A circle's centre is connected to its 50 cm long chord by a perpendicular that is 21 cm long. Find the circle's radius.

Correct — D. A perpendicular dropped from the centre onto a chord bisects that chord, so the 21 cm segment meets the 50 cm chord at its midpoint.

Half-chord = 50 ⁄ 2 = 25 cm
Perpendicular from centre = 21 cm

Radius, half-chord and perpendicular form a right triangle, with the radius as hypotenuse:
r² = 25² + 21²
= 625 + 441 = 1066
r = √1066 cm → option (d)

Why the others are wrong

• (a)
  √1068 overshoots by 2. The only sum on offer is 25² + 21² = 625 + 441 = 1066, so this is an addition slip, not a method error — recheck the two squares before the root.
• (b)
  √1065 is one short of the true total. 25² = 625 exactly, so a 624 anywhere in the working is the slip; 625 + 441 comes to 1066.
• (c)
  √1064 is 1066 − 2, the value you get by taking 21² as 439 rather than 441. All four options sit within 4 of each other, so the addition is the entire question.

One relation covers most chord questions. Drop a perpendicular of length d from the centre to a chord of length c, and it splits the chord into two halves of c⁄2.

That perpendicular, the half-chord and the radius form a right triangle whose hypotenuse is the radius, so r² = d² + (c⁄2)².

The relation runs in either direction: given r and d you get the chord, c = 2√(r² − d²); given c and d you get the radius, as here.

Key facts

• A perpendicular from the centre of a circle to a chord bisects that chord.
• For a chord of length c at perpendicular distance d from the centre, r² = d² + (c⁄2)².
• Here 25² = 625 and 21² = 441, so r = √1066 cm, roughly 32.65 cm.
• Run backwards, the same relation gives the chord: c = 2√(r² − d²).

Common traps

• Using the full 50 cm chord as a leg, which gives √2941 instead of √1066.
• Subtracting the squares — the radius is the hypotenuse here, so the squares add.
• Rushing 625 + 441 when three options sit within 2 of the right total.

No directly related past PYQ was found.