WAEC SSCE - General Mathematics - 1999

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Given, $y = \sqrt{ax-b}$. Squaring both sides, we get: $$y^2 = ax - b$$ Adding b on both sides, we get: $$ax = y^2 + b$$ Dividing both sides by a, we get: $$x = \frac{y^2 + b}{a}$$ Hence, the correct option is $\boxed{\textbf{(D)}\ x = \frac{y^2 + b}{a}}$.

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There are 11 people in total, and we know that some of them can speak English only, some of them can speak French only, and some of them can speak both English and French. Let's denote the number of people who can speak both languages as "x". From the information given, we know that: - 7 people can speak English, including those who can speak both languages (which means that some of these 7 people can also speak French) - 6 people can speak French, including those who can speak both languages (which means that some of these 6 people can also speak English) To find the number of people who can speak both languages, we can use the formula: Total = English only + French only + Both - Neither We know that there are no people who can speak neither language, so we can simplify the formula to: Total = English only + French only + Both Substituting the given values, we get: 11 = 7 - x + 6 - x + x 11 = 13 - x x = 2 So, there are 2 people who can speak both English and French. The probability of choosing a person who can speak both languages out of the 11 people is: Probability = Number of people who can speak both / Total number of people Probability = 2 / 11 Therefore, the answer is \(\frac{2}{11}\).

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To multiply 3.72 by 0.025, we can first ignore the decimal points and treat the numbers as if they were whole numbers, i.e., 372 and 25. We multiply these numbers to get 9300. Next, we need to determine the power of 10 that represents the decimal point in the product. To do this, we count the total number of decimal places in the two original numbers (three in 3.72 and two in 0.025), which gives us five decimal places. Therefore, our answer needs to be in scientific notation with a power of 10 that is negative five. Finally, we can insert the decimal point into the product and express the answer in scientific notation as follows: 3.72 x 0.025 = 0.093 = 9.3 x 10^(-2) Therefore, the answer is \(9.3\times 10^{-2}\).

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Since UV is a diameter of circle VSU, we know that ∠VUS = 90°. Also, since RST is tangent to circle VSU at S, then ∠VST = 90°. Therefore, ∠VUS + ∠VST = 90° + 90° = 180°. Since VSU is a straight line, then ∠SUV = 180° - ∠SVU - ∠VUS = 180° - 50° - 90° = 40°. Since RST is tangent to circle VSU at S, then ∠RST = ∠SVU = 50° (tangent and radius form a right angle). Finally, we can calculate ∠RSV using the fact that the angles of a triangle sum up to 180°: ∠RSV = 180° - ∠VUS - ∠SUV - ∠RST = 180° - 90° - 40° - 50° = 0° Therefore, the answer is (d) 40°. Note that this is a trick question, as ∠RSV is not defined in this case since R, S, and V are collinear.

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We can solve this problem using trigonometry. The angle of depression is the angle between the horizontal line and the line of sight from the top of the cliff to the boat. To find this angle, we need to first find the length of the line of sight. Let's call the angle of depression θ, the height of the cliff h, and the distance from the foot of the cliff to the boat d. From the problem statement, we know that h = 20m and d = 75m. We can use the tangent function to find the length of the line of sight: tan(θ) = h / d tan(θ) = 20 / 75 tan(θ) = 0.2667 To find θ, we need to take the inverse tangent (or arctangent) of both sides: θ = tan^-1(0.2667) Using a calculator, we get: θ ≈ 14.9° Therefore, the angle of depression of the boat from the top of the cliff is approximately 14.9°. Answer: 14.9°

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Amina started with m mangoes, ate 3 and shared the remainder equally with her brother Uche, which means she gave \(\frac{m-3}{2}\) mangoes to Uche. The statement "each had at least 10" means that the number of mangoes each of them had is greater than or equal to 10. Therefore, the correct inequality is: \[\frac{m-3}{2}\ge10\] This means that the number of mangoes Amina started with minus 3, divided by 2, must be greater than or equal to 10, which implies that each of them had at least 10 mangoes.

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We can simplify the given equation as follows: $$K\sqrt{28}+\sqrt{63}-\sqrt{7}=0$$ $$K\sqrt{4\cdot7}+\sqrt{9\cdot7}-\sqrt{7}=0$$ $$2K\sqrt{7}+3\sqrt{7}-\sqrt{7}=0$$ $$2K\sqrt{7}+2\sqrt{7}=0$$ $$2\sqrt{7}(K+1)=0$$ Since $\sqrt{7}$ is not zero, we can divide both sides by $2\sqrt{7}$ to get: $$K+1=0$$ Therefore, $K=-1$. So the answer is (B) -1.

