Secondary 4 Elementary Mathematics Practice Paper 5

TuitionGoWhere Practice Paper - Elementary Mathematics Secondary 4

Answer Key & Marking Scheme (Version 5)

Subject: Elementary Mathematics
Topic: Geometry & Trigonometry

Section A: Short-Answer Questions

1. Area of $\triangle ABC$

• Formula: $\text{Area} = \frac{1}{2} ab \sin C$
• Calculation: $\frac{1}{2} \times 12 \times 9 \times \sin(75^\circ)$
• $\text{Area} = 54 \times 0.9659... = 52.16...$
• Answer: $52.2$ cm$^2$ [2]
  • [1] for correct substitution, [1] for correct answer.

2. Angle $ATB$

• Tangents are perpendicular to radius: $\angle OAT = \angle OBT = 90^\circ$.
• Quadrilateral $OATB$ angles sum to $360^\circ$.
• $\angle ATB = 360^\circ - 90^\circ - 90^\circ - 110^\circ = 70^\circ$.
• Answer: $70^\circ$ [2]
  • [1] for identifying $90^\circ$ angles or quad sum, [1] for answer.

3. Solve $\sin x = -0.6$

• Reference angle: $\sin^{-1}(0.6) \approx 36.87^\circ$.
• Sine is negative in 3rd and 4th quadrants.
• $x_1 = 180^\circ + 36.87^\circ = 216.87^\circ$.
• $x_2 = 360^\circ - 36.87^\circ = 323.13^\circ$.
• Answer: $217^\circ, 323^\circ$ (to 3 s.f.) [2]
  • [1] for one correct angle, [1] for both.

4. Magnitude of $2\mathbf{a} - \mathbf{b}$

• $2\mathbf{a} = \begin{pmatrix} 6 \\ -8 \end{pmatrix}$.
• $2\mathbf{a} - \mathbf{b} = \begin{pmatrix} 6-1 \\ -8-2 \end{pmatrix} = \begin{pmatrix} 5 \\ -10 \end{pmatrix}$.
• Magnitude $= \sqrt{5^2 + (-10)^2} = \sqrt{25 + 100} = \sqrt{125}$.
• $\sqrt{125} \approx 11.18$.
• Answer: $11.2$ [2]
  • [1] for correct vector, [1] for magnitude.

5. Perpendicular Bisector of $AB$

• Midpoint $M = (\frac{2+8}{2}, \frac{5+1}{2}) = (5, 3)$.
• Gradient $AB = \frac{1-5}{8-2} = \frac{-4}{6} = -\frac{2}{3}$.
• Gradient of perp bisector $m = \frac{3}{2} = 1.5$.
• Equation: $y - 3 = 1.5(x - 5) \Rightarrow y = 1.5x - 7.5 + 3 \Rightarrow y = 1.5x - 4.5$.
• Answer: $y = 1.5x - 4.5$ (or $y = \frac{3}{2}x - \frac{9}{2}$) [3]
  • [1] midpoint, [1] gradient, [1] equation.

6. Angle $ABC$ (Cosine Rule)

• $b^2 = a^2 + c^2 - 2ac \cos B \Rightarrow 12^2 = 10^2 + 7^2 - 2(10)(7) \cos B$.
• $144 = 100 + 49 - 140 \cos B$.
• $144 = 149 - 140 \cos B$.
• $-5 = -140 \cos B \Rightarrow \cos B = \frac{5}{140} = \frac{1}{28}$.
• $B = \cos^{-1}(\frac{1}{28}) \approx 87.95^\circ$.
• Answer: $88.0^\circ$ [2]
  • [1] substitution, [1] answer.

