**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) Perimeter of the Metal Plate:

The perimeter of the metal plate consists of the lengths of the curved edges and the straight edge connecting the endpoints.

For the minor sector OABCD, the curved edge has an arc length equal to \(2r\) times the angle \(\alpha\). This is because the entire circumference of a circle with radius \(2r\) is \(2\pi \times 2r = 4\pi r\), and the angle \(\alpha\) is a fraction of the full circle. So, the arc length of the minor sector is \(2r \alpha\).

For the major sector OAED, the curved edge has an arc length equal to \(r\) times the angle \(\pi – \alpha\). Similarly, this is because the entire circumference of a circle with radius \(r\) is \(2\pi r\), and the angle \(\pi – \alpha\) is a fraction of the full circle. So, the arc length of the major sector is \(r (\pi – \alpha)\).

Finally, the straight edge connecting the endpoints has a length of \(2r\).

The total perimeter is the sum of these lengths:

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

(ii)Area of the Metal Plate:

The area of the metal plate consists of the sum of the areas of the shaded and unshaded regions.

For the minor sector OABCD, the area is given by the formula for the area of a sector: \(\frac{1}{2}r^2\alpha\).

Since the radius is \(2r\), this becomes \(\frac{1}{2}(2r)^2 \alpha = 2r^2 \alpha\).

For the major sector OAED, the area is similarly given by \(\frac{1}{2}r^2 (\pi – \alpha)\).

The total area is the sum of these areas:

\(A = 2r^2 \alpha + \frac{1}{2}r^2 (\pi – \alpha)\)

(iii) Finding \(\alpha\) in terms of \(\pi\):

Given that the shaded and unshaded areas are equal, you set up the equation:

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

Solving for \(\alpha\),

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

**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**

(i) To find the inverse function $f^{-1}(x)$, switch $x$ and $y$ in the definition of $f$ and solve for $y$:

\(y = \frac{2}{3 – 2x}\)

\(x = \frac{2}{3 – 2y}\)

\(3 – 2y = \frac{2}{x}\)

\(2y = 3 – \frac{2}{x}\)

\(y = \frac{3}{2} – \frac{1}{x}\)

So, $f^{-1}(x) = \frac{3}{2} – \frac{1}{x}$.

(ii) To find the value of $a$ for which $\operatorname{gf}(-1) = 3$, first, find $g(f(-1))$:

\(g(f(-1)) = g\left(\frac{2}{3 – 2(-1)}\right)\)

\(= g\left(\frac{2}{5}\right)\)

\(4\left(\frac{2}{5}\right) + a = 3\)

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

\(a = 3 – \frac{8}{5} = \frac{7}{5}\)

So, the value of $a$ is $\frac{7}{5}$.

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

This equation can be written as:

\(\frac{x – a}{4} = \frac{3}{2} – \frac{1}{x}\)

\(x – a = 6x – \frac{4}{x}\)

\(6x – x – a – \frac{4}{x} = 0\)

\(5x – a – \frac{4}{x} = 0\)

\(5x^2 – ax – 4 = 0\)

\((a + 6)^2 = a^2 + 12a + 36\)

So, the correct equation is \(a^2 + 12a + 36 = 16\) or \(a^2 + 12a + 20 = 0\).

We can use the quadratic formula to solve for \(a\):

\(a = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\)

In this case, \(a^2 + 12a + 20 = 0\), so \(a = \frac{-12 \pm \sqrt{12^2 – 4(1)(20)}}{2(1)}\):

\(a = \frac{-12 \pm \sqrt{144 – 80}}{2}\)

\(a = \frac{-12 \pm \sqrt{64}}{2}\)

\(a = \frac{-12 \pm 8}{2}\)

This gives two possible values for \(a\):

\(a_1 = \frac{-12 + 8}{2} = -2\)

\(a_2 = \frac{-12 – 8}{2} = -10\)

So, the correct solutions for \(a\) are \(a = -2\) or \(a = -10\).

**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) Express the perimeter of the shaded region in terms of \(r\), \(\Theta\), and \(\pi\).

The shaded region consists of three parts: the quarter circle \(AOB\), the arc \(AC\), and the segment \(ABC\). The perimeter (\(P\)) is the sum of the lengths of these three parts.

Quarter Circle \(AOB\):The circumference of a full circle is \(2\pi r\), so the quarter circle has a circumference of \(\frac{1}{4} \cdot 2\pi r = \frac{1}{2}\pi r\).

