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Area of a triangle · 3 marks

You may wonder, like earlier, is there a gap in our argument? What would we do if angle EFG is obtuse and the triangle were shaped like triangle EFG in the figure? Please work out the answer to this question.

EFG
Answer: Extend GF to H so that EH ⊥ GH (EH = h, HF = x). Then area(EFG) = area(EHG) − area(EHF) = 12(b + x)h − 12xh = 12bh. The formula still holds.

Step-by-step solution

Idea: When the foot of the height falls outside the base, write the triangle as the difference of two right-angled triangles, whose areas we already know.

EFGHhbx
  1. Since ∠EFG is obtuse, the perpendicular from E meets line GF outside FG. Extend GF to H with EH ⊥ GH. Let EH = h (the height), FG = b (the base) and HF = x.1 mark
  2. Right triangle EHG has legs h and b + x: area = 12(b + x)h. Right triangle EHF has legs h and x: area = 12xh. (A right triangle is half of a rectangle.)1 mark
  3. Triangle EFG = triangle EHG minus triangle EHF: area = 12(b + x)h − 12xh = 12bh. So the formula 12 × base × height works for obtuse triangles too.1 mark
Extend the base and use area(EFG) = area(EHG) − area(EHF) = ½(b + x)h − ½xh = ½bh. The formula ½ × base × height still holds.

Check: Numbers: b = 6, x = 2, h = 4: 12 × 8 × 4 − 12 × 2 × 4 = 16 − 4 = 12 = 12 × 6 × 4 ✓.

Answer to write in the exam

Produce GF to H with EH ⊥ GH; EH = h, FG = b, HF = x

ar(EHG) = 12(b + x)h, ar(EHF) = 12xh

ar(EFG) = ar(EHG) − ar(EHF)

= 12(b + x)h − 12xh

∴ ar(EFG) = 12bh

Common mistakes that cost marks

  • Taking the slant side EF as the height. The height is the perpendicular EH, which lies outside the triangle.
  • Adding the two right triangles instead of subtracting.
  • Using HG (b + x) as the base of triangle EFG. Its base is FG = b.

How this can come in the exam

MCQ (1 mark)

In triangle PQR, ∠Q is obtuse, QR = 9 cm, and the perpendicular from P to RQ produced is 4 cm. The area of triangle PQR is

  1. 13 cm2
  2. 18 cm2
  3. 36 cm2
  4. 26 cm2
Show answer

(B) 18 cm2
12 × 9 × 4 = 18 cm2; the height may lie outside the triangle.

Try one yourself

In an obtuse triangle the base is 10 cm. The perpendicular from the opposite vertex meets the base extended at a point 3 cm beyond the base, and is 6 cm long. Find the area by subtracting two right triangles.

Show answer

12 × 13 × 6 − 12 × 3 × 6 = 39 − 9 = 30 cm2 = 12 × 10 × 6 ✓.

More questions like this

All Perimeter and area questions · All maths questions

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