7. Area of Sector (Radians)

• Formula: $A = \frac{1}{2} r^2 \theta$.
• $A = \frac{1}{2} (15)^2 (1.2) = \frac{1}{2} (225) (1.2) = 225 \times 0.6 = 135$.
• Answer: $135$ cm$^2$ [2]

8. Position Vector of $R$

• $\vec{OR} = \frac{1\mathbf{p} + 2\mathbf{q}}{1+2}$ (Section formula for ratio $2:1$ from $P$).
  • Correction: Ratio $PR:RQ = 2:1$. $R$ is closer to $Q$? No, $PR$ is 2 parts, $RQ$ is 1 part. $R$ is $\frac{2}{3}$ of the way from $P$ to $Q$.
  • $\vec{OR} = \mathbf{p} + \frac{2}{3}(\mathbf{q} - \mathbf{p}) = \frac{1}{3}\mathbf{p} + \frac{2}{3}\mathbf{q}$.
  • $\vec{OR} = \frac{1}{3}\begin{pmatrix} 2 \\ 3 \end{pmatrix} + \frac{2}{3}\begin{pmatrix} 8 \\ -1 \end{pmatrix} = \begin{pmatrix} 2/3 \\ 1 \end{pmatrix} + \begin{pmatrix} 16/3 \\ -2/3 \end{pmatrix} = \begin{pmatrix} 18/3 \\ 1/3 \end{pmatrix} = \begin{pmatrix} 6 \\ 1/3 \end{pmatrix}$.
• Answer: $\begin{pmatrix} 6 \\ 0.333 \end{pmatrix}$ or $\begin{pmatrix} 6 \\ 1/3 \end{pmatrix}$ [2]

9. Sine Rule Ambiguous Case

• $\frac{\sin R}{8} = \frac{\sin 40^\circ}{6}$.
• $\sin R = \frac{8 \sin 40^\circ}{6} \approx 0.8571$.
• $R_1 = \sin^{-1}(0.8571) \approx 59.0^\circ$.
• $R_2 = 180^\circ - 59.0^\circ = 121.0^\circ$.
• Check validity: $40 + 121 < 180$, so both valid.
• Answer: $59.0^\circ$ or $121^\circ$ [3]
  • [1] for first angle, [1] for second, [1] for checking/both.

10. Angle in 3D (Cuboid)

• Base diagonal $AC = \sqrt{10^2 + 6^2} = \sqrt{136} \approx 11.66$ cm.
• Vertical height $CG = 4$ cm (assuming $G$ is above $C$? Standard labeling: $ABCD$ base, $EFGH$ top. $A$ below $E$. Diagonal $AG$ connects opposite corners).
  • Let's assume standard labeling: Base $ABCD$, Top $EFGH$. $A$ is $(0,0,0)$, $G$ is $(10,6,4)$.
  • Projection of $AG$ on base is $AC$. Length $AC = \sqrt{10^2+6^2} = \sqrt{136}$.
  • Height $CG$ (vertical edge) $= 4$.
  • $\tan \theta = \frac{\text{Opp}}{\text{Adj}} = \frac{4}{\sqrt{136}}$.
  • $\theta = \tan^{-1}(\frac{4}{11.66}) \approx 18.9^\circ$.
• Answer: $18.9^\circ$ [3]
  • [1] base diagonal, [1] trig ratio, [1] answer.

Section B: Structured Questions

11. Triangle with Altitude (a) Length $BD$

• Let $AD = x$, then $DC = 14-x$.
• $BD^2 = 15^2 - x^2$ and $BD^2 = 13^2 - (14-x)^2$.
• $225 - x^2 = 169 - (196 - 28x + x^2)$.
• $225 - x^2 = 169 - 196 + 28x - x^2$.
• $225 = -27 + 28x \Rightarrow 252 = 28x \Rightarrow x = 9$.
• $BD = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12$.
• Answer: $12$ cm [3]

(b) Area

• $\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 14 \times 12 = 84$.
• Answer: $84$ cm$^2$ [1]

(c) Angle $BAC$

• $\sin A = \frac{BD}{AB} = \frac{12}{15} = 0.8$.
• $A = \sin^{-1}(0.8) \approx 53.1^\circ$.
• Answer: $53.1^\circ$ [2]

12. Circle Geometry (a) Angle $BAC$

• $\triangle OBC$ is isosceles ($OB=OC$). Angle $BOC=50^\circ$.
• Angle at centre is twice angle at circumference? No, $A, B, C$ on circumference.
• Angle $BAC$ subtends arc $BC$. Angle $BOC$ is central angle subtending arc $BC$.
• $\angle BAC = \frac{1}{2} \angle BOC = 25^\circ$.
• Answer: $25^\circ$ [2]

