IBDP Maths analysis and approaches Topic: SL 3.4 :Length of an arc; area of a sector HL Paper 1

Question

A circular disc is cut into twelve sectors whose areas are in an arithmetic sequence.

The angle of the largest sector is twice the angle of the smallest sector.

Find the size of the angle of the smallest sector.

Answer/Explanation

Markscheme

METHOD 1

If the areas are in arithmetic sequence, then so are the angles.     (M1)

\( \Rightarrow {S_n} = \frac{n}{2}(a + l) \Rightarrow \frac{{12}}{2}(\theta + 2\theta ) = 18\theta \)     M1A1

\( \Rightarrow 18\theta = 2\pi \)     (A1)

\(\theta = \frac{\pi }{9}\)     (accept \(20^\circ \))     A1

[5 marks] 

METHOD 2

\({{\text{a}}_{12}} = 2{a_1}\)     (M1)

\(\frac{{12}}{2}({a_1} + 2{a_1}) = \pi {r^2}\)     M1A1

\(3{a_1} = \frac{{\pi {r^2}}}{6}\)

\(\frac{3}{2}{r^2}\theta = \frac{{\pi {r^2}}}{6}\)     (A1)

\(\theta = \frac{{2\pi }}{{18}} = \frac{\pi }{9}\)     (accept \(20^\circ \))     A1

[5 marks] 

METHOD 3

Let smallest angle = a , common difference = d

\(a + 11d = 2a\)     (M1)

\(a = 11d\)     A1

\({S_n} = \frac{{12}}{2}(2a + 11d) = 2\pi \)     M1

\(6(2a + a) = 2\pi \)     (A1)

\(18a = 2\pi \)

\(a = \frac{\pi }{9}\)     (accept \(20^\circ \))     A1

[5 marks]

Question

The diagram below shows two straight lines intersecting at O and two circles, each with centre O. The outer circle has radius R and the inner circle has radius r .

 

Consider the shaded regions with areas A and B . Given that \(A:B = 2:1\), find the exact value of the ratio \(R:r\) .

Answer/Explanation

Markscheme

\(A = \frac{\theta }{2}({R^2} – {r^2})\)     A1

\(B = \frac{\theta }{2}{r^2}\)     A1

from \(A:B = 2:1\) , we have \({R^2} – {r^2} = 2{r^2}\)     M1

\(R = \sqrt 3 r\)     (A1)

hence exact value of the ratio \(R:r{\text{ is }}\sqrt 3 :1\)     A1     N0

[5 marks]

Question

The logo, for a company that makes chocolate, is a sector of a circle of radius \(2\) cm, shown as shaded in the diagram. The area of the logo is \(3\pi {\text{ c}}{{\text{m}}^2}\).

Find, in radians, the value of the angle \(\theta \), as indicated on the diagram.[3]

a.

Find the total length of the perimeter of the logo.[2]

b.
Answer/Explanation

Markscheme

METHOD 1

\({\text{area}} = \pi {2^2} – \frac{1}{2}{2^2}\theta \;\;\;( = 3\pi )\)     M1A1

Note:     Award M1 for using area formula.

\( \Rightarrow 2\theta  = \pi  \Rightarrow \theta  = \frac{\pi }{2}\)     A1

Note:     Degrees loses final A1

METHOD 2

let \(x = 2\pi  – \theta \)

\({\text{area}} = \frac{1}{2}{2^2}x\;\;\;( = 3\pi )\)     M1

\( \Rightarrow x = \frac{3}{2}\pi \)     A1

\( \Rightarrow \theta  = \frac{\pi }{2}\)     A1

METHOD 3

Area of circle is \(4\pi \)     A1

Shaded area is \(\frac{3}{4}\) of the circle     (R1)

\( \Rightarrow \theta  = \frac{\pi }{2}\)     A1

[3 marks]

a.

\({\text{arc length}} = 2\frac{{3\pi }}{2}\)     A1

\({\text{perimeter}} = 2\frac{{3\pi }}{2} + 2 \times 2\)

\( = 3\pi  + 4\)     A1

[2 marks]

Total [5 marks]

b.

Question

The following diagram shows a sector of a circle where \({\rm{A\hat OB}} = x\) radians and the length of the \({\text{arc AB}} = \frac{2}{x}{\text{ cm}}\).

Given that the area of the sector is \(16{\text{ c}}{{\text{m}}^2}\), find the length of the arc \(AB\).

Answer/Explanation

Markscheme

\({\text{arc length}} = \frac{2}{x} = rx\;\;\;\left( { \Rightarrow r = \frac{2}{{{x^2}}}} \right)\)     M1

\(16 = \frac{1}{2}{\left( {\frac{2}{{{x^2}}}} \right)^2}x\;\;\;\left( { \Rightarrow \frac{2}{{{x^3}}} = 16} \right)\)     M1

Note:     Award M1s for attempts at the use of arc-length and sector-area formulae.

\(x = \frac{1}{2}\)     A1

\({\text{arc length}} = {\text{4 (cm)}}\)     A1

[4 marks]

Scroll to Top