MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 2)
Examiner: Vashi S Y
Instructions & Guidelines:
- This simulated workspace contains Questions 1 to 3 totaling exactly 100 marks.
- Always attempt writing out full steps in the workspace lines first before checking the solutions.
- 💡 Tip: Use your browser's print function (Ctrl/Cmd + P or click the Download / Print PDF button) to save this out as a clean, structured worksheet!
QUESTION 1: ANALYTICAL GEOMETRY
[27 Marks]In the Cartesian plane, the points $A(-1; 6)$, $B(3; 8)$, and $C(5; 0)$ form the vertices of a triangle. Point $M$ is the midpoint of $AC$.
Apply the midpoint formula using coordinates $A(-1; 6)$ and $C(5; 0)$:
$$M = \left( \frac{x_A + x_C}{2}; \frac{y_A + y_C}{2} \right)$$
$$M = \left( \frac{-1 + 5}{2}; \frac{6 + 0}{2} \right) = \left( \frac{4}{2}; \frac{6}{2} \right)$$
$$M = (2; 3)$$
- Correct $x$-coordinate calculation ($2$) (1 Mark)
- Correct $y$-coordinate calculation ($3$) (1 Mark)
Apply the distance formula to $A(-1; 6)$ and $C(5; 0)$:
$$AC = \sqrt{(x_C - x_A)^2 + (y_C - y_A)^2}$$
$$AC = \sqrt{(5 - (-1))^2 + (0 - 6)^2} = \sqrt{(6)^2 + (-6)^2}$$
$$AC = \sqrt{36 + 36} = \sqrt{72}$$
$$AC = 6\sqrt{2} \text{ units}$$
- Correct substitution into distance formula (1 Mark)
- Simplification under the square root to $\sqrt{72}$ (1 Mark)
- Final simplest surd conversion $6\sqrt{2}$ (1 Mark)
Apply the gradient formula for points $A(-1; 6)$ and $C(5; 0)$:
$$m_{AC} = \frac{y_C - y_A}{x_C - x_A}$$
$$m_{AC} = \frac{0 - 6}{5 - (-1)} = \frac{-6}{6}$$
$$m_{AC} = -1$$
- Correct formula written (1 Mark)
- Correct substitution of coordinates (1 Mark)
- Final simplified gradient of $-1$ (1 Mark)
Perpendicular lines have gradients that multiply to $-1$ ($m_1 \times m_2 = -1$):
$$m_{\perp} \times m_{AC} = -1 \implies m_{\perp} \times (-1) = -1 \implies m_{\perp} = 1$$
Substitute gradient $m = 1$ and point $B(3; 8)$ into standard equation:
$$y - 8 = 1(x - 3) \implies y - 8 = x - 3$$
$$y = x + 5$$
- Gradient of perpendicular line $m = 1$ (1 Mark)
- Substituting coordinates correctly (1 Mark)
- Correct standard form equation $y = x + 5$ (1 Mark)
Let $\theta$ be the angle of inclination of the line $AC$:
$$\tan \theta = m_{AC} \implies \tan \theta = -1$$
Calculate reference angle:
$$\theta_{\text{ref}} = \tan^{-1}(1) = 45^\circ$$
Since the gradient is negative, the angle of inclination is obtuse (lies in the second quadrant):
$$\theta = 180^\circ - 45^\circ$$
$$\theta = 135^\circ$$
- Setting up $\tan \theta = -1$ (1 Mark)
- Calculating reference angle of $45^\circ$ (1 Mark)
- Correct final angle of inclination $135^\circ$ (1 Mark)
Since $ABCD$ is a parallelogram, the diagonals $AC$ and $BD$ bisect each other. Thus, the midpoint of $BD$ is identical to the midpoint of $AC$, which is $M(2; 3)$:
$$x_M = \frac{x_B + x_D}{2} \implies 2 = \frac{3 + x_D}{2} \implies 4 = 3 + x_D \implies x_D = 1$$
$$y_M = \frac{y_B + y_D}{2} \implies 3 = \frac{8 + y_D}{2} \implies 6 = 8 + y_D \implies y_D = -2$$
$$D = (1; -2)$$
- Setting up $x$-midpoint equation and solving $x = 1$ (2 Marks)
