Grade 11 Maths · Trigonometry

Use the area rule

Find the area of any triangle as ½ × two sides × sin of the angle between them, for example ½ × 6 × 8 × sin 30° = 12.

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Area Rule grade 11: IntroductionKevinmathscience

Area of a Triangle - 1/2 absin(C) - GCSE Higher Maths1st Class Maths

2D Trigonometry: Example with areaKevinmathscience

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Question 6.3 · March 2021 · Provincial · 3 marks

In the sketch,

PS

is the median of

△ PQR

and thus

QS = SR = x

\hat{Q} = a

and

Q \hat{P} S = b

6.3 Hence, show that: Area of

△ PSR = \frac{x ^{2} s i n a \cdot s i n ( a + b )}{2 s i n b}

(3)

Work it out on paper first. Then mark yourself with the official memo, one tick at a time.

More past paper questions: use the area rule

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3. Question 7.3 · November 2024 · Paper 2 · Provincial · 3 marks
In the diagram, VP $=$ PQ $= 5$ units, R $\hat{Q}$ P $=$ S $\hat{P}$ V $= α$ , S $\hat{P}$ R $= β$ and PS $∥$ QR. SV is perpendicular to VP. $Diagram for this question$
7.3 Hence show that the area of $Δ$ RPS $= \frac{25. t a n α}{2}$ (3)
Show the memo (3 ticks)Memo
1. ✓
  Correct formula for area ruleArea $Δ$ RPS $= \frac{1}{2} (RP) (PS) sin β$
  1 mark · Use the area rule
2. ✓
  Substitution $= \frac{1}{2} (\frac{5 s i n α}{s i n β}) (\frac{5}{c o s α}) sin β$
  1 mark · Use the area rule
3. ✓
  Simplification $= \frac{25 s i n α}{2 c o s α}$
  $= \frac{25 t a n α}{2}$
  1 mark · Simplify an algebraic fraction
4. Question 4.3.2 · November 2024 · Paper 2 · Provincial · 6 marks
In the diagram, $Δ$ ABC is drawn with vertices A $(- 7; 0)$ , B $(3; 8)$ and C. P is the midpoint of line AB. Q is a point on line AC. AQC is a line on the $x$ -axis. E $(- \frac{1}{2}; 2)$ is a point on line PQ. PQ $∥$ BC $Diagram for this question$
4.3 Calculate the:
4.3.2 Area of trapezium PBCQ(6)
Show the memo (6 ticks)Memo
1. ✓
  AB = $164$ AB $= (3 + 7)^{2} + (8 - 0)^{2}$
  $= 164$
  1 mark · Calculate distance between two points
2. ✓
  Correct substitution in area of $Δ$ ABCArea $Δ$ ABC $= \frac{1}{2} \times 164 \times 16 sin 38, 65°$
  1 mark · Use the area rule
3. ✓
  63,99 $= 63, 99$
  1 mark · Use the area rule
4. ✓
  Correct substitution in area of $Δ$ APQArea $Δ$ APQ $= \frac{1}{2} \times \frac{164}{2} \times 8 sin 38, 65°$
  1 mark · Use the area rule
5. ✓
  16,00 $= 16, 00$
  1 mark · Use the area rule
6. ✓
  AnswerArea PBCQ $= 63, 99 - 16, 00 = 47, 99$
  1 mark · Calculate area or perimeter
5. Question 5.6 · November 2025 · Paper 1 · Provincial · 5 marks
The graphs of $f (x) = \frac{a}{x + p} + q$ and $g (x) = \frac{x}{3} + c$ are drawn below. C $(3; - 1)$ is the point of intersection between the asymptotes of $f$ . A and B are $x$ - and $y$ -intercepts of $f$ and $g$ respectively. E $(4; 2)$ is a point on $f$ in the first quadrant. $Diagram for this question$
5.6 Calculate the area of $△$ ABE if the length AB $= 210$ units.(5)
Show the memo (5 ticks)Memo
1. ✓
  $m_{A B}$ and $m_{E A}$ $m_{A B} = \frac{1}{3}$
  $m_{E A} = \frac{0 - 2}{6 - 4} = - 1$
  1 mark · Calculate gradient between two points
2. ✓
  Length of EAEA $= (4 - 6)^{2} + (2 - 0)^{2}$
  $= 22$
  1 mark · Calculate distance between two points
3. ✓
  $18, 43°$ and $45°$ $tan θ = \frac{1}{3}$ , $θ = 18, 43°$
  $∴$ B $\hat{A}$ O $= 18, 43°$ (vert. opp. $∠$ s)
  Ref. $∠$ : E $\hat{A}$ O $= 45°$ (from $m_{E A} = - 1$ )
  $∴$ E $\hat{A}$ B $= 63, 43°$
  1 mark · Calculate the inclination of a line
4. ✓
  Substitution in the area formulaArea $△$ ABE $= \frac{1}{2} \times$ EA $\times$ AB $\times sin$ E $\hat{A}$ B
  $= \frac{1}{2} \times (22) \times (210) sin 63, 43°$
  1 mark · Use the area rule
5. ✓
  Answer $= 8$ units $^{2}$
  1 mark · Use the area rule
6. Question 3.8 · November 2025 · Paper 2 · Provincial · 6 marks
In the diagram, W $(1; 6)$ , V, S and Q are the vertices of a quadrilateral WVSQ. P $(0; 2)$ is the point where line PQ intersects the $y$ -axis and R $(3; 0)$ is the point where line QR intersects the $x$ -axis. WV $∥$ QS and WU $=$ UV. PQRO is a rectangle. $Diagram for this question$
3.8 Determine the ratio: $\frac{Area of PQRO}{Area of Δ QUW}$ (6)
Show the memo (6 ticks)Memo
1. ✓
  WU $W U = \frac{101}{2}$
  1 mark · Calculate distance between two points
2. ✓
  UQ $U Q = (6 - 3)^{2} + (\frac{13}{2} - 2)^{2} = \frac{3 13}{2}$
  1 mark · Calculate distance between two points
3. ✓
  Correct substitution into area ruleArea of $Δ Q U W = \frac{1}{2} \times \frac{3 13}{2} \times \frac{101}{2} \times sin 50, 60°$
  1 mark · Use the area rule
4. ✓
  Area of triangle $= 10, 50$
  1 mark · Use the area rule
5. ✓
  Correct substitution in area of rectangleArea of PQRO $= 2 \times 3$
  1 mark · Calculate area or perimeter
6. ✓
  Answer $\frac{Area of PQRO}{Area of Δ Q U W} = \frac{2 \times 3}{10 , 50} = \frac{4}{7}$
  1 mark · Calculate area or perimeter

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