MATHWISE ACADEMY
GRADE 11 MATHEMATICS
June Practice Examination (Paper 2 - Test 4)
Examiner: Vashi S Y
QUESTION 1: ANALYTICAL GEOMETRY
[27 Marks]In the Cartesian plane, the points $A(-2; 5)$, $B(5; 4)$, and $C(4; -3)$ form the vertices of a triangle. Point $M$ is the midpoint of $AC$.
Apply the gradient formula for points $A(-2; 5)$ and $C(4; -3)$:
$$m_{AC} = \frac{y_C - y_A}{x_C - x_A}$$
$$m_{AC} = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6}$$
$$m_{AC} = -\frac{4}{3}$$
- Correct substitution of coordinates into gradient formula (1 Mark)
- Final simplified gradient ratio $-\frac{4}{3}$ (1 Mark)
Apply the midpoint formula using coordinates $A(-2; 5)$ and $C(4; -3)$:
$$M = \left( \frac{x_A + x_C}{2}; \frac{y_A + y_C}{2} \right)$$
$$M = \left( \frac{-2 + 4}{2}; \frac{5 + (-3)}{2} \right) = \left( \frac{2}{2}; \frac{2}{2} \right)$$
$$M = (1; 1)$$
- Correct $x$-coordinate substitution & calculation (1 Mark)
- Correct $y$-coordinate substitution & calculation (1 Mark)
- Final coordinate mapped pair $M(1; 1)$ (1 Mark)
To prove perpendicularity, the product of their gradients must equal $-1$ ($m_{BM} \times m_{AC} = -1$).
First, calculate gradient of $BM$ using points $B(5; 4)$ and $M(1; 1)$:
$$m_{BM} = \frac{1 - 4}{1 - 5} = \frac{-3}{-4} = \frac{3}{4}$$
We calculated the gradient of $AC$ to be $-\frac{4}{3}$ in Question 1.1:
$$m_{BM} \times m_{AC} = \frac{3}{4} \times \left(-\frac{4}{3}\right) = -1$$
Since the product is exactly $-1$, the line segment $BM$ is perpendicular to $AC$ ($BM \perp AC$).
- Calculating correct gradient $m_{BM} = \frac{3}{4}$ (1 Mark)
- Proving product of gradients is $-1$ (1 Mark)
- Correct final perpendicularity deduction with geometric reasoning (1 Mark)
We proved that the line perpendicular to $AC$ is the line $BM$ with gradient $m = \frac{3}{4}$ (from Question 1.3).
Substitute gradient $m = \frac{3}{4}$ and point $B(5; 4)$ into the point-slope linear equation:
$$y - y_B = m(x - x_B) \implies y - 4 = \frac{3}{4}(x - 5)$$
$$y - 4 = \frac{3}{4}x - \frac{15}{4} \implies y = \frac{3}{4}x - \frac{15}{4} + 4$$
$$y = \frac{3}{4}x + \frac{1}{4} \quad \text{or} \quad y = 0.75x + 0.25$$
- Identifying perpendicular gradient slope $m = \frac{3}{4}$ (1 Mark)
- Correct substitution of coordinates (1 Mark)
- Final standard linear equation form (1 Mark)
Let $\theta$ be the angle of inclination of line $AC$:
$$\tan \theta = m_{AC} \implies \tan \theta = -\frac{4}{3}$$
Calculate reference angle (using positive value):
$$\theta_{\text{ref}} = \tan^{-1}\left(\frac{4}{3}\right) \approx 53.13^\circ$$
Since the gradient is negative, the angle of inclination is obtuse (lies in the second quadrant):
$$\theta = 180^\circ - 53.13^\circ$$
$$\theta \approx 126.9^\circ$$
- Setting up correct angle formula relationship (1 Mark)
- Calculating reference angle $53.1^\circ$ (1 Mark)
- Correct final obtuse angle of inclination $126.9^\circ$ (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 $M(1; 1)$:
$$x_M = \frac{x_B + x_D}{2} \implies 1 = \frac{5 + x_D}{2} \implies 2 = 5 + x_D \implies x_D = -3$$
$$y_M = \frac{y_B + y_D}{2} \implies 1 = \frac{4 + y_D}{2} \implies 2 = 4 + y_D \implies y_D = -2$$
$$D = (-3; -2)$$
- Setting up $x$-midpoint equation and solving $x = -3$ (2 Marks)
- Setting up $y$-midpoint equation and solving $y = -2$ (2 Marks)
Parallelogram $ABCD$ consists of two identical triangles, $\triangle ABC$ and $\triangle ADC$. Thus, $\text{Area}_{ABCD} = 2 \times \text{Area}_{\triangle ABC}$.
