WAEC SSCE - General Mathematics - 2011

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The perpendicular bisectors of the sides of an acute-angled triangle intersect at a point inside the triangle. This point is called the circumcenter, which is equidistant from the three vertices of the triangle. To see why this is true, consider two sides of the triangle, and let their perpendicular bisectors intersect at a point O. Since O lies on the perpendicular bisector of each of the two sides, it is equidistant from the endpoints of each of those sides. Therefore, O is equidistant from two vertices of the triangle. Similarly, O is equidistant from the third vertex, so it must be the circumcenter of the triangle. Since the triangle is acute-angled, the circumcenter lies inside the triangle.

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The area of a rectangle is given by multiplying the length by the width. Therefore, the area of the garden is: Area = length × width Area = 18.6m × 12.5m Area = 232.5m² Rounding to three significant figures gives 233m². Therefore, the answer is (d) 233m².

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To find the equation of the line of symmetry of the graph, we need to identify the axis of symmetry. The axis of symmetry is a vertical line that passes through the vertex of the parabola. The vertex of the parabola is the point where the parabola changes direction, and it can be found by using the formula: x = -b / (2a) where a and b are the coefficients of the quadratic equation y = ax\(^2\) + bx + c. In the given equation y = x\(^2\) - 3x - 3, a = 1, b = -3, and c = -3. Substituting these values in the formula, we get: x = -(-3) / (2*1) = 3/2 = 1.5 Therefore, the line of symmetry is a vertical line passing through x = 1.5. So, the correct answer is (C) x = 1.5.

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In the given diagram, we can see that lines n and q are parallel and m is a transversal cutting them. Therefore, angles n and q are alternate interior angles and are equal, i.e., statement i is true. Also, we can see that lines n and p are parallel and q is a transversal cutting them. Therefore, angles n and p are corresponding angles and are equal. As the sum of the corresponding angles is equal to 180 degrees, we have n + p = 180 degrees, i.e., statement iii is also true. However, we cannot determine whether statement ii and iv are true or not based on the given information and the diagram. Therefore, the correct answer is (a) i and iii.

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If we can find the value of x and y, then we can calculate x/y by dividing x by y. Given: 27^x = 9^y We can rewrite 27 as 3³ and 9 as 3² to get: (3³)^x = (3²)^y Applying the power of a power rule, we get: 3^3x = 3^2y For two exponential expressions to be equal, their bases must be equal. Therefore: 3^3x = 3^2y implies 3x = 2y Dividing both sides by y, we get: \(\frac{x}{y} = \frac{2}{3}\) Therefore, the answer is \(\frac{2}{3}\).

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The volume of a cylinder can be calculated using the formula: V = πr²h, where r is the radius of the base and h is the height of the cylinder. Substituting the given values, we have: V = π(14)²(18) V = 11088 cm³ To convert cm³ to litres, we divide by 1000: V = 11088/1000 V = 11.088 litres Rounding to the nearest litre, we get: V ≈ 11 litres Therefore, the answer is 11.

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In the given diagram, we have a circle with center O and radii OP and OR. The reflex angle PQR is 240° and |PQ| = |QR|. We need to find the value of x. Since |PQ| = |QR|, we know that triangle PQR is an isosceles triangle. Therefore, the angles opposite to PQ and QR are equal. Let's denote the angle PQR by y. Then we have: 2y + 60° = 360° (sum of angles in a triangle) 2y = 300° y = 150° Therefore, each of the angles opposite to PQ and QR is equal to (180° - 150°)/2 = 15°. Now, consider the triangle OQP. We know that the sum of angles in a triangle is 180°. Therefore: ∠OQP + ∠QOP + ∠OPQ = 180° Since OP and OQ are radii, ∠QOP = ∠OPQ. Let's denote this angle by z. Then we have: z + z + 15° = 180° 2z = 165° z = 82.5° Finally, consider the triangle OXR. We know that the sum of angles in a triangle is 180°. Therefore: ∠OXR + ∠ORX + ∠ROX = 180° Since OR and OX are radii, ∠ORX = ∠ROX. Let's denote this angle by x. Then we have: x + x + 60° = 180° 2x = 120° x = 60° Therefore, the value of x is 60°. Answer: 60°.

