Circle with diameter AB. Through centre O, OC ⟂ AB. If chord AC = 7√2 cm, find the area of the circle (in cm²).

Introduction / Context:
Right triangles inside circles often arise when a chord is perpendicular to a diameter at the centre. This creates a right isosceles triangle using the two radii as equal legs and the given chord as the hypotenuse.

Given Data / Assumptions:

• AB is a diameter; O is its midpoint (centre).
• OC ⟂ AB at O; hence triangle AOC is right-angled at O.
• AC, a chord, equals 7√2 cm.

Concept / Approach:
In right triangle AOC, OA = OC = r (radii). Therefore AC is the hypotenuse with AC = r√2. From this, compute r, then area = πr².

Step-by-Step Solution:
AC = r√2 = 7√2 ⇒ r = 7 cm.Area = πr² = π * 49.Using π = 22/7, area = (22/7) * 49 = 154 cm².

Verification / Alternative check:
Using π ≈ 3.14, 3.14 * 49 ≈ 153.86, which is closest to 154 among the options.

Why Other Options Are Wrong:
24.5 and 49 correspond to r² or r values; 98 misses the factor of π.

Common Pitfalls:
Mistaking AC as a diameter; here AB is the diameter and AC is a chord right-angled at O.

Final Answer:
154