A. The length of BC is l. Show that the area of the green region enclosed between the two circles is (1/4)pi l².
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Let the radius of the outer circle be R and the inner circle be r.
The green region is the circular ring (annulus) between them:
Area = piR² – pir² = pi(R² – r²).
Since the chord BC touches the inner circle at A, OA is a radius to the point of contact, so OA is perpendicular to BC.
The perpendicular from the centre to a chord bisects the chord, so:
AB = AC = l / 2.
In right-angled triangle OAB:
OA² + AB² = OB²
r² + (l / 2)² = R²
R² – r² = (l / 2)² = l² / 4.
Substituting this into the area expression:
Area = pi(l² / 4) = (1/4)pi l².
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/