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To solve this system of equations, we can use the method of elimination. We will add the two equations together, which will eliminate the y variable: (3x - 2y) + (x + 2y) = 7 + (-3) Simplifying, we get: 4x = 4 Dividing both sides by 4, we get: x = 1 Substituting x = 1 into one of the equations, we get: 1 + 2y = -3 Solving for y, we get: y = -2 Therefore, the solution to the system of equations is x = 1, y = -2. So, the correct option is (a) x = 1, y = -2.

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In this problem, we are given that G varies directly as the square of H. This means that if H is multiplied by some factor, then G will be multiplied by the square of that factor. Mathematically, we can write this as: G ∝ H^2 where the symbol "∝" means "varies directly as". We are also given that G is 4 when H is 3. Using this information, we can write: 4 ∝ 3^2 To find H when G = 100, we can use the same relationship: G ∝ H^2 If we let the constant of proportionality be k, we can write: G = kH^2 To solve for k, we can use the initial condition where G is 4 when H is 3: 4 = k(3^2) Simplifying, we get: k = 4/9 Now we can use this value of k to find H when G is 100: 100 = (4/9)H^2 Multiplying both sides by 9/4, we get: 225 = H^2 Taking the square root of both sides, we get: H = 15 Therefore, the correct answer is (a) 15. In summary, we used the direct variation relationship between G and H^2 to find the constant of proportionality, and then used that constant and the given value of G to solve for H.

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The number 42.7531 can be written in expanded form as: 42 + 0.7 + 0.05 + 0.003 + 0.0001 The digit 3 is located in the thousandths place, which represents the decimal value of 0.001. So, the value of 3 in the number 42.7531 is equal to: 3 x 0.001 = 0.003 Therefore, the correct answer is option (A), which is 3 divided by 10000.

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To calculate the height of the oil in the container, we need to use the formula for the volume of a rectangular prism: Volume = length x breadth x height First, we need to convert the given volume from liters to cubic centimeters, since the dimensions of the container are in centimeters. 96 liters = 96,000 cubic centimeters Next, we can plug in the given values into the formula: 96,000 = 220 x 40 x height Solving for height, we get: height = 96,000 / (220 x 40) height ≈ 11.0 cm (rounded to the nearest cm) Therefore, the height of the oil in the container is approximately 11 cm. Note: When working with volume, it's important to make sure the units are consistent throughout the problem. In this case, we converted liters to cubic centimeters to match the dimensions of the container.

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The digit 3 in the number 42.7531 is in the thousandth place (the digit after the decimal point and three places to the right of it). Therefore, its value is \(\frac{3}{1000}\).

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In a regular polygon with n sides, each exterior angle measures 360/n degrees. We are given that in this polygon, each exterior angle is 45 degrees. Therefore, we can set up an equation: 360/n = 45 To solve for n, we can cross-multiply and simplify: 360 = 45n n = 360/45 n = 8 Therefore, the regular polygon in question has 8 sides. Answer: 8.

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We are given a quadratic equation x² + 7x + 10 = 0 and we need to find the smaller value of x that satisfies the equation. To solve the equation, we can factorize it by finding two numbers whose product is 10 and whose sum is 7. We can see that the two numbers are 2 and 5, since 2 × 5 = 10 and 2 + 5 = 7. So, we can write the equation as (x + 2)(x + 5) = 0. For this equation to be true, either (x + 2) = 0 or (x + 5) = 0. Therefore, we get x = -2 or x = -5. Since we are asked to find the smaller value of x, we choose x = -5 as the answer. Hence, the smaller value of x that satisfies the equation x² + 7x + 10 = 0 is -5.

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We can begin by grouping the first two terms and the last two terms together: am + bn = a(m) + b(n) = (a+b)n - bn an + bm = a(n) + b(m) = (a+b)m - am Now, we can substitute these expressions back into the original equation: am + bn - an - bm = [(a+b)n - bn] - [(a+b)m - am] We can simplify this expression by combining like terms: am + bn - an - bm = (a+b)n - bn - (a+b)m + am am + bn - an - bm = (a+b)n - (a+b)m + am - bn Finally, we can factor out the common factor of (a+b) from the first two terms and the common factor of (-1) from the last two terms: am + bn - an - bm = (a+b)(n-m) - (b-a)(n-m) Therefore, the answer is (a+b)(n-m) - (b-a)(n-m), which can be further simplified to (a-b)(m-n). Thus, the correct option is (a - b)(m - n).

