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I had this question in an exam:

In the triangle below, $AB = 12cm$, $BC = 19cm$ and $AC = 14cm$. Calculate the area of the triangle.

Exam question

The answer to this question finds the angle $A$ using the cosine rule and then uses this formula to find the area:

$$ \frac { 1 }{ 2 } ab \sin { A } $$

Why can't I just use Heron's formula, where the area of $\triangle ABC$ with perimeter $S$ is:

$$ \sqrt { s(s - a) (s - b) (s - c) } $$

This is a GCSE question.

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    $\begingroup$ what is what is gcse? $\endgroup$
    – abel
    Mar 8, 2015 at 14:24
  • $\begingroup$ @abel general certificate of secondary education, the UK equivalent to high school $\endgroup$ Mar 8, 2015 at 14:28
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    $\begingroup$ This appears to be a GCSE curriculum document. An Appendix provides a list of formulas a student is expected to know. It does seem curious that Heron's Formula is absent. But, then, so is $A = \frac12 bh$ ... except insofar as that's covered as a special case of $A=\frac12 ab \sin C$. Yet, it lists both the Law of Cosines and the Pythagorean Theorem. Go figure. Your question probably isn't answerable by anyone but the GCSE folks themselves. $\endgroup$
    – Blue
    Mar 8, 2015 at 14:37
  • $\begingroup$ I would ask them - if they would listen.. I guess I'll have to wait for me to be 18 for that.. $\endgroup$ Mar 8, 2015 at 14:39
  • $\begingroup$ I'm guessing (I don't know) cosine rule is more of a core skill to know in other areas? GCSE is designed to give a broad access to higher maths. $\endgroup$
    – Karl
    Mar 8, 2015 at 14:44

4 Answers 4

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I'm guessing this is because they are only asking the question to test the candidates ability to notice and apply trigonometry. Most candidates will not have been taught Heron's formula and so they are not expecting anyone to use it. I imagine that any correct answer would be awarded full marks regardless unless they specify which method to use as part of the question.

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  • $\begingroup$ I suppose so, but, this is wrong use of the trigonometric function $\endgroup$ Mar 8, 2015 at 14:34
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    $\begingroup$ I'm curious as to why you say it's a wrong use of the trig function? $\endgroup$
    – Karl
    Mar 8, 2015 at 14:48
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Bit late on this I know.
Heron's formula is fun but often painful and long; most of the time in GCSE it would be much more work. It is also off-syllabus as the GCSE course tries to teach you trigonometry - so there is no point in teaching you shortcuts for trigonometry right now as the cosine rule etc. will lead on to more important and difficult trigonometry in A Level maths, for example.

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Following my comment $$2bc\cos A =b^2+c^2-a^2$$

I'll use $X$ for the area, so that $4X=2bc \sin A$

Square these two and add:

$$16X^2+(b^2+c^2-a^2)^2=4b^2c^2$$ so that $$16X^2=4b^2c^2-(b^2+c^2-a^2)^2$$

You can use this directly to compute $X$ - possibly more easily than Heron.

If we factorise the right-hand side as the difference of two squares we get:

$$16X^2=(2bc+b^2+c^2-a^2)(2bc-b^2-c^2+a^2)=\left((a+b)^2-c^2\right)\left(a^2-(b-c)^2\right)$$

Use the difference of two squares on this to get $$16X^2=(a+b+c)(a+b-c)(a-b+c)(-a+b+c)$$ which easily reduces to Heron's formula - but why do the extra work?

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Seconding Karl's answer that the exam may not actually disallow Heron's formula (as claimed), due to the British exam boards practising "positive marking" these days:

From a 2018 GCSE mark scheme:

GENERIC MARKING PRINCIPLE 3:

Marks must be awarded positively:

• marks are awarded for correct/valid answers, as defined in the mark scheme. However, credit is given for valid answers which go beyond the scope of the syllabus and mark scheme, referring to your Team Leader as appropriate

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