1.05d Radians: arc length s=r*theta and sector area A=1/2 r^2 theta

\includegraphics{figure_4} In the diagram, \(ABCD\) is a parallelogram with \(AB = BD = DC = 10\) cm and angle \(ABD = 0.8\) radians. \(APD\) and \(BQC\) are arcs of circles with centres \(B\) and \(D\) respectively.

1. Find the area of the parallelogram \(ABCD\). [2]
2. Find the area of the complete figure \(ABQCDP\). [2]
3. Find the perimeter of the complete figure \(ABQCDP\). [2]

\includegraphics{figure_2} The diagram shows a triangle \(AOB\) in which \(OA\) is 12 cm, \(OB\) is 5 cm and angle \(AOB\) is a right angle. Point \(P\) lies on \(AB\) and \(OP\) is an arc of a circle with centre \(A\). Point \(Q\) lies on \(AB\) and \(OQ\) is an arc of a circle with centre \(B\).

1. Show that angle \(BAO\) is 0.3948 radians, correct to 4 decimal places. [1]
2. Calculate the area of the shaded region. [5]

\includegraphics{figure_2} In the diagram, \(OADC\) is a sector of a circle with centre \(O\) and radius 3 cm. \(AB\) and \(CB\) are tangents to the circle and angle \(ABC = \frac{1}{4}\pi\) radians. Find, giving your answer in terms of \(\sqrt{3}\) and \(\pi\),

1. the perimeter of the shaded region, [3]
2. the area of the shaded region. [3]

\includegraphics{figure_1} The diagram shows a major arc \(AB\) of a circle with centre \(O\) and radius 6 cm. Points \(C\) and \(D\) on \(OA\) and \(OB\) respectively are such that the line \(AB\) is a tangent at \(E\) to the arc \(CED\) of a smaller circle also with centre \(O\). Angle \(COD = 1.8\) radians.

1. Show that the radius of the arc \(CED\) is 3.73 cm, correct to 3 significant figures. [2]
2. Find the area of the shaded region. [4]

\includegraphics{figure_8} The diagram shows an isosceles triangle \(ACB\) in which \(AB = BC = 8\) cm and \(AC = 12\) cm. The arc \(XC\) is part of a circle with centre \(A\) and radius \(12\) cm, and the arc \(YC\) is part of a circle with centre \(B\) and radius \(8\) cm. The points \(A\), \(B\), \(X\) and \(Y\) lie on a straight line.

1. Show that angle \(CBY = 1.445\) radians, correct to \(4\) significant figures. [3]
2. Find the perimeter of the shaded region. [4]

\includegraphics{figure_6} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(\angle OAB\) is equal to \(x\) radians. The shaded region is bounded by \(AB\), \(AC\) and the circular arc with centre \(A\) joining \(B\) and \(C\). The perimeter of the shaded region is equal to half the circumference of the circle.

1. Show that \(x = \cos^{-1}\left(\frac{\pi}{4 + 4x}\right)\). [3]
2. Verify by calculation that \(x\) lies between 1 and 1.5. [2]
3. Use the iterative formula $$x_{n+1} = \cos^{-1}\left(\frac{\pi}{4 + 4x_n}\right)$$ to determine the value of \(x\) correct to 2 decimal places. Give the result of each iteration to 4 decimal places. [3]

\includegraphics{figure_6} The diagram shows points \(A\) and \(B\) on a circle with centre \(O\) and radius \(r\). The tangents to the circle at \(A\) and \(B\) meet at \(T\). The shaded region is bounded by the minor arc \(AB\) and the lines \(AT\) and \(BT\). Angle \(AOB\) is \(2\theta\) radians.

1. In the case where the area of the sector \(AOB\) is the same as the area of the shaded region, show that \(\tan \theta = 2\theta\). [3]
2. In the case where \(r = 8\) cm and the length of the minor arc \(AB\) is 19.2 cm, find the area of the shaded region. [3]

\includegraphics{figure_5} The diagram shows a triangle \(OAB\) in which angle \(OAB = 90°\) and \(OA = 5\) cm. The arc \(AC\) is part of a circle with centre \(O\). The arc has length 6 cm and it meets \(OB\) at \(C\). Find the area of the shaded region. [5]

\includegraphics{figure_6} In the diagram, \(A\) is a point on the circumference of a circle with centre \(O\) and radius \(r\). A circular arc with centre \(A\) meets the circumference at \(B\) and \(C\). The angle \(OAB\) is \(\theta\) radians. The shaded region is bounded by the circumference of the circle and the arc with centre \(A\) joining \(B\) and \(C\). The area of the shaded region is equal to half the area of the circle.

1. Show that \(\cos 2\theta = \frac{2\sin 2\theta - \pi}{4\theta}\). [5]
2. Use the iterative formula $$\theta_{n+1} = \frac{1}{2}\cos^{-1}\left(\frac{2\sin 2\theta_n - \pi}{4\theta_n}\right),$$ with initial value \(\theta_1 = 1\), to determine \(\theta\) correct to 2 decimal places, showing the result of each iteration to 4 decimal places. [3]

\includegraphics{figure_4} The triangle \(XYZ\) in Figure 4 has \(XY = 6\) cm, \(YZ = 9\) cm, \(ZX = 4\) cm and angle \(ZXY = a\). The point \(W\) lies on the line \(XY\). The circular arc \(ZW\), in Figure 4, is a major arc of the circle with centre \(X\) and radius 4 cm.