Arc \(AC\):The length of the arc \(AC\) can be found using the formula for the circumference of a circle and the fraction of the circle represented by the angle \(\Theta\):

\[ \text{Arc length} = \frac{\Theta}{360^\circ} \cdot 2\pi r \]

Segment \(ABC\):The length of \(CD\) is \(r\) (as \(CD\) is parallel to \(AO\)). The length of \(AB\) is \(r\Theta\) (as \(AB\) is a fraction of the circumference corresponding to the angle \(\Theta\)). So, the length of \(BC\) is \(r\Theta – r\).

\[ P = \frac{1}{2}\pi r + \frac{\Theta}{360^\circ} \cdot 2\pi r + (r\Theta – r) \]

\[ P = \frac{1}{2}\pi r + \frac{\Theta}{180^\circ} \cdot \pi r + r\Theta – r \]

(ii) Area \(A):

In this case, we’re finding the area of the shaded region, which consists of the quarter circle \(AOB\) and the segment \(ABC\).

Quarter Circle \(AOB\):

– The area of a quarter circle is given by \(\frac{1}{4}\pi r^2\).

\( \text{Area of quarter circle} = \frac{1}{4}\pi \cdot (5 \, \text{cm})^2 = \frac{1}{4}\pi \cdot 25 = \frac{25}{4}\pi \, \text{cm}^2 \)

Segment \(ABC\):

– The area of a segment can be found by subtracting the area of the corresponding triangle from the area of the sector.

Sector Area:

– The area of the sector is given by \(\frac{1}{2} r^2 \left(\frac{1}{2}\pi – \Theta\right)\).

\( \text{Sector area} = \frac{1}{2} \cdot (5 \, \text{cm})^2 \cdot \left(\frac{1}{2}\pi – 0.6\right) \)

Triangle Area:

– The triangle is formed by \(O\), \(C\), and a point on the circle. The area of the triangle is given by \(\frac{1}{2} \cdot r \cdot \cos \Theta \cdot \frac{1}{2} \cdot r \cdot \sin \Theta\).

\( \text{Triangle area} = \frac{1}{2} \cdot 5 \, \text{cm} \cdot \cos 0.6 \cdot \frac{1}{2} \cdot 5 \, \text{cm} \cdot \sin 0.6 \)

Segment Area:

– Subtract the triangle area from the sector area to get the area of the segment.

\( \text{Segment area} = \text{Sector area} – \text{Triangle area} \)

Now, add the area of the quarter circle and the area of segment \(ABC\) to find the total area \(A\):

\( A = \frac{25}{4}\pi + \left(\frac{1}{2} \cdot (5)^2 \cdot \left(\frac{1}{2}\pi – 0.6\right) – \frac{1}{2} \cdot 5 \cdot \cos 0.6 \cdot \frac{1}{2} \cdot 5 \cdot \sin 0.6\right) \)

\( A \approx 6.31 \, \text{cm}^2 \)

**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**

(i) Show that angle \(CBD = \frac{9\pi}{14}\) radians:

Given that \(\angle CAB = \frac{2\pi}{7}\), we can find \(\angle BAC\) by considering that \(\triangle ABC\) is isosceles with \(AB = AC\). Therefore,

\(\angle BAC = \frac{\pi – \angle CAB}{2} = \frac{5\pi}{14}\).

Now, since \(\angle ABD\) is a straight line, \(\angle CBD = \pi – \angle BAC = \frac{9\pi}{14}\).

(ii) Find the perimeter of the shaded region:

Using Sine Rule in \(\triangle ABC\):

– The sine rule for a triangle is given by \(\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}\).

– Apply the sine rule to \(\triangle ABC\) with \(AB = AC = 8\) cm and \(\angle CAB = \frac{2\pi}{7}\).

\( \sin \frac{\pi}{7} = \frac{\frac{1}{2} BC}{8} \)

\( BC = \frac{8}{\sin \frac{\pi}{7}} \)

Law of Cosines to find \(BC^2\)

– The law of cosines is given by \(c^2 = a^2 + b^2 – 2ab \cos C\), where \(a\), \(b\), and \(c\) are sides of a triangle opposite to angles \(A\), \(B\), and \(C\) respectively.

– Apply the law of cosines to \(\triangle ABC\) with \(AB = AC = 8\) cm and \(\angle CAB = \frac{2\pi}{7}\).

\( BC^2 = 8^2 + 8^2 – 2(8)(8) \cos \frac{2\pi}{7} \)

Solve for \(BC\):

\( BC = \sqrt{8^2 + 8^2 – 2(8)(8) \cos \frac{2\pi}{7}} \)

Calculation of \(BC\):

– Substitute the values and calculate \(BC\):

\( BC \approx 6.94 \)

Length of Arc \(CD\):

– Use the length of \(BC\) and the angle \(\frac{9\pi}{14}\) to find the length of arc \(CD\) in \(\circ B\):

\( \text{Length of arc } CD = BC \times \frac{9\pi}{14} \)

Length of Arc \(CB\):

– Use the radius \(AB\) and the angle \(\frac{2\pi}{7}\) to find the length of arc \(CB\) in \(\circ A\):

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

– Add the lengths of arc \(CD\), arc \(CB\), and the straight line segment \(AB\) to find the perimeter of the shaded region:

\( \text{Perimeter} = BC + \text{Length of arc } CD + \text{Length of arc } CB + AB \)

\( \text{Perimeter} \approx 6.94 + 14.02 + 7.18 = 28.1 \)

So, the perimeter of the shaded region is approximately \(28.1\) cm.