- Setting up $y$-midpoint equation and solving $y = -2$ (2 Marks)
To calculate the area, we can find the base ($AC$) and height ($BM$) because $BM \perp AC$ (since diagonal $BD \perp AC$ at midpoint $M$):
1. Length $AC$ base using $A(-1; 6)$ and $C(5; 0)$:
$$AC = \sqrt{(5 - (-1))^2 + (0 - 6)^2} = \sqrt{6^2 + (-6)^2} = \sqrt{72} = 6\sqrt{2} \text{ units}$$
2. Perpendicular height $BM$ using $B(3; 8)$ and $M(2; 3)$:
$$BM = \sqrt{(2 - 3)^2 + (3 - 8)^2} = \sqrt{(-1)^2 + (-5)^2} = \sqrt{1 + 25} = \sqrt{26} \text{ units}$$
3. Area of Parallelogram $ABCD = AC \times BM$ (since $ABCD$ consists of two congruent triangles $\triangle ABC$ and $\triangle ADC$):
$$\text{Area} = AC \cdot BM = \sqrt{72} \cdot \sqrt{26} = \sqrt{1872} = 12\sqrt{13}$$
$$\text{Area} \approx 43.27 \text{ square units}$$
- Noting perpendicular diagonal height relationship ($BM \perp AC$) (1 Mark)
- Correctly substituting & calculating $AC = \sqrt{72}$ (2 Marks)
- Correctly substituting & calculating $BM = \sqrt{26}$ (2 Marks)
- Applying parallelogram area formula (2 Marks)
- Substituting values and calculating correct final area (2 Marks)
QUESTION 2: TRIGONOMETRY
[26 Marks]Given: $5\tan \theta + 12 = 0$ and $90^\circ \le \theta \le 270^\circ$.
Isolate the trig ratio: $5\tan\theta = -12 \implies \tan \theta = -\frac{12}{5} = \frac{y}{x}$
Since $\tan \theta < 0$ (negative in Q2 & Q4) and the domain is $90^\circ \le \theta \le 270^\circ$ (encompasses Q2 & Q3), the angle $\theta$ lies in Quadrant 2.
Thus, $x$ is negative and $y$ is positive: $x = -5$ and $y = 12$.
Apply Pythagoras to solve for hypotenuse $r$ (always positive):
$$r^2 = x^2 + y^2 = (-5)^2 + 12^2 = 25 + 144 = 169 \implies r = 13$$
Write down trig ratios: $\sin \theta = \frac{y}{r} = \frac{12}{13}$ and $\cos \theta = \frac{x}{r} = -\frac{5}{13}$.
Evaluate the expression:
$$\cos\theta - \sin\theta = -\frac{5}{13} - \frac{12}{13} = -\frac{17}{13}$$
- Identifying Quadrant 2 with a sketch (2 Marks)
- Using Pythagoras to find $r = 13$ (1 Mark)
- Stating correct $\sin \theta = \frac{12}{13}$ ratio (1 Mark)
- Stating correct $\cos \theta = -\frac{5}{13}$ ratio (1 Mark)
- Final calculation subtraction $-\frac{17}{13}$ (1 Mark)
Method 1: Square Identity Solution (Recommended)
Recall the fundamental trigonometric identity: $\sin^2\theta + \cos^2\theta = 1$
Rearranging this yields: $1 - \sin^2\theta = \cos^2\theta$
From Question 2.1.1, we proved that $\cos\theta = -\frac{5}{13}$:
$$1 - \sin^2\theta = \cos^2\theta = \left(-\frac{5}{13}\right)^2$$
$$= \frac{25}{169}$$
Method 2: Direct Substitution Solution
Substitute the value of $\sin\theta = \frac{12}{13}$ directly into the given expression:
$$1 - \sin^2\theta = 1 - \left(\frac{12}{13}\right)^2$$
$$= 1 - \frac{144}{169}$$
$$= \frac{169 - 144}{169}$$
$$= \frac{25}{169}$$
- Identifying identity substitution or substituting $\sin\theta = \frac{12}{13}$ ratio (1 Mark)
- Applying the squaring operation correctly to yield $\frac{144}{169}$ (or state $\cos^2\theta = (-\frac{5}{13})^2$) (1 Mark)
- Final correct simplified fraction $\frac{25}{169}$ (1 Mark)
Apply reduction formulae and co-functions step-by-step:
- $\sin(180^\circ + x) = -\sin x \quad [\text{Q3 negative}]$
- $\cos(90^\circ - x) = \sin x \quad [\text{Co-function Q1}]$
- $\tan(180^\circ - x) = -\tan x \quad [\text{Q2 negative}]$
- $\sin(360^\circ - x) = -\sin x \quad [\text{Q4 negative}]$
Substitute back into the expression:
$$= \frac{(-\sin x) \cdot \sin x}{(-\tan x) \cdot (-\sin x)} = \frac{-\sin^2 x}{\tan x \cdot \sin x}$$
Rewrite $\tan x$ as $\frac{\sin x}{\cos x}$:
$$= \frac{-\sin^2 x}{\frac{\sin x}{\cos x} \cdot \sin x} = \frac{-\sin^2 x}{\frac{\sin^2 x}{\cos x}} = -\sin^2 x \cdot \frac{\cos x}{\sin^2 x}$$
$$= -\cos x$$
- Reduction $\sin(180^\circ + x) = -\sin x$ (1 Mark)
- Co-function $\cos(90^\circ - x) = \sin x$ (1 Mark)
- Reduction $\tan(180^\circ - x) = -\tan x$ (1 Mark)
- Reduction $\sin(360^\circ - x) = -\sin x$ (1 Mark)
- Quotient substitution $\tan x = \frac{\sin x}{\cos x}$ (1 Mark)
- Final simplified ratio $-\cos x$ (1 Mark)
Start with the Left Hand Side (LHS) and find the lowest common denominator, $\text{LCD} = \sin x(1 + \cos x)$:
$$\text{LHS} = \frac{\sin^2 x + (1 + \cos x)^2}{\sin x(1 + \cos x)}$$
Expand binomial numerator terms and collect like terms:
$$\text{LHS} = \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x(1 + \cos x)}$$
Substitute the fundamental square identity $\sin^2 x + \cos^2 x = 1$:
$$\text{LHS} = \frac{1 + 1 + 2\cos x}{\sin x(1 + \cos x)} = \frac{2 + 2\cos x}{\sin x(1 + \cos x)}$$
Extract the common factor $2$ in the numerator:
$$\text{LHS} = \frac{2(1 + \cos x)}{\sin x(1 + \cos x)}$$
$$\text{LHS} = \frac{2}{\sin x} = \text{RHS}$$
- Finding the common denominator $\sin x(1+\cos x)$ (1 Mark)
- Correctly expanding binomial $(1+\cos x)^2$ (1 Mark)
- Substituting the square identity $\sin^2 x + \cos^2 x = 1$ (1 Mark)
- Common factor grouping $2(1 + \cos x)$ (1 Mark)
- Canceling common bracket to show LHS = RHS (1 Mark)
Factorise the quadratic trinomial:
$$(3\tan\theta + 1)(\tan\theta - 1) = 0$$
Set each linear binomial factor equal to zero:
$$\tan \theta = -\frac{1}{3} \quad \text{or} \quad \tan \theta = 1$$
Determine General Solutions (period of tangent is $180^\circ$):
1. For $\tan \theta = -\frac{1}{3}$ (Reference Angle $\approx 18.43^\circ$):
$$\theta \approx -18.43^\circ + k \cdot 180^\circ \quad \text{or} \quad 161.57^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z}$$
2. For $\tan \theta = 1$ (Reference Angle $= 45^\circ$):
$$\theta = 45^\circ + k \cdot 180^\circ, \quad k \in \mathbb{Z}$$
- Correct binomial factorisation factors $(3\tan\theta + 1)(\tan\theta - 1) = 0$ (1 Mark)
- Correct values $\tan \theta = -\frac{1}{3}$ and $\tan \theta = 1$ (1 Mark)
- Calculating reference angle $18.43^\circ$ (1 Mark)
- General solution $\theta = -18.43^\circ + k \cdot 180^\circ$ (1 Mark)
- Calculating reference angle $45^\circ$ (1 Mark)
- General solution $\theta = 45^\circ + k \cdot 180^\circ$ with $k \in \mathbb{Z}$ (2 Marks)
QUESTION 3: EUCLIDEAN GEOMETRY
[47 Marks]In Circle geometry, you are required to reproduce the proofs of formal core theorems. Ensure your reasons match the exact, standard examination conditions.