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To solve the inequality 2x + 3 < 5x, we want to isolate x on one side of the inequality sign. We can do this by subtracting 2x from both sides of the inequality: 2x + 3 < 5x -2x -2x 3 < 3x Now we have 3 < 3x. To isolate x, we can divide both sides of the inequality by 3: 3 < 3x /3 /3 1 < x So the solution to the inequality 2x + 3 < 5x is x > 1. This means that any value of x that is greater than 1 will make the inequality true. For example, if x = 2, then 2(2) + 3 = 7, which is indeed less than 5(2) = 10. Therefore, the correct option is "x > 1".

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Given the quadratic equation: \(x^2 +15x + 50 = ax^2 + bx + c = 0\), where a, b, and c are constants. The question is asking which of the following statements is not true. We can first use the quadratic formula to find the values of x in terms of a, b, and c: $$x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ Comparing this with the given equation, we have: $$a = 1, \quad b = 15, \quad c = 50$$ Using the quadratic formula, we have: $$x = \dfrac{-15 \pm \sqrt{15^2 - 4(1)(50)}}{2(1)} = -5, -10$$ So, statement (a) x = -5 is true. To find out which statement is not true, we can check each option. Statement (b) x = 10: We can see that this is not a solution to the quadratic equation. Statement (c) x + 10 = 0: This can be rewritten as x = -10, which is a solution to the quadratic equation. Statement (d) bc = 750: Multiplying the coefficients of x gives: b*c = a*(-50) = -50a, so this statement is true. Therefore, the statement that is not true is (b) x = 10.

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To simplify the given expression, we need to factorize the numerator and denominator, then cancel out any common factors. The numerator can be factorized as the difference of two squares: 2 - 18m^2 = 2(1 - 9m^2) = 2(1 + 3m)(1 - 3m) The denominator is already factorized: 1 + 3m Now we can cancel out the common factor of (1 + 3m) from the numerator and denominator: \[\frac{2-18m^2}{1+3m} = \frac{2(1+3m)(1-3m)}{1+3m} = 2(1-3m)\] Therefore, the simplified form of the expression is 2(1-3m).

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The volume of a pyramid is given by the formula: V = 1/3 * Base Area * Height The base of a square pyramid is a square, so the Base Area is equal to the side length squared (B = s²). In this problem, the height is given as 15cm, and the volume is given as 80cm³. We are asked to find the length of a side of the base. Let's call that length "s". So we can set up an equation using the formula for the volume of a pyramid: 80 = 1/3 * s² * 15 To solve for s, we can start by simplifying the equation: 240 = s² * 15 Dividing both sides by 15: 16 = s² Taking the square root of both sides (and remembering to consider both the positive and negative square roots): s = ±4 Since we are dealing with the length of a side of a pyramid, the answer cannot be negative. Therefore, the length of a side of the base is 4cm (option C). Answer: 4.0cm

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To find $\sqrt{K}\backslash sqrt\{K\}$ given that $81\times 2^{2n-2}=K81\; \backslash times\; 2^\{2n-2\}\; =\; K$, we need to simplify the

expression step-by-step.

First,

rewrite $8181$ in terms of powers of $33$: $81=3^{4}81\; =\; 3^4$

So,

the given equation becomes: $K=3^4\times 2^{2n-2}K\; =\; 3^4\; \backslash times\; 2^\{2n-2\}$

Next,

we need to find $\sqrt{K}\backslash sqrt\{K\}$: $\sqrt{K}=\sqrt{3^4\times 2^{2n-2}}\backslash sqrt\{K\}\; =\; \backslash sqrt\{3^4\; \backslash times\; 2^\{2n-2\}\}$

Using the property of square roots $\sqrt{a\times b}=\sqrt{a}\times \sqrt{b}\backslash sqrt\{a\; \backslash times\; b\}\; =\; \backslash sqrt\{a\}\; \backslash times\; \backslash sqrt\{b\}$: $\sqrt{K}=\sqrt{3^4}\times \sqrt{2^{2n-2}}\backslash sqrt\{K\}\; =\; \backslash sqrt\{3^4\}\; \backslash times\; \backslash sqrt\{2^\{2n-2\}\}$

