\(\begin{aligned} &=\dfrac{1}{2}ab \sin C \\\\ &=\dfrac{1}{2}ac \sin B \\\\ &=\dfrac{1}{2}bc \sin A \end{aligned}\)
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| Heron’s formula: |
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| Area of a triangle |
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\(=\sqrt{s(s-a)(s-b)(s-c)}\)
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| where \(a, \,b\) and \(c\) are the sides of the triangle and |
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\(s=\dfrac{a+b+c}{2}\)
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Example:
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| Find the area of the following triangle. |
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\(\begin{aligned} &\text{Area of } \Delta ABC\\\\ &=\dfrac{1}{2}(16)(13) \sin 36^\circ \\\\ &=61.13 \text{ cm}^2 \end{aligned}\)
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The diagram shows a triangle \(ABC\).
Calculate the area of the triangle.
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| Use Heron's formula to calculate the area of the triangle. |
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\(\begin{aligned} s&=\dfrac{1}{2}(3.8+1.8+3) \\\\ &=4.3 \\\\ &\text{Area of }\Delta{ABC} \\\\ &=\sqrt{\begin{matrix}4.3(4.3-3.8)\\(4.3-1.8)(4.3-3)\end{matrix}} \\\\ &=2.643 \text{ cm}^2. \end{aligned}\)
