**Q****uestion**

A piece of wire of length 24cm is bent to form the perimeter of a sector of a circle of radius r cm.

**(i)Show that the area of the sector ,\(Acm^{2}\) is given by \(A=12r-r^{2}\)**

(ii)Express A in the form \(a-\left ( r-b \right )^{2}\),where a and b are constants.

**(iii)Given that r can vary,state the greatest value of A and find the corresponding angle of the sector.**

**▶️Answer/Explanation**

(i) Show that the area of the sector \(A\) is given by \(A = 12r – r^2\)

The perimeter of the sector is the length of the wire, which is equal to the sum of the arc length and the two radii.

\( \text{Perimeter} = \text{Arc Length} + 2r \)

Since the arc length is a fraction of the total circumference (which is \(2\pi r\)), we can express the arc length as \(\frac{\theta}{360} \times 2\pi r\), where \(\theta\) is the central angle of the sector.

So, we have the equation:

\( 24 = \frac{\theta}{360} \times 2\pi r + 2r \)

\( \theta = \frac{360}{\pi} (24 – 2r) \)

\( A = \frac{\theta}{360} \times \pi r^2 \)

\( A = \frac{360}{\pi} (24 – 2r) \times \frac{\pi}{360} \times \pi r^2 \)

\( A = (24 – 2r)r = 12r – r^2 \)

(ii) Express \(A\) in the form \(a – (r – b)^2\), where \(a\) and \(b\) are constants:

\( A = 12r – r^2 \)

\( A = -(r^2 – 12r + 36) + 36 \)

\( A = -(r – 6)^2 + 36 \)

So, \(A\) can be expressed in the form \(a – (r – b)^2\) with \(a = 36\) and \(b = 6\).

(iii) Given that \(r\) can vary, state the greatest value of \(A\) and find the corresponding angle of the sector:

The expression \(-(r – 6)^2 + 36\) indicates that \(A\) is maximized when \((r – 6)^2\) is minimized, which occurs when \(r = 6\). So, the greatest value of \(A\) is \(36\), and it occurs when \(r = 6\).

\( \theta = \frac{360}{\pi} (24 – 2r) \)

Substitute \(r = 6\):

\( \theta = \frac{360}{\pi} (24 – 2(6)) = \frac{360}{\pi} \times 12 \)

So, the corresponding angle \(\theta\) is \(120^\circ\).

Therefore, the greatest value of \(A\) is \(36\) square centimeters, and the corresponding angle of the sector is \(120^\circ\).

**Question**

The line with gradient -2 passing through the point P(3t,2t)intersects the x-axis at A and the y-axis at B.

**(i)Find the area of triangle AOB in terms of t.**

(ii)The line through P perpendicular to AB intersects the x-axis at C.

Show that the mid-point of PC lies on the line y=x.

**▶️Answer/Explanation**

(i) Find the area of triangle \(AOB\) in terms of \(t\):

Given the equation \(y – 2t = -2(x – 3t)\), we can simplify it to \(y + 2x = 8t\).

The x-intercept (\(A\)):

– Set \(y = 0\): \(0 + 2x = 8t\)

– Solve for \(x\): \(x = 4t\)

– So, the coordinates of \(A\) are \((4t, 0)\).

The y-intercept (\(B\)):

– Set \(x = 0\): \(y + 2(0) = 8t\)

– Solve for \(y\): \(y = 8t\)

– So, the coordinates of \(B\) are \((0, 8t)\).

Now, the area of triangle \(AOB\) is given by the formula:

\( \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \)

In this case, the base is the distance between points \(A\) and \(B\), which is \(4t\), and the height is the y-coordinate of \(B\), which is \(8t\).

\( \text{Area} = \frac{1}{2} \times 4t \times 8t = 16t^2 \)

So, the area of triangle \(AOB\) is \(16t^2\).

(ii) The line through \(P\) perpendicular to \(AB\) intersects the \(x\)-axis at \(C\).