Hint: Draw circle center $O$ with chord $AB$ and perpendicular $OM \perp AB$. Prove $AM = MB$.
Construction: Draw circle center $O$ with chord $AB$. Draw perpendicular line $OM \perp AB$ where $M$ is on $AB$. Join radii $OA$ and $OB$.
| Statement | Reason |
|---|---|
| In $\triangle OMA$ and $\triangle OMB$: | |
| $1. \quad OA = OB$ | Radii of circle |
| $2. \quad OM = OM$ | Common side |
| $3. \quad \hat{M}_1 = \hat{M}_2 = 90^\circ$ | Given ($OM \perp AB$) |
| $\therefore \triangle OMA \equiv \triangle OMB$ | Right-angle, Hypotenuse, Side (RHS) |
| $\dots \implies AM = MB$ | Corresponding sides of congruent triangles |
- Radii construction $OA$ and $OB$ (1 Mark)
- Radii equality $OA = OB$ with reason (1 Mark)
- Common side $OM$ (1 Mark)
- Equal perpendicular angles (1 Mark)
- Congruency statement ($\triangle OMA \equiv \triangle OMB$) with reason "RHS" (1 Mark)
- Deduction $AM = MB$ (1 Mark)
Circle Rider 1 Scenario: In the circle below, $O$ is the center. $A$, $B$, $C$, and $D$ lie on the circumference. $AB \parallel CD$, and $D\hat{A}C = 36^\circ$.
Since $AB$ is a diameter, the angle subtended by it at the circumference is a right-angle:
$$A\hat{C}B = 90^\circ \quad [\text{angle in a semi-circle}]$$
- Identifying angle subtended by diameter is $90^\circ$ (3 Marks)
- Correct reason: "angle in semi-circle" (2 Marks)
Cyclic Quad Scenario: In the diagram below, $ABCD$ is a cyclic quadrilateral. The tangent to the circle at point $A$ is $EAF$. $AB$ is produced to $G$. Let tangent angle $E\hat{A}D = 35^\circ$ and interior angle $A\hat{B}C = 115^\circ$.
Angles $A\hat{B}C$ and $C\hat{D}A$ are opposite angles of a cyclic quadrilateral:
$$A\hat{B}C + C\hat{D}A = 180^\circ \quad [\text{opp } \angle \text{s of cyclic quad}]$$
$$115^\circ + C\hat{D}A = 180^\circ$$
$$C\hat{D}A = 65^\circ$$
- Identifying cyclic quad opposite angle condition (1 Mark)
- Stating interior angle equation correctly (1 Mark)
- Correct calculation of $C\hat{D}A = 65^\circ$ (1 Mark)
- Correct CAPS reason: "opp angles of cyclic quad" (1 Mark)
Method 1: Straight Line Angles
Angles $A\hat{B}C$ and $C\hat{B}G$ lie on a straight line ($ABG$):
$$A\hat{B}C + C\hat{B}G = 180^\circ \quad [\angle\text{s on straight line}]$$
$$115^\circ + C\hat{B}G = 180^\circ \implies C\hat{B}G = 65^\circ$$
Method 2: Cyclic Quadrilateral Theorem
The exterior angle of a cyclic quad is equal to the interior opposite angle, thus:
$$C\hat{B}G = C\hat{D}A \quad [\text{ext angle of cyclic quad}]$$
$$C\hat{B}G = 65^\circ$$
- Identifying supplementary straight line or exterior relationship (1 Mark)
- Correct mathematical expression step (1 Mark)
- Correct calculation of size $C\hat{B}G = 65^\circ$ (1 Mark)
- Providing valid geometric reason (either "angles on straight line" or "ext angle of cyclic quad") (1 Mark)
Apply the tan-chord theorem (the angle between the tangent $EAF$ and chord $AD$ is equal to the angle subtended by the chord in the alternate segment):
$$A\hat{C}D = E\hat{A}D \quad [\text{tan-chord theorem}]$$
$$A\hat{C}D = 35^\circ$$
- Identifying correct equal angle relationship $A\hat{C}D = E\hat{A}D$ (2 Marks)
- Correct size $35^\circ$ (1 Mark)
- Correct CAPS reason: "tan-chord theorem" (1 Mark)
In $\triangle ACD$, the sum of interior angles is $180^\circ$:
$$C\hat{A}D + C\hat{D}A + A\hat{C}D = 180^\circ \quad [\text{sum of } \angle \text{s in } \triangle]$$
We calculated $C\hat{D}A = 65^\circ$ in Question 3.3.1 and $A\hat{C}D = 35^\circ$ in Question 3.3.3:
$$C\hat{A}D + 65^\circ + 35^\circ = 180^\circ$$
$$C\hat{A}D + 100^\circ = 180^\circ$$
$$C\hat{A}D = 80^\circ$$
- Identifying sum of interior angles equation is $180^\circ$ (1 Mark)
- Correct substitution of values ($65^\circ$ and $35^\circ$) (2 Marks)
- Correct final angle $C\hat{A}D = 80^\circ$ (1 Mark)
- Correct theorem reason stated ("sum of angles in triangle") (1 Mark)
Tangents from External Point Scenario: In the diagram below, tangents $PA$ and $PB$ are drawn to a circle from an external point $P$. $O$ is the center of the circle. Join radii $OA$ and $OB$. Let $A\hat{P}B = 50^\circ$.
