Question
Points A and B lie on the circumference of a circle of radius rcm with centre at O .
The sector OAB is shown on the following diagram. The angle AÔB is denoted as \(\theta\) and is measured in radians.
The perimeter of the sector is 10 cm and the area of the sector is 6.25 cm2 .
(a) Show that \(4r^{2}-20r+25=0\) .
(b) Hence, or otherwise, find the value of r and the value of \(\theta\) .
▶️Answer/Explanation
Ans:
(a) \(2r+r\theta =10\)
\(\frac{1}{2}r^{2}\theta =6.25\)
attempt to eliminate \(\theta\) to obtain an equation in r
correct intermediate equation in r
\(10-2r=\frac{25}{2r}\) OR \(\frac{10}{r}-2=\frac{25}{2r^{2}}\) OR \(\frac{1}{2}r^{2}(\frac{10}{r}-2)=6.25\) OR \(12.5+2r^{2}=10r\)
\(4r^{2}-20r+25=0\)
(b) attempt to solve quadratic by factorizing or use of formula or completing the square
\((2r-5)^{2}=0\) OR \(r=\frac{20\pm \sqrt{(-20)^{2}-4(4)(25)}}{2(4)}\left ( =\frac{20\pm \sqrt{400-400}}{8} \right )\)
\(r=\frac{5}{2}\)
attempt to substitute their value or r into their perimeter or area equation
\(\theta =\frac{10-2\left ( \frac{5}{2} \right )}{\left ( \frac{5}{2} \right )}\) OR \(\theta =\frac{25}{2\left ( \frac{5}{2} \right )^{2}}\)
\(\theta=2\)
Detailed Solution
(a) we know that the length of arc, l = \(r\times \theta \)
and perimeter of sector = r + r + \(r\times \theta \) = 10 cm
therefore \(\theta = \frac{10-2r}{r} \) ………………. eqn 1
Area of sector = \(\frac{\theta }{2\Pi } \Pi r^{2}\) = 6.25 cm2 = \(\frac{25}{4} \) ………………. eqn 2
= \(\frac{\theta }{2\ } \ r^{2}\)
substituting equation 1 in equation 2 we get
= \(\frac{10-2r}{2r}\times r^{2} = \frac{10r-2r^{2}}{2} = \frac{25}{4}\)
= \(2\times (10r-2r^{2}) = 25\)
Therefore \(4r^{2}-20r + 25 = 0 \) , hence proved
(b) Substituting the value of a =4, b = -20, c = 25 in the quadratic formula \(r = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)
we get, \( r = \frac{-\left ( -20 \right )\pm \sqrt{\left ( -20 \right )^{2}-4\times 4\times 25}}{2\times 4} \)
Therefore the value of \( r = \frac{5}{2}\)
Now, putting the value of r in equation 1 we get
\(\theta = \frac{10-2\times \frac{5}{2}}{\frac{5}{2}} \)
\(\theta=2\)
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 shows a tangent, (TP) , to the circle with centre O and radius r . The size of \({\rm{P\hat OA}}\) is \(\theta \) radians.
Find the area of triangle AOP in terms of r and \(\theta \) .[1]
Find the area of triangle POT in terms of r and \(\theta \) .[2]
Using your results from part (a) and part (b), show that \(\sin \theta < \theta < \tan \theta \) .[2]
▶️Answer/Explanation
Markscheme
area of \({\text{AOP}} = \frac{1}{2}{r^2}\sin \theta \) A1
[1 mark]
\({\text{TP}} = r\tan \theta \) (M1)
area of POT \( = \frac{1}{2}r(r\tan \theta )\)
\( = \frac{1}{2}{r^2}\tan \theta \) A1
[2 marks]
area of sector OAP \( = \frac{1}{2}{r^2}\theta \) A1
area of triangle OAP < area of sector OAP < area of triangle POT R1
\(\frac{1}{2}{r^2}\sin \theta < \frac{1}{2}{r^2}\theta < \frac{1}{2}{r^2}\tan \theta \)
\(\sin \theta < \theta < \tan \theta \) AG
[2 marks]
Question
From a vertex of an equilateral triangle of side \(2x\), a circular arc is drawn to divide the triangle into two regions, as shown in the diagram below.
Given that the areas of the two regions are equal, find the radius of the arc in terms of x.
▶️Answer/Explanation
Markscheme
area of triangle \( = \frac{1}{2}{(2x)^2}\sin \frac{\pi }{3}\) (M1)
\( = {x^2}\sqrt 3 \) A1
Note: A \(0.5 \times {\text{base}} \times {\text{height}}\) calculation is acceptable.
area of sector \({\text{ = }}\frac{\theta }{2}{r^2} = \frac{\pi }{6}{r^2}\) (M1)A1
area of triangle is twice the area of the sector
\( \Rightarrow 2\left( {\frac{\pi }{6}{r^2}} \right) = {x^2}\sqrt 3 \) M1
\( \Rightarrow r = x\sqrt {\frac{{3\sqrt 3 }}{\pi }} \,\,\,\,\,\)or equivalent A1
[6 marks]