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The actual length of the stick is 1.75m and the measured length is 1.80m. The error is the difference between the actual and measured length: 1.80m - 1.75m = 0.05m To find the percentage error, we divide the error by the actual length and multiply by 100%: \frac{0.05}{1.75} \times 100\% \approx 2.857\% \approx \frac{20}{7}\% Therefore, the percentage error is approximately \frac{20}{7}\%.

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Using the property that \(\log_{a}b = \frac{\log{b}}{\log{a}}\), we can simplify the given expression as follows: \[\frac{\log \sqrt{27}}{\log \sqrt{81}} = \frac{\log 27^{\frac{1}{2}}}{\log 81^{\frac{1}{2}}} = \frac{\frac{1}{2}\log 27}{\frac{1}{2}\log 81} = \frac{\log 3^3}{\log 3^4} = \frac{3\log 3}{4\log 3} = \frac{3}{4}\] Therefore, the simplified form of \(\frac{\log \sqrt{27}}{\log \sqrt{81}}\) is \(\frac{3}{4}\), and the correct option is (D).

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Let's denote the radius of the cylinder as r and its height as h. We are given that the height of the cylinder is equal to its radius, so h = r. We also know the volume of the cylinder, which is given by: V = \(\pi\)r\(^2\)h Substituting h = r, we get: V = \(\pi\)r\(^2\)r = \(\pi\)r\(^3\) We are given that the volume of the cylinder is 0.216 \(\pi\) m\(^3\). So, we can solve for r as follows: 0.216 \(\pi\) = \(\pi\)r\(^3\) r\(^3\) = 0.216 Taking the cube root of both sides, we get: r = 0.6 Therefore, the radius of the cylinder is 0.6 meters. So, the answer is 0.60m.

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To get from (2x - 3y) to (x - 2y), we need to subtract x from 2x and add 2y to -3y. Therefore, we need to add (x - 2y) - (2x - 3y) to (2x - 3y) to get (x - 2y). Simplifying (x - 2y) - (2x - 3y), we have: (x - 2y) - (2x - 3y) = x - 2y - 2x + 3y = -x + y Therefore, we need to add (-x + y) to (2x - 3y) to get (x - 2y). Simplifying (2x - 3y) + (-x + y), we have: (2x - 3y) + (-x + y) = 2x - 3y - x + y = x - 2y So, we need to add (-x + y) to (2x - 3y) to get (x - 2y). Therefore, the answer is (B) y - x.

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We know that: cos x = adjacent side/hypotenuse So, if cos x = 1/r, then adjacent side = 1 and hypotenuse = r. Using the Pythagorean theorem, we can find the opposite side: opposite side = √(hypotenuse^2 - adjacent side^2) = √(r^2 - 1) Finally, we can find the value of tan x: tan x = opposite side/adjacent side = √(r^2 - 1)/1 = √(r^2 - 1) Therefore, the answer is (d) \(\sqrt{r^2 - 1}\).

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When a quadratic equation has roots at x = a and x = b, it can be written in factored form as (x-a)(x-b) = 0. Therefore, from the given roots, the factors are (x - \(\frac{1}{2}\)) and (x + \(\frac{2}{3}\)). To get the quadratic equation, we can expand the factors by multiplying them together, which gives us: (x - \(\frac{1}{2}\))(x + \(\frac{2}{3}\)) = x² - \(\frac{1}{2}\)x + \(\frac{2}{3}\)x - \(\frac{1}{2}\)\(\frac{2}{3}\) = x² + \(\frac{1}{6}\)x - \(\frac{1}{3}\) Therefore, the correct option is 6x² + x - 2 = 0.

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To solve this problem, we need to use the formula for the volume of a rectangular container: Volume = Length x Breadth x Height We are given the length and breadth of the container, as well as the volume of the oil. We need to find the height of the oil. First, we need to convert the volume of the oil from litres to cubic centimeters, since the dimensions of the container are given in centimeters. 1 litre = 1000 cubic centimeters Therefore, 96 litres = 96 x 1000 = 96,000 cubic centimeters Now, we can plug in the values we have into the formula for the volume of the container: Volume = Length x Breadth x Height 96,000 = 220 x 40 x Height Simplifying, we get: Height = 96,000 / (220 x 40) Height = 2.18 To round off to the nearest centimeter, we need to look at the first decimal place. If the value in the first decimal place is 5 or more, we round up. If it is less than 5, we round down. In this case, the value in the first decimal place is 1, which is less than 5. Therefore, we round down to 2. So, the height of the oil in the container is approximately 2cm. Therefore, the correct answer is (a) 11cm.