Since $\sqrt{3^4}=3^2=9\backslash sqrt\{3^4\}\; =\; 3^2\; =\; 9$ and $\sqrt{2^{2n-2}}=2^{(2n-2)\mathrm{/}2}=2^{n-1}\backslash sqrt\{2^\{2n-2\}\}\; =\; 2^\{(2n-2)/2\}\; =\; 2^\{n-1\}$: $\sqrt{K}=9\times 2^{n-1}\backslash sqrt\{K\}\; =\; 9\; \backslash times\; 2^\{n-1\}$

Thus, the correct answer is: $\boxed{{\displaystyle 9\times 2^{n-1}}}\backslash boxed\{9\; \backslash times\; 2^\{n-1\}\}$ = 

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To simplify \(\frac{4}{x+1}-\frac{3}{x-1}\), we first need to find a common denominator for the two fractions. The common denominator is \((x+1)(x-1)\), so we can rewrite the expression as follows: $$\frac{4(x-1)}{(x+1)(x-1)} - \frac{3(x+1)}{(x+1)(x-1)}$$ Next, we can combine the two fractions by subtracting the second fraction from the first, giving us: $$\frac{4(x-1)-3(x+1)}{(x+1)(x-1)}$$ Simplifying the numerator, we get: $$\frac{4x-4-3x-3}{(x+1)(x-1)} = \frac{x-7}{x^2-1}$$ Therefore, the answer is $\frac{x-7}{x^2-1}$

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We know that the area of a triangle is given by the formula: Area = (1/2) x base x height Let the height of the triangle be h and the base be b. We are given that h:b = 1:3, which means that h = (1/3)b. Substituting this value of h in the formula for the area, we get: 216 = (1/2) x b x (1/3)b Multiplying both sides by 6 to get rid of the fraction, we get: 1296 = b^2 Taking the square root of both sides, we get: b = ±36 Since the base of a triangle cannot be negative, we take the positive value of b, which is b = 36. Therefore, the length of the base of the triangle is 36 cm, and the correct option is "36cm".

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The average speed of a car is given by the total distance traveled divided by the total time taken. Let's assume that the car traveled a distance of D km in the given time. Then, we can write: Total time taken = 1 hour + 2 hours = 3 hours Distance traveled in the first hour = x km Distance traveled in the next 2 hours = y × 2 = 2y km Total distance traveled = D = x + 2y km Therefore, the average speed of the car is: Average speed = Total distance / Total time = D / 3 = (x + 2y) / 3 km/h Hence, the answer is option (C): \(\frac{x + 2y}{3}kmh^{-1}\).

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In a right angle triangle, the square of the hypotenuse (the longest side) is equal to the sum of the squares of the other two sides. Using this theorem, we can write the following equation: x² = 8² + (x-2)² Expanding the brackets, we get: x² = 64 + x² - 4x + 4 Simplifying and rearranging, we get: 4x = 68 x = 17 Therefore, the value of x is 17, which is the second option.

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Given that \(\tan x = \frac{1}{\sqrt{3}}\), we can draw a right-angled triangle where the opposite side is equal to 1 and the adjacent side is equal to \(\sqrt{3}\), and the hypotenuse is equal to 2. Therefore, sin x = \(\frac{1}{2}\) and cos x = \(\frac{\sqrt{3}}{2}\). Now, cos x - sin x = \(\frac{\sqrt{3}}{2}\) - \(\frac{1}{2}\) = \(\frac{\sqrt{3}-1}{2}\). Hence, the correct option is \(\frac{\sqrt{3}-1}{2}\).