**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**

(i)Given:

– \(\angle BAX = \alpha\)

– \(OAB\) is a sector of a circle with center \(O\) and radius \(r\)

To prove:

\(\angle AOB = 2\alpha\)

Angles \(BAO\) and \(OBA\) are both right angles since \(AX\) is tangent at \(A\) to the arc \(AB\). So, \(BAO = OBA = \frac{\pi}{2} – \alpha\).

The angle \(AOB\) is the central angle of the sector \(OAB\), and it is equal to the sum of the angles at the center subtended by arcs \(BA\) and \(BO\). Therefore, \(AOB = \pi – (BAO + OBA) = 2\alpha\).

(ii)Given:

– \(\angle BAX = \alpha\)

– \(OAB\) is a sector of a circle with center \(O\) and radius \(r\)

The area of the sector \(OAB\) is given by \(\frac{1}{2}r^2(2\alpha) = r^2\alpha\).

The area of the triangle \(OAX\) is given by \(\frac{1}{2}r^2\sin(2\alpha)\).

The area of the shaded segment is the difference between the area of the sector and the area of the triangle: \(r^2\alpha – \frac{1}{2}r^2\sin(2\alpha)\).

(b)Given:

– \(\alpha = \frac{\pi}{6}\)

– \(r = 4\)

– \(\text{side length of } ABC = 4\)

The area of one segment (\(S\)) is\( \Rightarrow \) \(\frac{1}{2}r^2\left(\frac{\pi}{3}\right) – \frac{1}{2}r^2\sin\left(\frac{\pi}{3}\right)\).

The total area of the equilateral triangle \(ABCT\) is given by \(\frac{1}{2} \times 4^2 \times \sin\left(\frac{\pi}{3}\right) = 4\sqrt{3}\).

The area of the shaded region is then \(T – 3S\), which simplifies to \(16\sqrt{3} – 8\pi\).

### 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

(b) Find the exact perimeter of the plate.

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

**▶️Answer/Explanation**

(a) Showing that \( \text{Angle PCS} = \frac{2}{3}\pi \) radians:

\(\Rightarrow \) Using Midpoint H and Sine of Angle HCP

Let the midpoint of PQ be H.

Since PS and QS are radii of the circle, \( \angle HCP = \frac{1}{2} \times \frac{\pi}{3} = \frac{\pi}{6} \).

\( \Rightarrow \) Using Sine of Angle PSQ

\( \sin \angle PSQ = \frac{4}{8} = \frac{1}{2} \).

Therefore, \( \angle PSQ = \frac{\pi}{6} \).

\( \Rightarrow \) Using Cosine Rule

Apply the cosine rule to triangle PCQ to find \( \angle PCQ = \frac{\pi}{3} \).

\(\Rightarrow\) By Inspection

Recognize that triangle PCQ or PCT is equilateral, so \( \angle PCQ = \frac{\pi}{3} \).

Since \( \angle PCS \) is the central angle of the sector CPQ, \( \angle PCS = \pi – \frac{\pi}{6} – \frac{\pi}{6} = \frac{2}{3}\pi \).

(b) Finding the Exact Perimeter of the Plate:

The perimeter is the sum of the lengths of arcs PS and QR and the lengths of semicircles PQ and RS.

\( \text{Perimeter} = 2 \times \text{Arc Length PS} + \text{Length of Semicircle PQ} \)

\( \text{Perimeter} = 2 \times \frac{4}{3}\pi + 4\pi = \frac{28}{3}\pi \)

(c) Showing that the Area of the Plate is \( \left(\frac{20}{3}\pi+8\sqrt{3}\right) \, \text{cm}^2 \):

Area of Sector CPQ:

\( \text{Area of Sector CPQ} = \frac{1}{2} \times 4^2 \times \frac{\pi}{3} = \frac{8\pi}{3} \)

Area of Segment Beyond CPQ:

\( \text{Area of Segment Beyond CPQ} = \frac{8\pi}{3} – \frac{1}{2} \times 4^2 \times \sin\left(\frac{\pi}{3}\right) = \frac{8\pi}{3} – 4\sqrt{3} \)

Area of Small Semicircle:

\( \text{Area of Small Semicircle} = \pi \times 2 \)

Area of the Plate:

\( \text{Area of Plate} = \text{Area of Large Circle} – [2 \times \text{Area of Small Semicircle}] – [2 \times \text{Area of Segment Beyond CPQ}] \)

\( \text{Area of Plate} = \pi \times 4^2 – \pi \times 2^2 – 2 \times \left(\frac{8\pi}{3} – 4\sqrt{3}\right) = \frac{20\pi}{3} + 8\sqrt{3} \)