Circles & Sectors: Eduqas GCSE Higher November 2021 Paper 1 Q12

Higher November 2021 Paper 1 Q12

EduqasHigherCurrent spec9 marksCircles & Sectors Standard Form Volume & Surface Area

12.

(a)

Sector OAB of a circle with centre O, radius r cm and angle x° at O

Diagram not drawn to scale

The diagram shows a sector of the circle with centre O and radius \(r\) cm.
The length of the arc, AB, is \(\dfrac{1}{6} \times \pi \times r\).

Work out the value of \(x\). [3]

(b) A cone has a radius of 6 cm and a slant height of 52 cm.

Show that the curved surface area of the cone must be a multiple of \(13\pi\). [2]

(c) Jupiter is a planet.

The radius of Jupiter is \(7 \times 10^4\) km.
You may assume the radius of Jupiter is constant.

Work out the surface area of Jupiter.
Give your answer in the form \(k\pi\), where \(k\) is in standard form. [4]

………………………… km²

Mark scheme (a)

Scheme	Marks	Comments
\(\dfrac{x}{360} \times 2(\times \pi \times r) = \dfrac{1}{6}(\times \pi \times r)\) or \(\dfrac{\frac{1}{6} \times \pi \times r}{2 \times \pi \times r}\) oe	M1
\(x = \dfrac{1}{6} \times 360 \div 2\) or \(\dfrac{1}{12} \times 360\) oe	M1
\(x = 30\)	A1	If no marks award SC1 for a final answer of 60 obtained from use of \(\dfrac{x}{360}(\times \pi \times r) = \dfrac{1}{6}(\times \pi \times r)\)

Mark scheme (b)

Scheme	Marks	Comments
\(24 \times 13\pi\) or \(\pi \times 6 \times 4 \times 13\) oe	B2	Must involve \(\pi\) for B2 or B1; allow B2 for e.g. \(13(24\pi)\) or for \(312\pi \div 13\pi = 24\) B1 for (surface area cone =) \(\pi \times 6 \times 52\) or \(312\pi\)

Mark scheme (c)

Scheme	Marks	Comments
\(4\pi(7 \times 10^4)^2\) si	B1	Missing brackets may be recovered in further correct work
\(4\pi(49 \times 10^8)\) or \(4\pi(4\,900\,000\,000)\) oe	M2	M1 for \(4\pi(49 \times 10^n),\ n \ne 8\) or for \(4\pi(m \times 10^8),\ m \ne 49\)
\((1.96 \times 10^{10})\pi\)	A1	Allow \(1.96\pi \times 10^{10}\) If no marks, award SC1 for \(k\pi(49 \times 10^8),\ k \ne 4\)
	(9)