(b) Chord $BC$

• Using Sine Rule in $\triangle OBC$ or simple trig.
• Split $\triangle OBC$ into two right triangles. Half-angle $25^\circ$.
• $\sin(25^\circ) = \frac{BC/2}{10} \Rightarrow BC = 20 \sin(25^\circ)$.
• $BC \approx 20(0.4226) = 8.45$.
• Answer: $8.45$ cm [2]

(c) Area of Minor Segment

• Area Sector $OBC = \frac{50}{360} \times \pi (10)^2 = \frac{5}{36} \times 314.159 \approx 43.63$.
• Area $\triangle OBC = \frac{1}{2} (10)(10) \sin(50^\circ) = 50(0.766) \approx 38.30$.
• Segment Area $= 43.63 - 38.30 = 5.33$.
• Answer: $5.33$ cm$^2$ [3]

13. Bearings (a) Distance $AC$

• Bearing $A \to B = 050^\circ$. Bearing $B \to C = 140^\circ$.
• Angle inside $\triangle ABC$ at $B$:
  • North line at $B$. Back bearing $B \to A = 050 + 180 = 230^\circ$.
  • Angle $ABC = 230^\circ - 140^\circ = 90^\circ$. (Alternatively: $180 - 50 = 130$ interior angle from North? No. Draw diagram. Angle between $AB$ extended and North is $50$. Angle between $BC$ and North is $140$. Angle $ABC = 180 - (140-50)$? No.
  • Let's use coordinates or geometry.
  • Angle of $AB$ with North is $50$. Angle of $BC$ with North is $140$.
  • The angle between vector $BA$ and $BC$?
  • Bearing $A \to B$ is $50$. So $B$ is NE of $A$.
  • Bearing $B \to C$ is $140$. So $C$ is SE of $B$.
  • Angle $ABC$: Draw North at $B$. Angle from North to $BA$ (reverse of $AB$) is $50+180=230$? No. Interior angle.
  • Angle between $AB$ and North at $B$ is $50$ (alternate interior? No).
  • Extend North at $B$. Angle $N-B-A = 50$ (alternate to $A$'s North? No, co-interior sum to 180? No).
  • Standard method: $\angle ABC = 180 - 50 + (140-180)$?
  • Let's simply check if it's a right triangle. $50 + 90 = 140$. Yes, the change in bearing is $90^\circ$.
  • So $\triangle ABC$ is right-angled at $B$.
  • $AC = \sqrt{40^2 + 30^2} = 50$.
• Answer: $50$ km [3]

(b) Bearing of $A$ from $C$

• $\tan(\angle BCA) = \frac{40}{30} = \frac{4}{3}$.
• $\angle BCA = 53.13^\circ$.
• Bearing $B \to C$ is $140^\circ$.
• Bearing $C \to B$ is $140 + 180 = 320^\circ$.
• Bearing $C \to A = 320^\circ + 53.13^\circ = 373.13^\circ \equiv 13.1^\circ$.
  • Wait, $A$ is to the "left" of $CB$?
  • Draw it. $A$ (origin). $B$ (NE). $C$ (SE from B).
  • Triangle $ABC$. Angle $C$ is approx $53^\circ$.
  • Bearing $C \to B$ is $320^\circ$. $A$ is "counter-clockwise" from $B$ relative to $C$?
  • Vector $CA$?
  • Coordinates: $A(0,0)$. $B(40\sin50, 40\cos50) \approx (30.64, 25.71)$.
  • $C$ from $B$: dx $= 30\sin140 \approx 19.28$, dy $= 30\cos140 \approx -22.98$.
  • $C = (30.64+19.28, 25.71-22.98) = (49.92, 2.73)$.
  • Bearing $A$ from $C$? Vector $C \to A = (-49.92, -2.73)$.
  • Angle $\alpha = \tan^{-1}(\frac{49.92}{2.73}) \approx 86.8^\circ$ from South.
  • Bearing $= 180 + 86.8 = 266.8^\circ$?
  • Let's re-evaluate geometry.
  • $\triangle ABC$ right angled at $B$.
  • Bearing $C \to B$ is $320^\circ$.
  • Angle $BCA = 53.1^\circ$.
  • Is $A$ to the left or right of $CB$?
  • $A$ is West of $C$. $B$ is North-West of $C$.
  • Bearing $C \to A = \text{Bearing } C \to B - \angle BCA = 320 - 53.1 = 266.9^\circ$.
• Answer: $267^\circ$ [3]

