A, B, C and D are points on the circumference of a circle, centre O - OCR - GCSE Maths - Question 8 - 2018 - Paper 1

Using the sine rule in triangle BCD:

$\frac{CD}{\sin(\text{angle CBD})} = \frac{BD}{\sin(\text{angle BCD})}$

Substituting the known values:

$\frac{6.4}{\sin(62°)} = \frac{BD}{\sin(90°)}$

Thus,

$BD = 6.4 \times \sin(90°) / \sin(62°)$

Calculating gives

$BD = 6.4 / \sin(62°) \approx 7.75 cm$

Since BD is the diameter, the radius (r) can be determined as follows:

$r = \frac{BD}{2} = \frac{7.75}{2} \approx 3.875 cm$

The area (A) of the circle is given by the formula:

$A = \pi r^2$

Substituting the radius we found:

$A = \pi (3.875)^2 \approx \pi (15.015625) \approx 47.12 cm^2$