The slope of the line AB is the negative reciprocal of the slope of the line passing through P:

\(m_{AB} \cdot m_{PC} = -1\)

\((-2) \cdot m_{PC} = -1\)

\(m_{PC} = \frac{1}{2}\)

Now, the line through P with slope \(\frac{1}{2}\) has the equation in point-slope form:

\(y – 2t = \frac{1}{2}(x – 3t)\)

\(y = \frac{1}{2}x + \frac{5}{2}t\)

When this line intersects the \(x\)-axis (\(y = 0\)), we can find point C:

\(0 = \frac{1}{2}x + \frac{5}{2}t\)

\(x = -5t\)

This gives us the coordinates of point C as \((-5t, 0)\).

Now, the mid-point of PC is the average of the x-coordinates and y-coordinates:

\(M\left(\frac{-5t + 3t}{2}, \frac{0 + 2t}{2}\right)\)

\(M(-t, t)\)

To show that this point lies on the line \(y = x\), we can check if \(y = x\) for these coordinates:

\(t = -t\)

This is true, so the mid-point of PC, \(M(-t, t)\), lies on the line \(y = x\).

**Question **

The diagram shows triangle ABC in which AB is [perpendicular to BC .The length of AB is 4cm and angle CAB is \(\alpha\) radians.The arc DE with cent6re A and radius 2cm meets AC at D and AB at E .Find in terms of \(\alpha \)

(i)the area of the shaded region.

(ii)the perimeter of the shaded region.

**Answer/Explanation**

(i)Area \(\Delta =\frac{1}{2}\times 4\times 4\tan \alpha \)

Area Sector\(=\frac{1}{2}\times 2^{2}\alpha \)

Shaded area \(=8\tan \alpha -2\alpha \)

(ii)\(DC=\frac{4}{\cos \alpha }-2\)

arc DE=\(2\alpha\)

Perimeter =\(\frac{4}{\cos \alpha }+4\tan \alpha +2\alpha\)

**Question**

In the diagram,OAB is a sector of a circle with centre O and radius 8cm.Angle BOA is \(\alpha\) radian OAC is a semicircle with diameter OA.The area of the semicircle OAC is twice the area of the sector OAB.

(i)Find \(\alpha\) in terms of \(\pi \)

(ii)Find the perimeter of the complete figure in terms of \(\pi \).

**Answer/Explanation**

(i)\((OAB)=\frac{1}{2}\times 8^{2}\alpha ,(OAC)=\frac{1}{2}\times \pi \times 4^{2}\)

\(\alpha =\frac{\pi }{8}\)

(ii)\(8+8\times their \alpha +\frac{1}{2}\times 8\times \pi \)

\(8+5\pi\)

**Question**

In the diagram, ABC is an equilateral triangle of side 2cm.The mid-point of BC is Q.An arc of a circle with centre A touches BC at Q ,and meets AB at P and AC at R.Find the total area of the shaded regions,giving your answer in terms of \(\pi \) and \(\sqrt{3}\).

**Answer/Explanation**

AQ(or r)\(=\sqrt{3}\)

Area \(\Delta =\sqrt{3}\) (or area \(\Delta AQC=\frac{\sqrt{3}}{2}\)

Area sector \(APR=\frac{1}{2}\left ( \sqrt{3} \right )^{2}\times \frac{\pi }{3}=\frac{\pi }{2}\)

Shaded region \(=\sqrt{3}-\frac{\pi }{2}\)

**Question**

In the diagram OCA and ODB are radii of a circle with centre O and radius 2r cm.Angle \(AOB=\alpha \) radians.CD and AB are arcs of circle with centre O and radii r cm and 2rcm respectively.The perimeter of the shaded region ABDC is 4.4r cm.

(i)Find the value of \(\alpha \).

(ii)It is given that the area of the shaded region is \(30cm^{2}\).Find the value of r.

**Answer/Explanation**

(i)\(2r\alpha +r\alpha +2r=4.4r\)

\(\alpha =0.8\)

(ii)\(\frac{1}{2}\left ( 2r \right )^{2}0.8-\frac{1}{2}\left ( r^{2} \right )0.8=30\)

\(\frac{3}{2}r^{2}\times 0.8=30\rightarrow r=5\)

**Question **

The diagram shows an isosceles triangle ABC in which AC=16cm and AB=BC=10cm.The circular arcs BE and BD have centers at A and C respectively ,where D and E lie on AC.

