A chord of the larger among two concentric circles is of length 20 cm and it is tangent to the smaller circle. What is the area (in cm²) of the annular portion between the two circles?

Correct — B. Let the outer radius be R, the inner radius r, and the shared centre O.

The chord touches the smaller circle at M, so OM is perpendicular to the chord and OM = r.

A perpendicular from the centre bisects the chord:
half-chord = 20 ⁄ 2 = 10 cm

In right triangle OMA: R² = r² + 10²
so R² − r² = 100

Annulus area = πR² − πr² = π(R² − r²) = 100π cm² → option (b)

Why the others are wrong

• (a)
  The option image reads 164π, which needs R² − r² = 164 and so a chord of 2√164 ≈ 25.6 cm. The paper gives 20 cm, fixing the difference at 100.
• (c)
  175π would follow from R² − r² = 175. Only the difference of the squares is knowable here, and the 20 cm tangent chord pins it to 10² = 100.
• (d)
  122π suggests hunting for R and r separately. You cannot — infinitely many pairs fit, and every one of them has R² − r² = 100, so the ring has the same area.

Two standard circle facts meet in one right triangle.

A perpendicular dropped from the centre to a chord bisects that chord, so the half-chord is 10 cm. And a tangent is perpendicular to the radius at the point of contact, so the perpendicular distance from O to this chord is exactly the inner radius r.

That makes r, 10 and R the two legs and the hypotenuse of a right triangle: R² = r² + 100.

The annulus area π(R² − r²) therefore depends only on the chord, never on the individual radii — which is why the question can withhold them.

Key facts

• A perpendicular from the centre to a chord bisects that chord.
• A tangent is perpendicular to the radius at the point of contact, so the inner radius is the perpendicular distance from the centre to this chord.
• The area of an annulus is π(R² − r²).
• For a tangent chord of length 2a, R² − r² = a², so a 20 cm chord gives 100π cm².

Common traps

• Using the full 20 cm as the leg and answering 400π.
• Trying to pin down R and r individually when only their difference of squares is determined.
• Substituting 22⁄7 for π when the options are left in terms of π.

No directly related past PYQ was found.