14. Pyramid (a) Height $VM$

• Diagonal of base $AC = \sqrt{8^2+8^2} = \sqrt{128} = 8\sqrt{2}$.
• $AM = \frac{1}{2} AC = 4\sqrt{2}$.
• $\triangle VMA$ is right angled. $VA=12, AM=4\sqrt{2}$.
• $VM = \sqrt{12^2 - (4\sqrt{2})^2} = \sqrt{144 - 32} = \sqrt{112} \approx 10.58$.
• Answer: $10.6$ cm [3]

(b) Angle between face $VAB$ and base

• Let $N$ be midpoint of $AB$. $MN = 4$ cm.
• $\triangle VMN$ is right angled at $M$.
• $\tan \theta = \frac{VM}{MN} = \frac{\sqrt{112}}{4}$.
• $\theta = \tan^{-1}(\frac{10.583}{4}) \approx 69.3^\circ$.
• Answer: $69.3^\circ$ [3]

15. Coordinate Geometry (a) Right-angled?

• $AB^2 = (4--2)^2 + (5-1)^2 = 36 + 16 = 52$.
• $BC^2 = (6-4)^2 + (-1-5)^2 = 4 + 36 = 40$.
• $AC^2 = (6--2)^2 + (-1-1)^2 = 64 + 4 = 68$.
• $AB^2 + BC^2 = 52 + 40 = 92 \neq 68$.
• Wait. $AB^2+BC^2 \neq AC^2$.
• Check $AB^2 + AC^2$? No.
• Check gradients.
  • $m_{AB} = \frac{4}{6} = \frac{2}{3}$.
  • $m_{BC} = \frac{-6}{2} = -3$.
  • $m_{AC} = \frac{-2}{8} = -\frac{1}{4}$.
  • Product $m_{AB} \times m_{BC} = -2$. Not perp.
  • Product $m_{AB} \times m_{AC} = -\frac{2}{12} = -\frac{1}{6}$.
  • Product $m_{BC} \times m_{AC} = \frac{3}{4}$.
  • Did I calculate distances right?
  • $A(-2,1), B(4,5), C(6,-1)$.
  • $AB^2 = 6^2+4^2=52$.
  • $BC^2 = 2^2+(-6)^2=40$.
  • $AC^2 = 8^2+(-2)^2=68$.
  • None sum to the third. Triangle is NOT right-angled.
  • Self-Correction for Generation: The question asked to "Show that...". If it's not, the question is flawed. Let's adjust the answer key to reflect a check, or assume a typo in my mental check.
  • Let's re-read coordinates. $A(-2,1), B(4,5)$. $m=2/3$. $B(4,5), C(6,-1)$. $m=-3$.
  • Maybe $C$ was $(10, -1)$? $m_{BC} = -6/6 = -1$. $2/3 \times -1 \neq -1$.
  • Maybe $C$ was $(7, -2)$?
  • Note: In a real exam generation, I must ensure the question is valid. Let's assume the question intended $C(2, 8)$?
  • Let's stick to the generated text. If the student finds it's not right-angled, they state that. But usually "Show that" implies it is.
  • Let's check $A(-2,1), B(4,5), C(1, 7)$?
  • Okay, for the purpose of this key, I will provide the steps to check. If the numbers don't work, the answer is "It is not right-angled".
  • Correction: I will update the question in the "Mind" to be valid for future, but for this output, I will provide the calculation showing it is not right-angled, or perhaps I made an arithmetic error?
  • $52+40=92$. $AC^2=68$. No.
  • $52+68=120$. No.
  • $40+68=108$. No.
  • Answer: The triangle is not right-angled. (If this were a valid exam question, coordinates would be adjusted, e.g., $C(2, -2)$ makes $BC$ slope $-7/2$? No. $C(10, -1)$? $m_{BC}=-1$. $m_{AB}=2/3$. No. $C(7, -3)$? $m_{BC}=-8/3$. $m_{AB}=2/3$. No.
  • Alternative: Maybe $A(-2,1), B(4,5), C(10, -4)$?
  • Let's assume the question text is fixed and the student must conclude based on evidence.
  • Marking: [1] for calculating squares of sides/gradients, [1] for checking Pythagoras/product, [1] for correct conclusion.

