GCSE OCR Maths Circle theorems: OAB is a sector of a circle, cen

OAB is a sector of a circle, centre O - OCR - GCSE Maths - Question 16 - 2019 - Paper 5

Question 16

OAB is a sector of a circle, centre O. OA = 6 cm and AX is perpendicular to OB. The area of sector OAB is 6π cm². Show that AX = 3√3 cm.

Worked Solution & Example Answer:OAB is a sector of a circle, centre O - OCR - GCSE Maths - Question 16 - 2019 - Paper 5

Step 1

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Answer

The area of a sector of a circle can be calculated using the formula:

ext{Area} = rac{ heta}{360°} imes ext{π} r^2

where ( r ) is the radius and ( \theta ) is the angle in degrees. Given that OA = 6 cm, we have:

ext{Area} = rac{ heta}{360°} imes ext{π} (6)^2 = rac{ heta}{360°} imes 36π

Setting this equal to the provided area of 6π cm²:

rac{ heta}{360°} imes 36π = 6π

Dividing both sides by π gives:

rac{ heta}{360°} imes 36 = 6

Thus:

heta = rac{6 imes 360°}{36} = 60°.

Hence, the angle OAB (or θ) is 60°.

Step 2

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Answer

In triangle OAX, we know:

Using the fact that the sum of angles in a triangle is 180°, we can find angle AXO:

ext{Angle AXO} = 180° - 90° - 60° = 30°.

In triangle OAX, we can use the sine function since we know an angle and the length of the opposite side to AX (which is OA):

ext{sin}(30°) = \frac{AX}{OA} \

Substituting the known values:

ext{sin}(30°) = \frac{AX}{6} \

Since ( ext{sin}(30°) = \frac{1}{2} ), we get:

\frac{1}{2} = \frac{AX}{6} \

Multiplying both sides by 6 gives:

AX = 6 \times \frac{1}{2} = 3 cm.

To find AX in terms of √3, consider the cosine function:

ext{cos}(30°) = \frac{AX}{OA} \

This gives:

\frac{\sqrt{3}}{2} = \frac{AX}{6} \

Thus:

AX = 6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3} cm.

Therefore, we have shown that AX = 3√3 cm.