We proved in Question 1.3 that $BM$ is perpendicular to $AC$, so $BM$ is the perpendicular height on base $AC$.
1. Calculate base length $AC$ using points $A(-2; 5)$ and $C(4; -3)$:
$$AC = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10\text{ units}$$
2. Calculate perpendicular height $BM$ using points $B(5; 4)$ and $M(1; 1)$:
$$BM = \sqrt{(1 - 5)^2 + (1 - 4)^2} = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\text{ units}$$
3. Calculate Area:
$$\text{Area}_{ABCD} = 2 \times \left( \frac{1}{2} \times \text{base} \times \text{perpendicular height} \right)$$
$$\text{Area} = AC \times BM = 10 \times 5$$
$$\text{Area} = 50 \text{ square units}$$
- Identifying perpendicular diagonal height relationship ($BM \perp AC$) (1 Mark)
- Correctly substituting & calculating $AC = 10$ (2 Marks)
- Correctly substituting & calculating $BM = 5$ (2 Marks)
- Applying parallelogram area formula (2 Marks)
- Final exact area value $50\text{ units}^2$ (1 Mark)
QUESTION 2: TRIGONOMETRY
[27 Marks]Given: $\sin \theta = -\frac{15}{17}$ and $90^\circ \le \theta \le 270^\circ$.
Isolate and map trig ratio values: $\sin\theta = -\frac{15}{17} = \frac{y}{r}$ where $r = 17$ (always positive) and $y = -15$.
Since $\sin \theta < 0$ (negative in Q3 & Q4) and the domain is $90^\circ \le \theta \le 270^\circ$, the angle $\theta$ lies in Quadrant 3.
Apply Pythagoras to solve for $x$ (negative in Quadrant 3):
$$x^2 + y^2 = r^2 \implies x^2 + (-15)^2 = 17^2$$
$$x^2 + 225 = 289 \implies x^2 = 64 \implies x = -8$$
Substitute ratios: $\cos \theta = \frac{x}{r} = -\frac{8}{17}$ and $\tan \theta = \frac{y}{x} = \frac{-15}{-8} = \frac{15}{8}$.
Calculate final value:
$$\cos\theta + \tan\theta = -\frac{8}{17} + \frac{15}{8} = \frac{-64 + 255}{136} = \frac{191}{136} \approx 1.40$$
- Identifying Quadrant 3 with a sketch (2 Marks)
- Using Pythagoras to find $x = -8$ (1 Mark)
- Correct $\cos \theta = -\frac{8}{17}$ ratio (1 Mark)
- Correct $\tan \theta = \frac{15}{8}$ ratio (1 Mark)
- Final calculation sum fraction $\frac{191}{136}$ (1 Mark)
Apply the square identity relationship $\sec^2\theta - 1 = \tan^2\theta$:
We know $\tan\theta = \frac{15}{8}$ from Question 2.1.1:
$$\sec^2 \theta - 1 = \tan^2 \theta = \left(\frac{15}{8}\right)^2 = \frac{225}{64}$$
$$= \frac{225}{64} \quad \text{or} \quad \approx 3.52$$
- Identifying identity substitution or ratio substitution (1 Mark)
- Final exact calculated ratio $\frac{225}{64}$ (1 Mark)
Apply reduction formulae and co-functions step-by-step:
- $\cos(90^\circ + x) = -\sin x \quad [\text{Co-function Q2 negative}]$
- $\sin(180^\circ + x) = -\sin x \quad [\text{Q3 negative}]$
- $\tan(360^\circ - x) = -\tan x \quad [\text{Q4 negative}]$
- $\cos(180^\circ - x) = -\cos x \quad [\text{Q2 negative}]$
Substitute back into the expression:
$$= \frac{(-\sin x) \cdot (-\sin x)}{(-\tan x) \cdot (-\cos x)} = \frac{\sin^2 x}{\tan x \cdot \cos x}$$
Substitute quotient identity $\tan x = \frac{\sin x}{\cos x}$:
$$= \frac{\sin^2 x}{\left(\frac{\sin x}{\cos x}\right) \cdot \cos x} = \frac{\sin^2 x}{\sin x}$$
$$= \sin x$$
- Co-function $\cos(90^\circ + x) = -\sin x$ (1 Mark)
- Reduction $\sin(180^\circ + x) = -\sin x$ (1 Mark)
- Reduction $\tan(360^\circ - x) = -\tan x$ (1 Mark)
- Reduction $\cos(180^\circ - x) = -\cos x$ (1 Mark)
- Quotient substitution $\tan x = \frac{\sin x}{\cos x}$ (1 Mark)
- Final simplified ratio $\sin x$ (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 $\theta = 45^\circ + k \cdot 180^\circ$ (1 Mark)
- Stating parameter constant constraint $k \in \mathbb{Z}$ (1 Mark)
QUESTION 3: EUCLIDEAN GEOMETRY
[46 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) |
| $\therefore AM = MB$ | 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. Perpendicular segment $OM$ is drawn to chord $AB$. Let radius $OA = 15\text{ cm}$ and segment $OM = 9\text{ cm}$.