Apply the tangent-radius theorem (tangent is perpendicular to radius at point of contact):
$$O\hat{A}P = 90^\circ \quad [\text{radius } \perp \text{ tangent}]$$
$$O\hat{B}P = 90^\circ \quad [\text{radius } \perp \text{ tangent}]$$
- Correct size $O\hat{A}P = 90^\circ$ (1 Mark)
- Correct size $O\hat{B}P = 90^\circ$ (1 Mark)
- Providing correct CAPS reason: "radius $\perp$ tangent" (2 Marks)
Consider the quadrilateral $OAPB$. The sum of angles in a quadrilateral is $360^\circ$:
$$A\hat{O}B + O\hat{A}P + A\hat{P}B + O\hat{B}P = 360^\circ \quad [\text{sum of } \angle \text{s of quad}]$$
Substitute $O\hat{A}P = 90^\circ$ and $O\hat{B}P = 90^\circ$ (from Question 3.4.1), and $A\hat{P}B = 50^\circ$:
$$A\hat{O}B + 90^\circ + 50^\circ + 90^\circ = 360^\circ$$
$$A\hat{O}B + 230^\circ = 360^\circ$$
$$A\hat{O}B = 130^\circ$$
Alternative: Since opposite angles $O\hat{A}P + O\hat{B}P = 180^\circ$, quadrilateral $OAPB$ is a cyclic quad. Thus, $A\hat{O}B = 180^\circ - 50^\circ = 130^\circ \quad [\text{opp angles of cyclic quad}]$.
- Identifying sum of angles of quad is $360^\circ$ (or opposite cyclic quad condition) (1 Mark)
- Correct substitution of values (1 Mark)
- Correct final angle $A\hat{O}B = 130^\circ$ (1 Mark)
- Stating correct geometric reason ("sum of angles in quad") (1 Mark)
We know that $OP$ bisects the angle $A\hat{P}B$ because tangents from an external point to a circle are symmetrical about $OP$:
$$A\hat{P}O = \frac{1}{2} A\hat{P}B = \frac{1}{2} (50^\circ) = 25^\circ \quad [\text{tangents from ext point symmetrical}]$$
In right-angled $\triangle OAP$ (since $O\hat{A}P = 90^\circ$ from Question 3.4.1):
$$\tan(A\hat{P}O) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OA}{AP}$$
$$\tan(25^\circ) = \frac{7}{AP}$$
Isolate $AP$:
$$AP = \frac{7}{\tan(25^\circ)} \approx \frac{7}{0.4663076...}$$
$$AP \approx 15.01\text{ cm}$$
- Identifying bisection angle $A\hat{P}O = 25^\circ$ (2 Marks)
- Correct CAPS reason stated for symmetry bisection (2 Marks)
- Using the correct trigonometric ratio ($\tan$) in right-angled $\triangle OAP$ (2 Marks)
- Substituting values correctly into ratio formula (2 Marks)
- Isolating $AP = \frac{7}{\tan(25^\circ)}$ correctly (2 Marks)
- Final calculated and correctly rounded value of $15.01\text{ cm}$ (1 Mark)