(i)Show that angle BAC=0.6435radians,correct to 4 decimal places.

(ii)Find the area of the shaded region.

**Answer/Explanation**

\((i)\cos A=\frac{8}{10}\rightarrow A=0.6435\)(i)\(\cos A=\frac{8}{10}\rightarrow A=0.6435\)

(ii)EITHER:

\(\Delta ABC=\frac{1}{2}\times 16\times 6\) or \(\frac{1}{2}\times 10\times 16\sin 0.6435=48\)

Area 1 sector \(\frac{1}{2}\times 10^{2}\times 0.6435\)

Shaded area=\(2\times \)their sector -their \(\Delta ABC\)

OR:\(\Delta BDE=12\),\(\Delta ABC=30\)

Sector =32.18

\(2\times segment+\Delta BDE\)

=16.4

**Question **

The diagram shows a sector OAB of a circle with centre O and radius r. Angle AOB is θ radians.

The point C on OA is such that BC is perpendicular to OA. The point D is on BC and the circular arc AD has centre C.

(i) Find AC in terms of r and θ. [1]

(ii) Find the perimeter of the shaded region ABD when \(\Theta =\frac{1}{3}\pi \) and r=4, giving your answer as an exact value.

**Answer/Explanation**

(i)\(AC=r-r\cos \Theta \)

(ii)\(arc AB=\frac{4\pi }{3}\)

arc AD\(=\frac{\pi }{2}\times \)their \(AC=\frac{\pi }{2}\times (4-4\cos \frac{\pi }{3})=\pi \)

\(BD=4\sin \frac{\pi }{3}-their \(AC=2\sqrt{3}-2\)\)

Perimeter=\(\frac{7\pi }{3}+2\sqrt{3}-2\)

**Question**

The diagram shows a metal plate made by fixing together two pieces, OABCD (shaded) and OAED (unshaded). The piece OABCD is a minor sector of a circle with centre O and radius 2r. The piece OAED is a major sector of a circle with centre O and radius r. Angle AOD is \(\alpha\) radians. Simplifying your answers where possible, find, in terms of \( \pi \) ,\(\alpha\) and r.**(i) the perimeter of the metal plate.****(ii) the area of the metal plate.****It is now given that the shaded and unshaded pieces are equal in area.**

(iii) Find \(\alpha\) in terms of \(\pi \)

**Answer/Explanation**

(i)\(r(2\pi -\alpha)+2r\alpha+2r\)

\(2\pi r+r\alpha +2r\)

(ii)\( \frac{1}{2}(2r)^{2}\alpha +\pi r^{2}-\frac{1}{2}r^{2}\alpha \)

\(\frac{3r^{2}\alpha }{2}+\pi r^{2}\)

(iii)\(\pi r^{2}-\frac{1}{2}r^{2}\alpha =2r^{2}\alpha \)

\(\alpha =\frac{2}{5}\pi\)

**Question**

In the diagram, AB is an arc of a circle with centre O and radius 4 cm. Angle AOB is radians. The point D on OB is such that AD is perpendicular to OB. The arc DC, with centre O, meets OA at C.

(i) Find an expression in terms of \(\alpha\) for the perimeter of the shaded region ABDC.

(ii) For the case where \(\alpha =\frac{1}{6}\pi \) find the area of the shaded region ABDC, giving your answer in the form k, where k is a constant to be determined.(

**Answer/Explanation**

i)Arc AB=\(4\alpha \)

Arc DC\(=(4\cos \alpha )\alpha \)

AC(orDB)\(=4-4\cos \alpha \)

Perimeter\(=4\alpha \cos \alpha +4\alpha +8-8\cos \alpha \)

(ii)OD=\(4\cos \frac{\pi }{6}=2\sqrt{3}\)

Shaded area \(=\left [ \frac{1}{2} \times 4^{2}\times \frac{\pi }{6}\right ]\left [ -\frac{1}{2} (2\sqrt{3})^{2}\times \frac{\pi }{6}\right ]\)

\(\frac{\pi }{3}\)

**Question**

Relative to an origin O, the position vector of A is 3i + 2j − k and the position vector of B is 7i − 3j + k.