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To solve the problem, we can use the formula for the volume of a cylinder: V = πr\(^2\)h where V is the volume of the water in the cylinder, r is the radius of the cylinder, and h is the height of the water in the cylinder. We know that water is flowing into the cylinder at a rate of 5π cm\(^3\) per second, which means that the volume of the water in the cylinder is increasing at a rate of 5π cm\(^3\) per second. We want to find the height of the water in the cylinder at the end of 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in volume of the water in the cylinder after 9 seconds is: ΔV = 5π x 9 ΔV = 45π So, the volume of the water in the cylinder after 9 seconds is: V = πr\(^2\)h + 45π Since the radius of the cylinder is 3cm, we can substitute this value into the formula: V = π x 3\(^2\) x h + 45π V = 9πh + 45π We know that the volume of the water in the cylinder after 9 seconds is equal to the height of the water multiplied by the area of the base of the cylinder (which is πr\(^2\)): V = πr\(^2\)h 9πh + 45π = π x 3\(^2\) x h 9πh + 45π = 9πh 45π = 0 This equation has no solution, which means that something went wrong in our calculations. We made a mistake in assuming that the volume of the water in the cylinder was increasing at a constant rate of 5π cm\(^3\) per second. In reality, the rate of increase of the volume of water is not constant, because the height of the water in the cylinder is also increasing. To calculate the correct height of the water in the cylinder after 9 seconds, we need to integrate the rate of increase of the volume of water over time. The rate of increase of the volume of water at any time t is given by: dV/dt = πr\(^2\) dh/dt We know that the radius of the cylinder is 3cm and the rate of flow of water is 5π cm\(^3\) per second. So, we can write: dV/dt = π x 3\(^2\) x dh/dt 5π = 9π dh/dt dh/dt = 5/9 cm/s This means that the height of the water in the cylinder is increasing at a rate of 5/9 cm/s. We want to find the height of the water in the cylinder after 9 seconds. Let's use h to represent the height of the water in the cylinder after 9 seconds. The increase in height of the water in the cylinder after 9 seconds is: Δh = (5/9) x 9 Δh = 5 So, the height of the water in the cylinder after 9 seconds is: h = 5 + 0 h = 5cm Therefore, the correct answer is 5cm.

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(a) \(3y-2x=21\) and \(4y+5x=5\). Multiply the first by \(5\) and the second by \(2\): \[15y-10x=105,\qquad 8y+10x=10.\] Adding: \(23y=115\Rightarrow y=5\). Then \(3(5)-2x=21\Rightarrow-2x=6\Rightarrow x=-3\). So \(x=-3,\;y=5\).

(b) Card \(6\) wins \(\$60\); cards \(1\text{–}5\) lose \(\$10\times(\text{number})\).

• (i) A loss occurs whenever cards \(1,2,3,4,5\) are picked: \[P(\text{losing})=\frac{5}{6}.\]
• (ii) Losing exactly \(\$20\) in two picks: the outcomes per card are \(-10,-20,-30,-40,-50,+60\). The only pair of picks totalling \(-\$20\) is card \(1\) followed by card \(1\) \((-10-10=-20)\). \[P=\frac16\times\frac16=\frac{1}{36}.\]

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In \(\triangle PBC\), the right angle is at \(C\), \(|BC|=5\text{ cm}\), \(\angle PBC=37^{\circ}\), and \(P\) is the midpoint of \(AC\).

(a) \[\tan37^{\circ}=\frac{PC}{BC}\Rightarrow PC=5\tan37^{\circ}=5\times0.7536=3.768\text{ cm}.\] Since \(P\) is the midpoint, \(|AC|=2\times PC=7.536\approx\) 7.54 cm (3 s.f.).

(b) In \(\triangle ABC\), \[\tan(\angle ABC)=\frac{AC}{BC}=\frac{7.536}{5}=1.5072\Rightarrow\angle ABC=56.44^{\circ}.\] Therefore \[\angle PBA=\angle ABC-\angle PBC=56.44^{\circ}-37^{\circ}\approx19.4^{\circ}.\]

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Reading the diagram. The wall \(TQ\) is vertical and the ground \(QF\) is horizontal, meeting at the right angle \(Q\). The ladder \(TF = 10\text{ m}\) makes \(70^\circ\) with the horizontal at \(F\). When the foot is pulled out \(1\text{ m}\) to \(F'\), the top slides down to \(T'\), with \(T'F' = 10\text{ m}\) still.

(a) Height reached up the wall. In right triangle \(TQF\):

(b)(i) Distance the top slides down. First the original foot distance:

New foot distance: \(|QF'| = 3.420 + 1 = 4.420\text{ m}.\) New height, from \(|T'F'| = 10\):

Distance slid down:

(b)(ii) Angle \(\angle QF'T'\). In right triangle \(QF'T'\):

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(a) Completing the table of values

For \(y = 9\cos x + 5\sin x\), evaluate at the missing angles (to one decimal place):

\(x = 0^\circ:\; 9(1)+5(0)=9.0\)
\(x = 60^\circ:\; 9(0.5)+5(0.866)=4.5+4.33=8.8\)
\(x = 90^\circ:\; 9(0)+5(1)=5.0\)
\(x = 180^\circ:\; 9(-1)+5(0)=-9.0\)

(b) Graph of \(y = 9\cos x + 5\sin x\)

Using a scale of 2 cm to 30° on the x-axis and 2 cm to 1 unit on the y-axis, plot the eight points from the table and join them with a smooth curve for \(0^\circ \le x \le 210^\circ\).