Apply the Theorem of Pythagoras to right-angled $\triangle OMA$:
$$OA^2 = OM^2 + AM^2 \quad [\text{Pythagoras}]$$
$$15^2 = 9^2 + AM^2 \implies 225 = 81 + AM^2$$
$$AM^2 = 144 \implies AM = 12\text{ cm}$$
Since $OM \perp AB$, the line drawn from the center perpendicular to the chord bisects the chord:
$$AB = 2 \times AM = 2 \times 12 = 24\text{ cm} \quad [\text{line from centre } \perp \text{ chord bisects chord}]$$
$$AB = 24\text{ cm}$$
- Correct Pythagoras substitution & calculation of $AM^2 = 144$ (2 Marks)
- Calculating $AM = 12\text{ cm}$ (1 Mark)
- Stating perpendicular line bisect chord property with reason (1 Mark)
- Correct total chord length $AB = 24\text{ cm}$ (1 Mark)
To prove congruence between $\triangle OAM$ and $\triangle OBM$:
Compare elements in $\triangle OAM$ and $\triangle OBM$:
- $OA = OB \quad [\text{radii}]$
- $OM = OM \quad [\text{common side}]$
- $AM = MB \quad [\text{proved in Question 3.2.1; line from centre } \perp \text{ chord}]$
$$\therefore \triangle OAM \equiv \triangle OBM \quad [\text{S.S.S}]$$
- Radii statement $OA = OB$ with reason (1 Mark)
- Common side $OM = OM$ (1 Mark)
- Equal segments $AM = MB$ with previous proof reason (1 Mark)
- Formal congruency conclusion statement (1 Mark)
- Correct congruence code reason (S.S.S) (1 Mark)
Cyclic Quad Scenario: In the diagram below, $ABCD$ is a cyclic quadrilateral with center $O$. Radii $OA$ and $OB$ are joined. Tangent $EAF$ is drawn to the circle at point $A$. 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)
- Correct calculation of $C\hat{D}A = 65^\circ$ (1 Mark)
- Correct CAPS reason: "opp angles of cyclic quad" (1 Mark)
Angles $A\hat{B}C$ and $C\hat{B}G$ lie on a straight line ($ABG$):
$$A\hat{B}C + C\hat{B}G = 180^\circ \quad [\angle\text{s on straight line}]$$
$$115^\circ + C\hat{B}G = 180^\circ$$
$$C\hat{B}G = 65^\circ$$
Alternative: The exterior angle of a cyclic quad is equal to the interior opposite angle, thus $C\hat{B}G = C\hat{D}A = 65^\circ$ [ext angle of cyclic quad]. Both methods are correct.
- Stating size $C\hat{B}G = 65^\circ$ (2 Marks)
- Providing valid geometric reason (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 (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 $360^\circ$ (1 Mark)
- Correct substitution of values (1 Mark)
- Correct final angle $A\hat{O}B = 130^\circ$ (1 Mark)
- Stating correct geometric reason (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 $OP$:
$$A\hat{P}O = \frac{1}{2} A\hat{P}B = \frac{1}{2} (50^\circ) = 25^\circ$$
In right-angled $\triangle OAP$ (since $O\hat{A}P = 90^\circ$):
$$\tan(A\hat{P}O) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OA}{AP}$$
$$\tan(25^\circ) = \frac{7}{AP}$$
Isolate $AP$:
$$AP = \frac{7}{\tan(25^\circ)} \approx \frac{7}{0.4663076...}$$
$$AP \approx 15.01\text{ cm}$$
- Identifying bisection angle $A\hat{P}O = 25^\circ$ with correct reason (2 Marks)
- Using the correct trigonometric ratio ($\tan$) in $\triangle OAP$ (2 Marks)
- Substituting values correctly into ratio formula (1 Mark)
- Isolating and solving for $AP$ correctly (1 Mark)
- Final rounded value of $15.01\text{ cm}$ (1 Mark)