(i) Show that angle OAB is a right angle. **(ii) Find the area of triangle OAB.**

**Answer/Explanation**

(i)AB or BA \(=\pm \left [ (7i-3j+k)-(3i+2j-k) \right ]=\pm \left ( 4i-5j+2k \right )\)

\((AO.AB)=\pm (12-10-2)\) (allow as column if total given )

=0 hence OAB=\(90^{\circ}\)

(ii)\(\left | OA \right |=\sqrt{9+4+1}=\sqrt{14}\)

\(\left | AB \right |=\sqrt{25+16+4}=\sqrt{45}\)

Area\( \Delta =\frac{1}{2}\sqrt{14}\sqrt{45}=12.5\)

**Question**

The function f is defined by \(f:x\rightarrow \frac{2}{3-2x}\) for \(x\epsilon R\) ,\(x\neq \frac{3}{2}\)

(i) Find an expression for \(f^{-1}\left ( x \right )\).

The function g is defined by \(g:x\rightarrow 4x+a\) for \(x\epsilon R\) ,where a is ac constant.

**(ii)Find the value of a for which gf(-1)=3**

**(iii) Find the possible values of a given that the equation\( f^{-1}\left ( x \right )=g^{-1}\left ( x \right )\) has two equal roots.**

**Answer/Explanation**

\(f:x\rightarrow \frac{2}{3-2x}\) \(g:x\rightarrow 4x+a\),

(i)\(y=\frac{2}{3-2x}\rightarrow y\left ( 3-2x \right )=2\rightarrow 3-2x=\frac{2}{y}\)

\(\rightarrow 2x=3-\frac{2}{y}\rightarrow f^{-1}\left ( x \right )=\frac{3}{2}-\frac{1}{x}\)

(ii)\(gf(-1)=3f(-1)=\frac{2}{5}\)

\(\frac{8}{5}+a=3\rightarrow a=\frac{7}{5}\)

(iii)\(g^{-1}\left ( x \right )=\frac{x-a}{4}=f^{-1}\left ( x \right )\)

\(\rightarrow x^{2}-x\left ( a+6 \right )+4(=0)\)

Solving \left ( a+6 \right )^{2}=16 or \(a^{2}+12a+20(=0)\)

\(\rightarrow a=-2\) or -10

**Question.**

In the diagram, AOB is a quarter circle with centre O and radius r. The point C lies on the arc AB and the point D lies on OB. The line CD is parallel to AO and angle \(AOC = \Theta \)radians.

(i) Express the perimeter of the shaded region in terms of r,\(\Theta \)and \(\Pi\).**(ii) For the case where r = 5 cm and \(\Theta = 0.6\), find the area of the shaded region.**

**Answer/Explanation**

(i) \(CD=r\cos \Theta ,BD=r-r\sin \Theta\)oe

\(Arc CB=r\left ( \frac{1}{2}\pi -\Theta \right )oe\)

\(\rightarrow P=r\cos \Theta +r-r\sin \Theta +r\left ( \frac{1}{2}\pi -\Theta \right )oe\)

(ii) Sector \(=\frac{1}{2}.5^{2}.\left ( \frac{1}{2} \pi -0.6\right )\)

Triangle\(=\frac{1}{2}.5\cos 0.6.5\sin 0.6\)

→Area=6.31

(or \(\frac{1}{4}\)circle-traingle-sector)

**Question.**

**(a)**

In Fig. 1, OAB is a sector of a circle with centre O and radius r. AX is the tangent at A to the arc AB and angle BAX = α.**(i)** Show that angle AOB = 2α.**(ii) Find the area of the shaded segment in terms of r and α.**

**(b) **

**In Fig. 2, ABC is an equilateral triangle of side 4 cm. The lines AX, BX and CX are tangents to the equal circular arcs AB, BC and CA. Use the results in part (a) to find the area of the shaded region, giving your answer in terms of л and √3.**

**Answer/Explanation**

Ans:**(i) **\(BAO=OBA=\frac{\Pi}{2}-\alpha \)

\(AOB=\Pi-(\frac{\Pi}{2}-\alpha)-(\frac{\Pi}{2}-\alpha)=2\alpha\)

**(ii)**\(\frac{1}{2}r^{2}(2\alpha )-\frac{1}{2}r^{2}sin(2\alpha )\)

**(b) **

**Question**

The diagram shows a three-dimensional shape in which the base OABC and the upper surface DEFG

are identical horizontal squares. The parallelograms OAED and CBFG both lie in vertical planes.