(c) Solving from the graph

(i) \(9\cos x + 5\sin x = 0\). This is where the curve cuts the x-axis (\(y = 0\)). Reading the graph, the curve crosses between \(x=110^\circ\) and \(x=120^\circ\), at:

(Check: \(\tan x = -\dfrac{9}{5}=-1.8\Rightarrow x = 180^\circ-61^\circ = 119^\circ.\))

(ii) \(9\cos x + 5\sin x = 3.5\). Draw the horizontal line \(y = 3.5\) and read where it meets the curve. It meets the descending part of the curve at:

(Check, using \(9\cos x+5\sin x = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}=10.3\) and \(\alpha = \tan^{-1}\frac{5}{9}=29^\circ\): \(10.3\cos(x-29^\circ)=3.5\Rightarrow \cos(x-29^\circ)=0.340\Rightarrow x-29^\circ=70^\circ\Rightarrow x\approx 99^\circ.\))

(d) Maximum value of y

The highest point of the curve occurs near \(x = 30^\circ\). Since \(y = R\cos(x-\alpha)\) with \(R=\sqrt{9^2+5^2}\):

The maximum value of \(y\) is 10.3 (occurring at \(x \approx 29^\circ\)).

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Reading the diagram. \(ABCD\) is a trapezium with \(AD \parallel BC\) and \(\angle ABC = 90^\circ\). From the figure: \(|AD| = 15\text{ cm}\) (top), \(|BD| = 17\text{ cm}\) (diagonal), \(|BC| = 9\text{ cm}\) (bottom).

(a) \(|AB|\). Because \(AD \parallel BC\) and \(\angle ABC = 90^\circ\), the side \(AB\) is perpendicular to both parallels, so \(\angle DAB = 90^\circ\). Triangle \(ABD\) is right-angled at \(A\):

(b) Area of \(\triangle BCD\). The perpendicular distance between the parallel sides \(AD\) and \(BC\) equals \(|AB| = 8\text{ cm}\), so the height of \(\triangle BCD\) on base \(BC\) is \(8\text{ cm}\):

(c) \(|CD|\). Place \(B=(0,0)\), \(C=(9,0)\), \(A=(0,8)\), \(D=(15,8)\). Then

(d) Perimeter of the trapezium.

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(a) Cylindrical pipe, length \(28\text{ m}=2800\text{ cm}\), internal radius \(r=3.5\text{ cm}\), external radius \(R=5\text{ cm}\).

(i) Volume of metal used.

The metal is the hollow shell between the outer and inner cylinders:

The volume of metal is \(112200\text{ cm}^3\).

(ii) Volume of water the pipe holds when full.

This is the inner cylinder's volume:

Convert to litres using \(1\text{ litre}=1000\text{ cm}^3\):

The pipe holds \(107.8\) litres (to 1 d.p.).

(b) Tangent \(MP\) at \(M\), chord \(LN\parallel MP\): show \(\triangle LMN\) is isosceles.

By the tangent-chord (alternate segment) theorem, the angle between tangent \(MP\) and chord \(MN\) equals the angle in the alternate segment standing on \(MN\):

Since \(LN\parallel MP\) and \(MN\) is a transversal, alternate angles are equal:

From (1) and (2):

In triangle \(LMN\) the base angles at \(L\) and \(N\) are equal, so the sides opposite them are equal, i.e. \(|MN|=|ML|\).

Therefore triangle \(LMN\) is isosceles. (Q.E.D.)

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Let the cost be \(C=a+bn\), where \(a\) is the constant part and \(b\) the cost per pupil.

\(50\) pupils: \(a+50b=15705\). \(40\) pupils: \(a+40b=13305\). Subtracting: \(10b=2400\Rightarrow b=240\); then \(a=15705-50(240)=3705\). So \[C=3705+240n.\]

(a) For \(n=44\): \[C=3705+240(44)=3705+10560=\$14{,}265.\]

(b) Revenue \(=360n\). To run without a loss, \(360n\ge3705+240n\Rightarrow120n\ge3705\Rightarrow n\ge30.875\). The least whole number of pupils is 31.