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If Esther was initially facing S 20° W and then turned 90° clockwise, she would end up facing in a new direction. To determine the new direction, we can add 90° to her initial direction. When we add 90° to S 20° W, we rotate the direction clockwise by 90°, which means the new direction will be to the right of the initial direction. To find the new direction, we need to subtract the initial angle from 90°: 90° - 20° = 70° Therefore, Esther is facing in the direction of N 70° W after turning 90° clockwise from her initial direction of S 20° W.

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To calculate the value of D1.00 in naira, we can use the given exchange rate of N112.00 to D14.95. We can find the value of 1 D in Naira by dividing N112.00 by the equivalent value of D14.95. So, 1 D = N112.00/D14.95 To simplify this, we can first convert D14.95 to its decimal equivalent by dividing by 100: D14.95 = 14.95/100 = 0.1495 Now we can substitute this value into the equation: 1 D = N112.00/0.1495 Simplifying this expression, we get: 1 D = N748.16 Therefore, the value of D1.00 in Naira is N748.16. Answer: 7.49.

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(a) Numerator: \(\tfrac{1}{2}\text{ of }\tfrac{1}{4} \div \tfrac{1}{3} = \tfrac{1}{8}\div\tfrac{1}{3} = \tfrac{1}{8}\times3 = \tfrac{3}{8}.\)

Denominator: \(\tfrac{1}{6} - \tfrac{3}{4} + \tfrac{1}{2} = \tfrac{2 - 9 + 6}{12} = -\tfrac{1}{12}.\)

\[\frac{\tfrac{3}{8}}{-\tfrac{1}{12}} = \tfrac{3}{8}\times(-12) = -\tfrac{9}{2} = -4\tfrac{1}{2}.\]

(b) The bar denotes a negative characteristic, so \(\bar{1}.6741 = -1 + 0.6741 = -0.3259\). Squaring \(\sqrt{x}\):

\[x = \left(10^{\bar{1}.6741}\right)^2 = 10^{\,2(-0.3259)} = 10^{-0.6518} = 10^{\bar{1}.3482}.\]

The antilog of \(0.3482\) is \(2.229\), and the characteristic \(\bar{1}\) places one zero after the point:

\[x = 0.2229 \approx 0.223.\]

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(a) In the quadrilateral \(PT \parallel QS\), with \(\angle PTQ = 42^{\circ}\), \(\angle TSQ = 38^{\circ}\) and \(\angle QSR = 30^{\circ}\).

(i) Find \(x = \angle QTS\).

Since \(PT \parallel QS\), \(\angle PTS = \angle TSQ = 38^{\circ}\) (alternate angles). The angles at \(T\) on the straight line give:

\[ 42^{\circ} + x + 38^{\circ} = 180^{\circ}. \]

\[ 80^{\circ} + x = 180^{\circ} \quad\Rightarrow\quad x = 100^{\circ}. \]

(ii) Find \(y = \angle POT\).

Since \(PT \parallel QS\), \(\angle SQT = \angle PTQ = 42^{\circ}\) (alternate angles). In triangle \(QRS\), \(\angle SQR = 180^{\circ} - 90^{\circ} - 30^{\circ} = 60^{\circ}\) (sum of angles in a triangle). The angles on the straight line at \(Q\) give:

\[ y + 42^{\circ} + 60^{\circ} = 180^{\circ}. \]

\[ y + 102^{\circ} = 180^{\circ} \quad\Rightarrow\quad y = 78^{\circ}. \]

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(b) \(PQRS\) is a circle centre \(O\), with \(\angle POQ = 150^{\circ}\), \(\angle QSR = 40^{\circ}\) and \(\angle SQP = 45^{\circ}\). Find \(\angle RQS\).

The angle at the centre is twice the angle at the circumference on the same arc \(PQ\):

\[ \angle QSP = \tfrac{1}{2}\times 150^{\circ} = 75^{\circ}. \]

In triangle \(SPQ\), the sum of the angles is \(180^{\circ}\):

\[ \angle QPS + 75^{\circ} + 45^{\circ} = 180^{\circ} \quad\Rightarrow\quad \angle QPS = 60^{\circ}. \]

\(PQRS\) is a cyclic quadrilateral, so opposite angles are supplementary:

\[ \angle QRS = 180^{\circ} - \angle QPS = 180^{\circ} - 60^{\circ} = 120^{\circ}. \]

In triangle \(QRS\), the sum of the angles is \(180^{\circ}\):

\[ \angle RQS = 180^{\circ} - (\angle QRS + \angle QSR) = 180^{\circ} - (120^{\circ} + 40^{\circ}) = 20^{\circ}. \]

\(\angle RQS = 20^{\circ}.\)

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(a) Square both sides to remove the root:

\[t^2 = \frac{pq}{r} - r^2 q = q\left(\frac{p}{r} - r^2\right) = q\left(\frac{p - r^3}{r}\right).\]

Make q the subject:

\[q = \frac{r\,t^2}{p - r^3}.\]

(b) Write both sides to base 3: \(9^{(1-x)} = 3^{2(1-x)}\) and \(27^{y} = 3^{3y}\). Equating indices,

\[2(1 - x) = 3y \Rightarrow 2 - 2x = 3y. \quad (1)\]

Also \(x - y = -\tfrac{3}{2}\Rightarrow x = y - \tfrac{3}{2}. \quad (2)\)

Substitute (2) into (1):

\[2 - 2\left(y - \tfrac{3}{2}\right) = 3y \Rightarrow 2 - 2y + 3 = 3y \Rightarrow 5 = 5y \Rightarrow y = 1.\]

\[x = 1 - \tfrac{3}{2} = -\tfrac{1}{2}.\]

Answer: \(x = -\tfrac{1}{2},\ y = 1\).

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(a) Surface areas of similar spheres are in the ratio of the squares of their radii:

\[\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{9}{49}\Rightarrow \frac{r_{1}}{r_{2}}=\frac{3}{7}\]

The smaller radius \(r_{1}=12\) cm corresponds to the ratio part 3, so one part \(=\dfrac{12}{3}=4\) cm, and the bigger radius \(r_{2}=7\times4=28\) cm.

\[V=\frac{4}{3}\pi r_{2}^{3}=\frac{4}{3}(3.142)(28)^{3}=\frac{4}{3}(3.142)(21952)\approx 91964\text{ cm}^{3}\]

Volume of bigger sphere \(\approx 91{,}964\text{ cm}^{3}\).

(b) Take \(X\) as origin, West as the negative \(x\)-direction. \(Y=(-3,0)\). North-West is the bearing \(315^{\circ}\), i.e. direction \((-\sin45^{\circ},\cos45^{\circ})\), so

\[Z=(-3,0)+5(-\sin45^{\circ},\cos45^{\circ})=(-3-3.536,\ 3.536)=(-6.536,\ 3.536)\]

(i) Distance from \(X\):

\[XZ=\sqrt{6.536^{2}+3.536^{2}}=\sqrt{42.72+12.50}=\sqrt{55.22}\approx 7.4\ \text{km}\approx 7\text{ km}\]

(ii) \(Z\) lies to the North and West of \(X\). The angle west of due North is \(\tan^{-1}\dfrac{6.536}{3.536}=\tan^{-1}1.848=61.6^{\circ}\).

\[\text{Bearing of }Z\text{ from }X=360^{\circ}-61.6^{\circ}\approx 298^{\circ}\]

Bearing \(\approx 298^{\circ}\).

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(a) Pie chart

The whole grant of \($1{,}300\) is represented by the full circle, \(360^{\circ}\). Each item's sector angle is found from \(\text{angle}=\dfrac{\text{percentage}}{100}\times360^{\circ}\), which is the same as \(\dfrac{\text{amount}}{1300}\times360^{\circ}\).

Item	Percentage	Amount ($)	Sector angle
Magazine subscriptions	10%	130	\(\frac{130}{1300}\times360^{\circ}=36^{\circ}\)
New books	35%	455	\(\frac{455}{1300}\times360^{\circ}=126^{\circ}\)
Repair of damaged books	15%	195	\(\frac{195}{1300}\times360^{\circ}=54^{\circ}\)
New furniture	30%	390	\(\frac{390}{1300}\times360^{\circ}=108^{\circ}\)
Staff training	10%	130	\(\frac{130}{1300}\times360^{\circ}=36^{\circ}\)
Total	100%	1300	\(360^{\circ}\)

Drawing a circle and marking each sector in turn with the angles above gives the following pie chart:

(b) Percentage increase of books over furniture

Amount for new books \(=35\%\) of \(1300 = \$455\).

Amount for new furniture \(=30\%\) of \(1300 = \$390\).

The increase in the amount is \(455-390=\$65\), and this is expressed as a percentage of the furniture amount:

\[\text{Percentage increase}=\frac{455-390}{390}\times100\%=\frac{65}{390}\times100\%=16.67\%\approx 17\%\]

The books allocation is about \(17\%\) more than the furniture allocation (to the nearest whole number).

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(a) \(P\propto\dfrac{Q}{R^{2}}\Rightarrow P=\dfrac{kQ}{R^{2}}\).

Using \(P=1,\ Q=8,\ R=2\): \(1=\dfrac{8k}{4}=2k\Rightarrow k=\tfrac{1}{2}\).

When \(P=3,\ R=5\): \(3=\dfrac{\tfrac{1}{2}\,Q}{25}=\dfrac{Q}{50}\Rightarrow Q=150\).

\(Q=150\)

(b)(i) \(A(20^{\circ}N,60^{\circ}E)\) and \(B(20^{\circ}N,20^{\circ}E)\) are on the same parallel of latitude, so the flight is along the parallel \(20^{\circ}N\). Difference in longitude \(=60-20=40^{\circ}\).

\[\text{Distance}=\frac{40}{360}\times2\pi R\cos20^{\circ}=\frac{40}{360}\times2(3.142)(6400)(0.9397)\approx 4199\text{ km}\]\[\text{Average speed}=\frac{4199}{6}\approx 700\text{ km/h (3 s.f.)}\]

(ii) Flying due North from \(B\) is along a meridian, where distance \(=\dfrac{\theta}{360}\times2\pi R\), \(\theta\) the change in latitude.

\[420=\frac{\theta}{360}\times2(3.142)(6400)\Rightarrow \theta=\frac{420\times360}{40217.6}\approx 3.76^{\circ}\]

Starting at \(20^{\circ}N\) and moving north, latitude of \(C=20+3.76\approx 24^{\circ}N\).

Latitude of C \(\approx 24^{\circ}N\).

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(a) Evaluating \(2\cos x + 3\sin x\)

Given \(\sin x = 0.6 = \dfrac{3}{5}\) with \(0^\circ \le x \le 90^\circ\), x is acute, so \(\cos x\) is positive.

\[\cos x = \sqrt{1 - \sin^2 x} = \sqrt{1 - (0.6)^2} = \sqrt{1 - 0.36} = \sqrt{0.64} = 0.8 = \frac{4}{5}\]

Then

\[2\cos x + 3\sin x = 2\left(\frac{4}{5}\right) + 3\left(\frac{3}{5}\right) = \frac{8}{5} + \frac{9}{5} = \frac{17}{5}\]

(b) Semi-circle inscribed in isosceles triangle ABC

From the diagram, the diameter WZ lies along the top side AB, the centre O is on AB, and the semicircular arc is tangent to the two equal sides. The apex C is below, with \(|OC| = 30\ \text{cm}\) measured along the axis of symmetry, and \(\angle ACB = 130^\circ\).

By symmetry (AC = BC), the line CO bisects angle C, so the half-angle at C is

\[\frac{130^\circ}{2} = 65^\circ\]

(i) Radius of the circle. The radius drawn to the point where the arc touches a side is perpendicular to that side. In the right-angled triangle formed by O, C and the tangent point, OC is the hypotenuse and the radius r is opposite the \(65^\circ\) angle:

\[r = |OC|\sin 65^\circ = 30 \times 0.9063 = 27.19\]

\[r \approx 27.2\ \text{cm}\]

(ii) Area of the shaded portion. The shaded region is the part of triangle ABC lying outside the semicircle:

\[\text{Shaded} = \text{Area of } \triangle ABC - \text{Area of semicircle}\]

The centre O lies on AB, so OC = 30 cm is the height of the triangle from C to AB. The half-base is

\[\frac{1}{2}|AB| = 30\tan 65^\circ = 30 \times 2.1445 = 64.34\ \text{cm}\]

\[|AB| = 128.67\ \text{cm}\]

\[\text{Area of } \triangle ABC = \frac{1}{2}\times |AB| \times \text{height} = \frac{1}{2}\times 128.67 \times 30 = 1930.1\ \text{cm}^2\]

Area of the semicircle, with \(\pi = \dfrac{22}{7}\) and \(r = 27.2\):

\[\frac{1}{2}\pi r^2 = \frac{1}{2}\times \frac{22}{7}\times (27.2)^2 = \frac{1}{2}\times \frac{22}{7}\times 739.84 = 1162.6\ \text{cm}^2\]

Therefore

\[\text{Shaded area} = 1930.1 - 1162.6 = 767.5\ \text{cm}^2\]

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Score	1	2	3	4	5	6
Frequency	2	5	x	11	9	10

Find x. Total \(N=37+x\). Since \(P(3)=0.26\):

\[ \frac{x}{37+x}=0.26 \Rightarrow x=9.62+0.26x \Rightarrow 0.74x=9.62 \Rightarrow x=13 \]

So frequencies are \(2,5,13,11,9,10\) with \(N=50\).

(a) Median. With \(N=50\), the median is the mean of the 25th and 26th values. Cumulative frequencies: 2, 7, 20, 31, ... Both the 25th and 26th fall at score 4, so median = 4.

(b) Standard deviation. First the mean:

\[ \Sigma fx = 2+10+39+44+45+60 = 200,\quad \bar{x}=\frac{200}{50}=4 \] \[ \Sigma fx^2 = 2+20+117+176+225+360 = 900 \] \[ \text{Variance}=\frac{\Sigma fx^2}{N}-\bar{x}^2=\frac{900}{50}-4^2=18-16=2 \] \[ \text{SD}=\sqrt{2}\approx 1.41 \]

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(a) Perimeter of the sector. First find the angle \(\theta\) from the area:

\[\text{Area} = \frac{\theta}{360}\pi r^2 \Rightarrow 280 = \frac{\theta}{360}\times\frac{22}{7}\times21^2.\]

\[\frac{22}{7}\times441 = 1386,\qquad \frac{\theta}{360} = \frac{280}{1386} = 0.2020.\]

Arc length \(= \dfrac{\theta}{360}\times2\pi r = 0.2020\times\left(2\times\tfrac{22}{7}\times21\right) = 0.2020\times132 = 26.7\text{ cm}.\)

\[\text{Perimeter} = \text{arc} + 2r = 26.7 + 42 = 68.7\text{ cm}.\]

(b) Vertical angle of the cone. When bent, the arc becomes the base circumference and the sector radius becomes the slant height \(l = 21\) cm. Base radius R:

\[2\pi R = 26.7 \Rightarrow R = \frac{26.7}{2\times\tfrac{22}{7}} = 4.24\text{ cm}.\]

Half the vertical angle \(\alpha\) satisfies \(\sin\alpha = \dfrac{R}{l} = \dfrac{4.24}{21} = 0.2020\), so \(\alpha = 11.66^\circ\).

\[\text{Vertical angle} = 2\alpha = 23.3^\circ \approx 23^\circ.\]

Answer Details

(a) Divide the algebraic fractions.

Dividing by a fraction is the same as multiplying by its reciprocal:

\[\frac{x^{2}-4}{x^{2}+x}\div\frac{x^{2}-4x+4}{x+1}=\frac{x^{2}-4}{x^{2}+x}\times\frac{x+1}{x^{2}-4x+4}\]

Factorise each expression fully:

\[x^{2}-4=(x-2)(x+2),\qquad x^{2}+x=x(x+1),\qquad x^{2}-4x+4=(x-2)^{2}\]

Substitute the factorised forms:

\[=\frac{(x-2)(x+2)}{x(x+1)}\times\frac{x+1}{(x-2)^{2}}\]

Cancel the common factors \((x+1)\) and one \((x-2)\):

\[=\frac{x+2}{x(x-2)}=\frac{x+2}{x^{2}-2x}\]

(b) Using the graph.

Reading the marked points from the diagram, the curve \(y=ax^{2}+bx+c\) cuts the x-axis at \(L(-1,0)\) and \(N(2,0)\) and cuts the y-axis at \(M(0,2)\). The straight line \(y=mx+k\) passes through \(L(-1,0)\) and cuts the curve again at \(H(1.5,\,1.25)\). The reconstructed graph below shows these features.

(ii) Values of \(a\), \(b\), \(c\) and the equation of the curve. (Solved first, as the roots in part (i) depend on this equation.)