The point M is the mid-point of AF. Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. The position vectors of A and D are given by \(\vec{OA}=8i\) and \(\vec{OD}=3i+10k\)

**(i) Express each of the vectors \(\vec{AM} , \vec{GM}\) in the terms of i, j, k**

**(ii)** Use a scalar product to find angle GMA correct to the nearest degree.

**Answer/Explanation**

(i) \(\vec{AM} = 1.5i + 4j + 5k\)

\(\vec{GM}= 6.5i – 4j − 5

k\)

(b)

**Question **

**(a) The third and fourth terms of a geometric progression are 48 and 32 respectively. Find the sum****to infinity of the progression.**

(b) Two schemes are proposed for increasing the amount of household waste that is recycled each

week.

Scheme A is to increase the amount of waste recycled each month by 0.16 tonnes.

Scheme B is to increase the amount of waste recycled each month by 6% of the amount recycled in the previous month.

The proposal is to operate the scheme for a period of 24 months. The amount recycled in the

first month is 2.5 tonnes.For each scheme, find the total amount of waste that would be recycled over the 24- month period.

**Answer/Explanation**

**Question**

In the diagram, CXD is a semicircle of radius 7 cm with centre A and diameter CD. The straight line

YABX is perpendicular to CD, and the arc CYD is part of a circle with centre B and radius 8 cm. Find

the total area of the region enclosed by the two arcs.

**Answer/Explanation**

\(\angle CBA=sin^{-1}(\frac{7}{8})=1.0654\) OR

\(\angle CBD =cos^{-1} (\frac{-17}{32})=2.13\)

Sector BCYD \( =0.5\times 8^{2}\times 1.0654\)

or sector CBY\( =0.5\times 8^{2}\times 1.0654\)

\(\Delta BCD=7\times \sqrt{8^{2}-7^{2}} \)

Semi-circle CXD \(\frac{\Pi }{2}\times 7^{2}=76.9(77)\)

Total area = their68.19

‒ their27.11 + their76.97 = 118.0–118.1

### Question

The diagram shows a sector *CAB* which is part of a circle with centre *C*. A circle with centre *O* and

radius *r* lies within the sector and touches it at *D*, *E* and *F*, where *COD* is a straight line and angle

*ACD* is 1 radians.

(a) Find *CD* in terms of *r* and sin 1. [3]

It is now given that r = 4 and \(\theta =\frac{1}{6}\pi \)

(b) Find the perimeter of sector *CAB* in terms of π. [3]

(c) Find the area of the shaded region in terms of π and \(\sqrt{3}\) [4]

**Answer/Explanation**

Ans

10 (a) \(\left ( \sin \theta =\frac{r}{OC}\rightarrow \right )OC=\frac{r}{\sin \theta }\)

\(CD=r+\frac{r}{\sin \theta }\)

10 (b) Radius of arc \(AB=4+\frac{4}{\sin \frac{\pi }{6}}=4+8=12\)

\((Arc AB=)their \ 12\times \frac{2\pi }{6}\ or \ \left ( \frac{1}{2}AB= \right )\left ( their \ 12\times \frac{\pi }{6} \right )\)

Perimeter = 24+ 4π

10 (c) \(Area \ FOC=\frac{1}{2}\times 4\times their\ OC\times \sin \frac{\pi }{3}\)

\(8\sqrt{3}\)

Area sector FOE \(\frac{1}{2}\times \frac{2\pi }{3}\times 4^{2}=\frac{16\pi }{3}\)

\(Shaded\ area=16\sqrt{3}-\frac{16\pi }{3}\)

**Alternative method for question 10(c)**

\(FC=\sqrt{(their\ OC)^{2}-4^{2}}

\(Area\ FOC=\frac{1}{2}\times 4\times 4\sqrt{3}=8\sqrt{3}\)

Area of half sector \(FOE=\frac{1}{2}\times \frac{\pi }{3}\times 4^{2}=\frac{8\pi }{3}\)

\(Shaded \ area=16\sqrt{3}-\frac{16\pi }{3}\)

*Question*.