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(a) \[\sin105^{\circ}=\sin(60^{\circ}+45^{\circ})=\sin60^{\circ}\cos45^{\circ}+\cos60^{\circ}\sin45^{\circ}=\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}+\frac12\cdot\frac{\sqrt2}{2}=\frac{\sqrt6+\sqrt2}{4}.\]

(b) The odd house numbers \(11,13,15,\dots\) form an AP with \(a=11,\;d=2\). Sum of \(n\) terms: \[\frac{n}{2}\bigl(2(11)+(n-1)2\bigr)=n(n+10)=551\Rightarrow n^{2}+10n-551=0.\] \((n+29)(n-19)=0\Rightarrow n=19\). There are 19 houses.

(c) GP with \(t_{1}=2\), \(t_{3}=\tfrac29\): \(ar^{2}=\tfrac29\) and \(a=2\Rightarrow r^{2}=\tfrac19\Rightarrow r=\tfrac13\). The common ratio is \(\tfrac13\). Fifth term: \[t_{5}=ar^{4}=2\left(\frac13\right)^{4}=\frac{2}{81}.\]

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(a) Let the cost price be \(C\). Marked to gain \(20\%\): marked \(=1.2C\). Sold at \(10\%\) reduction: \[0.9\times1.2C=1.08C=864\Rightarrow C=\frac{864}{1.08}=\text{N}800.\] The original cost was N800.

(b) Length \((x+5)\), width \((x-2)\), diagonal \((x+6)\). By Pythagoras: \[(x+5)^{2}+(x-2)^{2}=(x+6)^{2}.\] \[x^{2}+10x+25+x^{2}-4x+4=x^{2}+12x+36\Rightarrow x^{2}-6x-7=0\Rightarrow(x-7)(x+1)=0.\] Taking the positive value, \(x=7\).

(ii) Area \(=(x+5)(x-2)=12\times5=60\text{ m}^{2}.\)

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(a) Value of x. Total frequency \(=5+(x-1)+x+9+4+1=18+2x\). Sum of \(fx\):

Since mean \(=5\):

So the frequencies are \(5, 10, 11, 9, 4, 1\) with total \(N=40\).

(b)(i) Mode. The highest frequency (11) is at mark 5, so mode = 5.

(ii) Median. With \(N=40\), the median is the mean of the 20th and 21st values. Cumulative frequencies: 5, 15, 26, ... The 20th and 21st both fall at mark 5, so median = 5.

(c) P(at least 7 marks). Marks 7 and 8 give \(4+1=5\):

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Scale: 2 cm to 2 units on both axes, i.e. 1 unit = 1 cm. So a length of 5 cm on the paper is the same as 5 units on the graph, and a side of the required rhombus measures 5 units.

Notice first that P(-5, 3) and Q(2, 3) have the same y-coordinate (3), so PQ is a horizontal line of length \(|2-(-5)| = 7\) units = 7 cm.

(a) The straight line PQ

P(-5, 3) and Q(2, 3) are plotted and joined by a straight line. This is the horizontal segment shown in blue at the level \(y = 3\).

(b) Locus L of points equidistant from P and Q

The set of points equidistant from P and Q is the perpendicular bisector of PQ. It is constructed by opening the compass to more than half of PQ, drawing equal arcs centred on P and on Q above and below the line, and joining their points of intersection. This bisector cuts PQ at its midpoint

and runs vertically, so the locus is the line \(x = -1.5\), shown dashed and labelled L.

(c) Locating R and S so that PRQS is a rhombus of side 5 cm

In the rhombus PRQS the four equal sides are PR, RQ, QS and SP, each 5 cm. Hence each of R and S must be 5 cm from P and 5 cm from Q. Setting the compass to 5 cm (= 5 units) and drawing arcs centred on P and on Q, the arcs intersect on the locus L at two points: R above PQ and S below PQ. Because PR = RQ = QS = SP = 5 cm, the figure PRQS is a rhombus, and its diagonals PQ and RS cross at right angles at M.

(d) Results

(i) Coordinates of R and S. R and S lie on L, so each has \(x = -1.5\). Taking R\((-1.5,\,y)\) with PR = 5:

Therefore, reading from the construction,

(ii) Area of the rhombus. The diagonals of a rhombus bisect each other at right angles, so the area is half the product of the diagonals. Here the diagonals are PQ and RS: