Angle AOC is made by drawing lines from the points A and C on the circle which meet at the centre (O).

The circle theorem states that the angle at the centre is double the angle at the circumference.
Therefore angle AOC is double angle ABC.

ed-igcse-arcs-and-sectors-vh-q5-diagram-1

$Angle A O C = 2 \times 75 ° = 150 °$

[1]

Form an expression for the area of the sector AOC using the formula for area of a sector $= \frac{θ}{360 °} \times π r^{2}$ and substituting the values into the formula, letting $θ = 150$ and $r = r$ .

$Area_{sector} = \frac{150}{360} \times π \times r^{2} = \frac{5}{12} π r^{2}$

[1]

Form an expression for the area of the triangle AOC using the formula for area of a triangle $= \frac{1}{2} a b \sin θ$ and substituting the values into the formula, letting $θ = 150$ and $r = r$ .

$Area_{triangle} = \frac{1}{2} \times r \times r \times \sin 150 = \frac{1}{4} r^{2}$

[1]

Form an expression for the area of the shaded region by subtracting the area of the triangle from the area of sector AOC.

$Area_{shaded} = \frac{5}{12} π r^{2} - \frac{1}{4} r^{2}$

Form an equation by setting the expression for the area of the shaded region equal to 200 cm².

$\frac{5}{12} π r^{2} - \frac{1}{4} r^{2} = 200$

[1]

Factorise out r².

$r^{2} (\frac{5}{12} π - \frac{1}{4}) = 200$

Isolate r².

$r^{2} = \frac{200}{(\frac{5}{12} π - \frac{1}{4})}$

Solve to find r by taking the square root of both sides.

$r = \sqrt{\frac{200}{(\frac{5}{12} π - \frac{1}{4})}} = 13.74256 . . .$

Round to 3 significant figures.

radius = 13.7 cm(3 s.f.) [1]
Answers in the range 13.7 to 13.8 accepted

$□$	$A B$ is parallel to $D C$
$□$	$A C$ bisects angle $B C D$
$□$	$A C$ bisects angle $B A D$

Circle Theorems (Edexcel IGCSE Maths)

Exam Questions