The diagram shows a metal plate ABC in which the sides are the straight line AB and the arcs AC and BC. The line AB has length 6 cm. The arc AC is part of a circle with centre B and radius 6 cm, and the arc BC is part of a circle with centre A and radius 6 cm.

**(a) **Find the perimeter of the plate, giving your answer in terms of π.

**(b) **Find the area of the plate, giving your answer in terms of π and \(\sqrt{3}\).

**Answer/Explanation**

**(a) ** Recognise that at least one of angles A, B, C is \(\frac{\Pi }{3}\)

One arc 6 × their \(\frac{\pi }{3}\) leading to two arcs 2× 6× their \(\frac{\Pi }{3}\)

\(Perimeter = 6 +4\Pi\)

**(b) ** \(Sector =\frac{1}{6}\times 6^{2}\times their \left ( \frac{\pi }{3} \right )\)

\(\frac{1}{2}\times \left ( 6^{2} \right )\times their \left ( \frac{\pi }{3} \right ) – \frac{1}{2}\times \left ( 6^{2} \right )\times sin \left ( their \left ( \frac{\pi }{3} \right ) \right ) + 6\pi \left [ =6\pi -9\sqrt{3} + 6\pi \right ]\)

\(Area = 12\pi – 9\sqrt{3}\)

**Question**

**Question**

The diagram shows a sector AOB which is part of a circle with centre O and radius 6cm and with angle AOB = 0.8 radius. The point C on OB is such that AC is perpendicular to OB. The are CD is part of circle with centre O, where D lies on OA.

Find the area of the shaded region.

**Answer/Explanation**

**Ans:**

OC = 6cos0.8 = 4.18(0)

Area sector OCD = \(\frac{1}{2}(their4.18)^2 \times 0.8\)

Required are = their ΔOCA – their sector OCD

2.01

**Question**

**Question**

The diagram shows a sector ABC which is part of a circle of radius a. The points D and E lie on AB and AC respectively and are such that AD = AE = ka, where k<1. The line DE divides the sector into two regions which are equal in area.

(a) For the case where angle \(BAC=\frac{1}{6}\pi\) radians, find k correct to 4 significant figures.

(b) For the general case in which angle \(BAC=\theta\) radians, where \(0<\theta<\frac{1}{2}\pi\), it is given that \(\frac{\theta}{sin\theta}>1\).

Find the set of possible values of k.

**Answer/Explanation**

**Ans:**

(a) \(ΔADE=\frac{1}{2}(ka)^2 sin\frac{\pi}{6}\)

\(\frac{1}{4}k^2a^2\)

Sector \(ABC=\frac{1}{2}a^2\frac{\pi}{6}\)

\(2\times \frac{1}{4}k^2a^2=\frac{1}{2}a^2\frac{\pi}{6}\)

\(k=(\sqrt{\frac{\pi}{6}})=0.7236\)

(b) \(2\times\frac{1}{2}sin \theta = \frac{1}{2} a^2 \theta\)

\(k^2=\frac{\theta}{2sin\theta}\)

\(k^2>\frac{1}{2}\) leading to \(\frac{1}{\sqrt{2}}<k<1\)

### Question.

In the diagram, ABC is a semicircle with diameter AC, centre O and radius 6 cm. The length of the arc AB is 15 cm. The point X lies on AC and BX is perpendicular to AX.

Find the perimeter of the shaded region BXC.

**Answer/Explanation**

Angle AOB = 15 ÷ 6 = 2.5 radians

Angle BOC = π – 2.5 (FT on angle AOB)

BC = 6(π – 2.5) (BC = 3.850)

sin (π – 2.5) = BX ÷6 (BX = 3.59)

Either OX = 6cos (π – 2.5) or Pythagoras (OX = 4.807)

XC = 6 – OX (XC = 1.193)→P = 8.63