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

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.

                                                                                      

In the diagram, AB = AC = 8 cm and angle CAB \(=\frac{2\Pi }{7}\)radians. The circular arc BC has centre A, the

circular arc CD has centre B and ABD is a straight line.
(i) Show that angle \(CBD = \frac{9\Pi }{14}\)radians.

(ii) Find the perimeter of the shaded region.

Answer/Explanation

4(i)\(ABC=\frac{\pi }{2}-\frac{\pi }{7}=\frac{5\pi }{14}\)  \( CBD=\pi -\frac{5\pi }{14}=\frac{9\pi }{14}\)

(ii) \(\sin \frac{\pi }{7}=\frac{\frac{1}{2}BC}{8}\) or \(\frac{BC}{\sin \frac{2\pi }{7}}=\frac{8}{\sin \frac{5\pi }{14}}\) or \(BC^{2}=8^{2}+8^{2}-2\left ( 8 \right )\left ( 8 \right )\cos \frac{2\pi }{7}\)

BC=6.94(2)

arc CD = their \(6.94\times \frac{9\pi }{14}\)

arc CB=\(8\times \frac{2\pi }{7}\)

perimeter=6.94+14.02+7.18=28.1

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 symmetrical metal plate. The plate is made by removing two identical pieces
     from a circular disc with centre C. The boundary of the plate consists of two arcs PS and QR of the
     original circle and two semicircles with PQ and RS as diameters. The radius of the circle with centre
     C is 4 cm, and PQ = RS = 4 cm also.
    (a) Show that angle PCS   \(=\frac{2}{3}\pi \)  radians                                                                                    [2]

    (b) Find the exact perimeter of the plate.                                                                                                              [3] 

    (c) Show that the area of the plate is \((\frac{20}{3}\pi +8\sqrt{3})cm^{2}\)

Answer/Explanation

Ans

8 (a) Either Let midpoint of PQ be H: sin HCP \(\frac{2}{4}\Rightarrow Angle \ HCP \ =\frac{\pi }{6}\)

          Or sin PSQ \(=\frac{4}{8}\Rightarrow Angle \ PSQ \ =\frac{\pi }{6}\)

          Or using cosine rule: angle PCQ \(=\frac{\pi }{3}\)

          Or by inspection: triangle PCQ or PCT is equilateral so angle PCQ \(=\frac{\pi }{3}\)

          Angle PCS \(=\pi -\frac{\pi }{6}-\frac{\pi }{6}=\frac{2}{3}\pi \)

8 (b)  \(Perimeter=2\times 4\times \frac{2\pi }{3}\ or \ 8\pi -\frac{8\pi }{3}\)

           \( +2π × 2\)

             \(\frac{28\pi }{3}\)

8(c) Area sector CPQ \(=\frac{1}{2}\times 4^{2}\times \frac{\pi }{3}=\frac{8\pi }{3}\)

        Area of segment of large circle beyond CPQ 

       \(\frac{8\pi }{3}-\frac{1}{2}\times 4^{2}\times \sin \left ( \frac{\pi }{3} \right )=\frac{8\pi }{3}-4\sqrt{3}\)

        Area of small semicircle =π× 2        or area of small circle = 2 π×2

        Area of plate = Large circle – [2 ×] small semicircle – [2 ×] segment area 

        \(\pi \times 4^{2}-\pi \times 2^{2}-2x\left ( \frac{8\pi }{3}-4\sqrt{3}\right )=\frac{20\pi }{3}+8\sqrt{3}\)

        Alternative method for Question 8(c)

        Alternative method for Question 8(c)  =\(\frac{1}{2}\times 4^{2}\times \frac{2\pi }{3}=\frac{16\pi }{3}\)

        Area of triangle PCQ \(=\frac{1}{2}\times 4^{2}\times \sin \frac{\pi }{3}=4\sqrt{3}\)

      Area of small semicircle =π× 2            or area of circle =  π×22

      Area of plate = [2 ×] large sector + [2 ×] triangle – [2 ×] small semicircle

      \(2\left ( \frac{16\pi }{3} \right )+2(4\sqrt{3})-\pi \times 2^{2}=\frac{20\pi }{3}+8\sqrt{3}\)