In the figure 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 14πl2.
Answer: OA ⊥ BC (radius to the point of contact) and A is the midpoint of BC, so R2 = r2 + (l2)2. Ring = πR2 − πr2 = π(l2)2 = 14πl2.
Step-by-step solution
Idea: The ring’s area depends only on R2 − r2, and the right triangle OAB gives exactly that difference.
- Let the radii be R (OB) and r (OA). BC touches the smaller circle at A, so OA ⊥ BC. The perpendicular from the centre to a chord bisects it, so BA = l2.1 mark
- In right △OAB: OB2 = OA2 + AB2 ⇒ R2 − r2 = l24.1 mark
- Green ring = πR2 − πr2 = π(R2 − r2) = 14πl2. Shown.1 mark
Area of the ring = π(R² − r²) = π(l/2)² = ¼πl².
Check: R = 5, r = 3: half-chord 4, l = 8; ring = π(25 − 9) = 16π = 14π × 64 ✓.
Answer to write in the exam
OA ⊥ BC (tangent ⊥ radius) ⇒ BA = AC = l2 (perpendicular from centre bisects chord)
In right △OAB: R2 = r2 + l24
Area of ring = πR2 − πr2 = π(R2 − r2)
∴ Area = 14πl2
Common mistakes that cost marks
- Using π(R − r)2 for the ring. It is πR2 − πr2.
- Taking AB = l instead of l2.
- Not giving the reason why OA ⊥ BC.
How this can come in the exam
MCQ (1 mark)
A chord of length 16 cm of the larger of two concentric circles touches the smaller one. The area of the ring between them is (use π = 3.14)
- 50.24 cm2
- 100.48 cm2
- 200.96 cm2
- 803.84 cm2
Show answer
(C) 200.96 cm2
Ring = 14πl2 = 14 × 3.14 × 256 = 3.14 × 64 = 200.96 cm2.
Try one yourself
The area of a ring between two concentric circles is 616 cm2. Find the length of a chord of the outer circle that touches the inner circle. (Use π = 227.)
Show answer
14πl2 = 616 ⇒ l2 = 784 ⇒ l = 28 cm.
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