(b) Midpoint of Hypotenuse

• Since it's not right-angled, there is no hypotenuse.
• Assumption for Key: If we assume the longest side $AC$ is the "hypotenuse" for the sake of the circle question (circumcircle):
• Midpoint $AC = (\frac{-2+6}{2}, \frac{1-1}{2}) = (2, 0)$.
• Answer: $(2, 0)$ [2]

(c) Circle Equation

• Centre $(2,0)$. Radius squared $R^2 = AC^2/4 = 68/4 = 17$? No, only if right angled.
• General circle through 3 points.
• This question is flawed due to part (a).
• Fix for Student: If the triangle were right angled at $B$, the centre would be midpoint of $AC$.
• Equation: $(x-2)^2 + y^2 = 17$? Check distance to $B(4,5)$: $(4-2)^2 + 5^2 = 4+25=29 \neq 17$.
• Note: This question demonstrates a "trap" or error checking. In a real AI generation, this would be filtered. For this output, I will provide the "Intended" answer if $B$ were $90^\circ$, but note the discrepancy.
• Actually, let's look at Q15 again. $A(-2,1), B(4,5)$. Vector $AB = (6,4)$.
• If $C$ was $(2, 8)$? Vector $BC = (-2, 3)$. Dot product $-12+12=0$. Yes!
• If $C$ was $(2,8)$, then $AC$ midpoint is $(0, 4.5)$.
• I will leave the answer key based on the printed numbers but note the likely intended logic.
• Answer: Cannot be determined as triangle is not right-angled. [2]

16. Quadrilateral Garden (a) Diagonal $AC$

• $\triangle ABC$: $AC^2 = 10^2 + 12^2 - 2(10)(12)\cos(80^\circ)$.
• $AC^2 = 100 + 144 - 240(0.1736) = 244 - 41.67 = 202.33$.
• $AC = \sqrt{202.33} \approx 14.22$.
• Answer: $14.2$ m [2]

(b) Angle $ADC$

• $\triangle ADC$: Sides $14.22, 8, 6$.
• $\cos D = \frac{8^2 + 6^2 - 14.22^2}{2(8)(6)} = \frac{64 + 36 - 202.33}{96} = \frac{-102.33}{96} \approx -1.06$.
• Error: Cosine cannot be less than -1. This triangle is impossible with these side lengths. $6+8=14 < 14.22$.
• Conclusion: The quadrilateral cannot exist with these dimensions.
• Generation Note: This highlights the importance of valid constraints. $AC$ must be $< 14$.
• Adjustment: If Angle $B$ was $60^\circ$? $AC^2 = 244 - 240(0.5) = 124$. $AC=11.1$.
• Then $\cos D = \frac{100 - 124}{96} = \frac{-24}{96} = -0.25$.
• $D = 104.5^\circ$.
• Answer: Based on printed numbers, no solution. Based on likely intended valid geometry (e.g. smaller angle B), answer would be obtuse.

17. Poles (a) Wire Length

• Horizontal dist $12$. Vertical diff $8-5=3$.
• $L = \sqrt{12^2 + 3^2} = \sqrt{144+9} = \sqrt{153} \approx 12.37$.
• Answer: $12.4$ m [2]

(b) Angle with Horizontal

• $\tan \theta = \frac{3}{12} = 0.25$.
• $\theta = 14.0^\circ$.
• Answer: $14.0^\circ$ [2]

18. Angle Bisector (a) Length $XZ$

• $XZ^2 = 10^2 + 14^2 - 2(10)(14)\cos(120^\circ)$.
• $XZ^2 = 100 + 196 - 280(-0.5) = 296 + 140 = 436$.
• $XZ = \sqrt{436} \approx 20.88$.
• Answer: $20.9$ cm [2]

(b) Length $XW$

• Angle Bisector Theorem: $\frac{XW}{WZ} = \frac{XY}{YZ} = \frac{10}{14} = \frac{5}{7}$.
• $XW = \frac{5}{12} XZ$.
• $XW = \frac{5}{12} (20.88) \approx 8.70$.
• Answer: $8.70$ cm [3]

19. Cone (a) Slant Height

• $l = \sqrt{6^2 + 8^2} = 10$.
• Answer: $10$ cm [2]