1. Show that, to 3 significant figures, \(a = 2.22\) radians. [2]
2. Find the area, in cm\(^2\), of the major sector \(XZWX\). [3]

The region, shown shaded in Figure 4, is to be used as a design for a logo. Calculate

1. the area of the logo [3]
2. the perimeter of the logo. [4]

\includegraphics{figure_1} Figure 1 shows the triangle \(ABC\), with \(AB = 8\) cm, \(AC = 11\) cm and \(\angle BAC = 0.7\) radians. The arc \(BD\), where \(D\) lies on \(AC\), is an arc of a circle with centre \(A\) and radius 8 cm. The region \(R\), shown shaded in Figure 1, is bounded by the straight lines \(BC\) and \(CD\) and the arc \(BD\). Find

1. the length of the arc \(BD\), [2]
2. the perimeter of \(R\), giving your answer to 3 significant figures, [4]
3. the area of \(R\), giving your answer to 3 significant figures. [5]

\includegraphics{figure_2} In Figure 2 \(OAB\) is a sector of a circle, radius 5 m. The chord \(AB\) is 6 m long.

1. Show that \(\cos A\hat{O}B = \frac{7}{25}\). [2]
2. Hence find the angle \(A\hat{O}B\) in radians, giving your answer to 3 decimal places. [1]
3. Calculate the area of the sector \(OAB\). [2]
4. Hence calculate the shaded area. [3]

A circle \(C\) has centre \(M\) \((6, 4)\) and radius 3.

1. Write down the equation of the circle in the form $$(x - a)^2 + (y - b)^2 = r^2.$$ [2]

\includegraphics{figure_3} Figure 3 shows the circle \(C\). The point \(T\) lies on the circle and the tangent at \(T\) passes through the point \(P\) \((12, 6)\). The line \(MP\) cuts the circle at \(Q\).

1. Show that the angle \(TMQ\) is 1.0766 radians to 4 decimal places. [4]

The shaded region \(TPQ\) is bounded by the straight lines \(TP\), \(QP\) and the arc \(TQ\), as shown in Figure 3.

1. Find the area of the shaded region \(TPQ\). Give your answer to 3 decimal places. [5]

\includegraphics{figure_1} Figure 1 shows a logo \(ABD\). The logo is formed from triangle \(ABC\). The mid-point of \(AC\) is \(D\) and \(BC = AD = DC = 6\) cm. \(\angle BCA = 0.4\) radians. The curve \(BD\) is an arc of a circle with centre \(C\) and radius 6 cm.

1. Write down the length of the arc \(BD\). [1]
2. Find the length of \(AB\). [3]
3. Write down the perimeter of the logo \(ABD\), giving your answer to 3 significant figures. [1]

\includegraphics{figure_2} Triangle \(ABC\) has \(AB = 9\) cm, \(BC = 10\) cm and \(CA = 5\) cm. A circle, centre \(A\) and radius 3 cm, intersects \(AB\) and \(AC\) at \(P\) and \(Q\) respectively, as shown in Fig. 3.

1. Show that, to 3 decimal places, \(\angle BAC = 1.504\) radians. [3]

Calculate,

1. the area, in cm\(^2\), of the sector \(APQ\), [2]
2. the area, in cm\(^2\), of the shaded region \(BPQC\), [3]
3. the perimeter, in cm, of the shaded region \(BPQC\). [4]

\includegraphics{figure_4} Figure 1 shows the cross-sections of two drawer handles. Shape \(X\) is a rectangle \(ABCD\) joined to a semicircle with \(BC\) as diameter. The length \(AB = d\) cm and \(BC = 2d\) cm. Shape \(Y\) is a sector \(OPQ\) of a circle with centre \(O\) and radius \(2d\) cm. Angle \(POQ\) is \(\theta\) radians. Given that the areas of the shapes \(X\) and \(Y\) are equal,

1. prove that \(\theta = 1 + \frac{1}{4}\pi\). [5]

Using this value of \(\theta\), and given that \(d = 3\), find in terms of \(\pi\),

1. the perimeter of shape \(X\), [2]
2. the perimeter of shape \(Y\). [3]
3. Hence find the difference, in mm, between the perimeters of shapes \(X\) and \(Y\). [2]

\includegraphics{figure_6} Figure 1 shows a gardener's design for the shape of a flower bed with perimeter \(ABCD\). \(AD\) is an arc of a circle with centre \(O\) and radius 5 m. \(BC\) is an arc of a circle with centre \(O\) and radius 7 m. \(OAB\) and \(ODC\) are straight lines and the size of \(\angle AOD\) is \(\theta\) radians.