Substitute the coordinates of \(L\), \(M\) and \(N\) into \(y=ax^{2}+bx+c\):

\[L(-1,0):\;\;a(-1)^{2}+b(-1)+c=0\;\Rightarrow\;a-b+c=0\quad(1)\]\[M(0,2):\;\;a(0)^{2}+b(0)+c=2\;\Rightarrow\;c=2\quad(2)\]\[N(2,0):\;\;a(2)^{2}+b(2)+c=0\;\Rightarrow\;4a+2b+c=0\quad(3)\]

Put \(c=2\) from (2) into (1) and (3):

\[a-b=-2\quad(1a),\qquad 4a+2b=-2\quad(3a)\]

From (1a), \(b=a+2\). Substitute into (3a):

\[4a+2(a+2)=-2\;\Rightarrow\;6a+4=-2\;\Rightarrow\;6a=-6\;\Rightarrow\;a=-1\]

Then \(b=a+2=1\) and \(c=2\). Therefore

\[a=-1,\qquad b=1,\qquad c=2,\]

and the equation of the curve is

\[y=-x^{2}+x+2.\]

(i) Roots of \(ax^{2}+bx+c=mx+k\).

The solutions are the x-coordinates of the points where the line meets the curve, namely \(L\) and \(H\). First find the line \(y=mx+k\) through \(L(-1,0)\) and \(H(1.5,1.25)\):

\[m=\frac{1.25-0}{1.5-(-1)}=\frac{1.25}{2.5}=0.5\]

Using \(L(-1,0)\): \(0=0.5(-1)+k\Rightarrow k=0.5\), so \(y=0.5x+0.5\).

Now equate the curve and the line:

\[-x^{2}+x+2=0.5x+0.5\]\[-x^{2}+0.5x+1.5=0\;\Rightarrow\;x^{2}-0.5x-1.5=0\]

Solve using the formula \(x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=1,\,b=-0.5,\,c=-1.5\):

\[x=\frac{0.5\pm\sqrt{(-0.5)^{2}-4(1)(-1.5)}}{2}=\frac{0.5\pm\sqrt{0.25+6}}{2}=\frac{0.5\pm\sqrt{6.25}}{2}=\frac{0.5\pm2.5}{2}\]\[x=\frac{0.5+2.5}{2}=1.5\qquad\text{or}\qquad x=\frac{0.5-2.5}{2}=-1\]

The roots are \(x=-1\) and \(x=1.5\), which are exactly the x-coordinates of \(L\) and \(H\) read from the graph.

(iii) Line of symmetry of the curve.

The line of symmetry of \(y=ax^{2}+bx+c\) is \(x=-\dfrac{b}{2a}\). With \(a=-1\) and \(b=1\):

\[x=-\frac{1}{2(-1)}=\frac{1}{2}\]

The line of symmetry is \(x=0.5\), the vertical line midway between the roots \(-1\) and \(2\) (shown dashed on the graph, passing through the maximum point \((0.5,\,2.25)\)).

Answer Details

Construction (ruler and a pair of compasses only).

(a) Rhombus PQRS of side 7 cm with ∠PQR = 60°.

1. Draw the base QR = 7 cm.
2. At Q construct an angle of 60°: with centre Q and a convenient radius draw an arc cutting QR; with the same radius and centre at that cut, step one arc across the first to fix the 60° direction. Along this direction mark off QP = 7 cm.
3. With centre P, radius 7 cm, and centre R, radius 7 cm, draw two arcs meeting at S. Join RS and SP. PQRS is the required rhombus (all four sides 7 cm).

(b) Locating the point X.

1. Bisect ∠PQR. This bisector from Q is the locus of points equidistant from the lines PQ and QR.
2. Construct the perpendicular bisector of QR. This is the locus of points equidistant from Q and R.
3. The two loci intersect at X (shown as dashed construction lines below).

(c) Measuring |XR|.

Because X lies on the bisector of the 60° angle at Q, the line QX makes 30° with QR. Because X also lies on the perpendicular bisector of QR, the foot M of that bisector is the midpoint of QR, so \(QM = \tfrac{7}{2} = 3.5\text{ cm}\). In right-angled triangle QMX (right angle at M):

\[ QX = \frac{QM}{\cos 30^\circ} = \frac{3.5}{0.8660} = 4.04\text{ cm}. \]

Since X is equidistant from Q and R, \(|XR| = |XQ|\). Therefore

\[ |XR| \approx 4.0\text{ cm}. \]

Measured directly on the accurate construction, \(|XR| \approx 4.0\text{ cm}\).