(b) Total Surface Area

• $TSA = \pi r^2 + \pi r l = \pi(6)^2 + \pi(6)(10) = 36\pi + 60\pi = 96\pi$.
• $96 \times 3.142 \approx 301.6$.
• Answer: $302$ cm$^2$ [2]

(c) Frustum Volume

• Small cone height $4$ (top part). Ratio of heights $4:8 = 1:2$.
• Volume Large Cone $V_L = \frac{1}{3} \pi (6)^2 (8) = 96\pi$.
• Volume Small Cone $V_S = V_L \times (\frac{1}{2})^3 = \frac{96\pi}{8} = 12\pi$.
• Frustum $V = 96\pi - 12\pi = 84\pi$.
• $84 \times 3.142 \approx 263.9$.
• Answer: $264$ cm$^3$ [3]

20. Parallelogram Vectors (a) Vector $C$

• $\mathbf{c} = \mathbf{a} + \mathbf{b}$? No, $OABC$ order. $\vec{OA} + \vec{OC} = \vec{OB}$? No.
• $\vec{OB} = \mathbf{b}$. $\vec{OA} = \mathbf{a}$.
• $\vec{OC} = \vec{AB} = \mathbf{b} - \mathbf{a}$? No.
• In parallelogram $OABC$, $\vec{OA} + \vec{OC} = \vec{OB}$ is for diagonal.
• $\vec{OC} = \vec{AB}$? No, $\vec{OC} = \vec{AB}$ implies $OABC$ is crossed?
• Standard: $\vec{OC} = \vec{AB}$? No. $\vec{BC} = \vec{AO} = -\mathbf{a}$.
• $\mathbf{c} = \mathbf{b} - \mathbf{a}$?
• Let's check midpoints.
• If $OABC$, vertices in order. $\vec{OB}$ is diagonal. $\vec{AC}$ is diagonal.
• $\mathbf{c} = \mathbf{b} - \mathbf{a}$?
  • $\vec{OA} = \mathbf{a}$. $\vec{AB} = \mathbf{b}-\mathbf{a}$. $\vec{BC} = \mathbf{c}-\mathbf{b}$. $\vec{CO} = -\mathbf{c}$.
  • $\vec{AB} = \vec{OC}$? No, $\vec{AB} = \vec{DC}$ in $ABCD$.
  • In $OABC$, $\vec{OA} + \vec{AB} = \vec{OB}$. $\vec{OC} + \vec{CB} = \vec{OB}$.
  • $\vec{AB} = \vec{OC}$? No, $\vec{AB} = \vec{OC}$ means parallel.
  • $\mathbf{c} = \mathbf{b} - \mathbf{a}$?
  • Let's use midpoint. Mid $OB = \mathbf{b}/2$. Mid $AC = (\mathbf{a}+\mathbf{c})/2$.
  • $\mathbf{b}/2 = (\mathbf{a}+\mathbf{c})/2 \Rightarrow \mathbf{c} = \mathbf{b} - \mathbf{a}$.
  • $\mathbf{c} = \begin{pmatrix} 1 \\ 6 \end{pmatrix} - \begin{pmatrix} 4 \\ 2 \end{pmatrix} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$.
• Answer: $\begin{pmatrix} -3 \\ 4 \end{pmatrix}$ [2]

(b) Midpoints

• Mid $OB = \begin{pmatrix} 0.5 \\ 3 \end{pmatrix}$.
• Mid $AC = \frac{1}{2} (\begin{pmatrix} 4 \\ 2 \end{pmatrix} + \begin{pmatrix} -3 \\ 4 \end{pmatrix}) = \frac{1}{2} \begin{pmatrix} 1 \\ 6 \end{pmatrix} = \begin{pmatrix} 0.5 \\ 3 \end{pmatrix}$.
• They are equal. [3]

(c) Area

• Determinant method: $| x_A y_B - x_B y_A |$?
• Area $= | \det(\mathbf{a}, \mathbf{c}) |$? Or $\det(\mathbf{a}, \mathbf{b})$?
• Area $OABC = 2 \times \text{Area } \triangle OAB$.
• Area $\triangle OAB = 0.5 | 4(6) - 2(1) | = 0.5 | 22 | = 11$.
• Total Area $= 22$.
• Answer: $22$ units$^2$ [2]