1. Find, in terms of \(\theta\), an expression for the area of the flower bed. [3]

Given that the area of the flower bed is 15 m\(^2\),

1. show that \(\theta = 1.25\), [2]
2. calculate, in m, the perimeter of the flower bed. [3]

The gardener now decides to replace arc \(AD\) with the straight line \(AD\).

1. Find, to the nearest cm, the reduction in the perimeter of the flower bed. [2]

\includegraphics{figure_8} The shape of a badge is a sector \(ABC\) of a circle with centre \(A\) and radius \(AB\), as shown in Fig 1. The triangle \(ABC\) is equilateral and has a perpendicular height 3 cm.

1. Find, in surd form, the length \(AB\). [2]
2. Find, in terms of \(\pi\), the area of the badge. [2]
3. Prove that the perimeter of the badge is \(\frac{2\sqrt{3}}{3}(\pi + 6)\) cm. [3]

\includegraphics{figure_10} Figure 1 shows the cross-section \(ABCD\) of a chocolate bar, where \(AB\), \(CD\) and \(AD\) are straight lines and \(M\) is the mid-point of \(AD\). The length \(AD\) is 28 mm, and \(BC\) is an arc of a circle with centre \(M\). Taking \(A\) as the origin, \(B\), \(C\) and \(D\) have coordinates \((7, 24)\), \((21, 24)\) and \((28, 0)\) respectively.

1. Show that the length of \(BM\) is 25 mm. [1]
2. Show that, to 3 significant figures, \(\angle BMC = 0.568\) radians. [3]
3. Hence calculate, in mm\(^2\), the area of the cross-section of the chocolate bar. [5]

Given that this chocolate bar has length 85 mm,

1. calculate, to the nearest cm\(^3\), the volume of the bar. [2]

\includegraphics{figure_1} Figure 1 shows a sketch of a segment \(PQRP\) of a circle with centre \(O\) and radius \(5\) cm. Given that • angle \(PQR\) is \(\theta\) radians • \(\theta\) is increasing, from \(0\) to \(\pi\), at a constant rate of \(0.1\) radians per second • the area of the segment \(PQRP\) is \(A\) cm²

1. show that $$\frac{dA}{d\theta} = K(1 - \cos \theta)$$ where \(K\) is a constant to be found. [2]
2. Find, in cm²s⁻¹, the rate of increase of the area of the segment when \(\theta = \frac{\pi}{3}\) [4]

The diagram shows a sector \(OAB\) of a circle with centre \(O\) and radius \(r\) cm. \includegraphics{figure_6} The angle \(AOB\) is \(1.2\) radians. The area of the sector is \(33.75\) cm\(^2\). Find the perimeter of the sector. [6]

\includegraphics{figure_1} The shape of a badge is a sector \(ABC\) of a circle with centre \(A\) and radius \(AB\), as shown in Fig 1. The triangle \(ABC\) is equilateral and has a perpendicular height 3 cm.

1. Find, in surd form, the length \(AB\). [2]
2. Find, in terms of \(\pi\), the area of the badge. [2]
3. Prove that the perimeter of the badge is \(\frac{2\sqrt{3}}{3}(\pi + 6)\) cm. [2]

\includegraphics{figure_3} Fig. 3 Triangle ABC has AB = 9 cm, BC = 10 cm and CA = 5 cm. A circle, centre A and radius 3 cm, intersects AB and AC at P and Q respectively, as shown in Fig. 3.

1. Show that, to 3 decimal places, ∠BAC = 1.504 radians. [3]

Calculate,

1. the area, in cm², of the sector APQ, [2]
2. the area, in cm², of the shaded region BPQC, [3]
3. the perimeter, in cm, of the shaded region BPQC. [4]

END

\includegraphics{figure_1} Figure 1 shows the cross-section ABCD of a chocolate bar, where AB, CD and AD are straight lines and M is the mid-point of AD. The length AD is 28 mm, and BC is an arc of a circle with centre M. Taking A as the origin, B, C and D have coordinates (7, 24), (21, 24) and (28, 0) respectively.

1. Show that the length of BM is 25 mm. [1]
2. Show that, to 3 significant figures, \(\angle BMC = 0.568\) radians. [3]
3. Hence calculate, in mm², the area of the cross-section of the chocolate bar. [5]

Given that this chocolate bar has length 85 mm,

1. calculate, to the nearest cm³, the volume of the bar. [2]

\includegraphics{figure_1} Figure 1 shows the cross-sections of two drawer handles. Shape \(X\) is a rectangle \(ABCD\) joined to a semicircle with \(BC\) as diameter. The length \(AB = d\) cm and \(BC = 2d\) cm. Shape \(Y\) is a sector \(OPQ\) of a circle with centre \(O\) and radius \(2d\) cm. Angle \(POQ\) is \(\theta\) radians. Given that the areas of the shapes \(X\) and \(Y\) are equal,

1. prove that \(\theta = 1 + \frac{1}{2}\pi\). [5]

Using this value of \(\theta\), and given that \(d = 3\), find in terms of \(\pi\),

1. the perimeter of shape \(X\), [2]
2. the perimeter of shape \(Y\). [3]
3. Hence find the difference, in mm, between the perimeters of shapes \(X\) and \(Y\). [2]