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larisa [96]
3 years ago
15

Given: ΔABC, AB = BC = AC = a. Find: The area of ΔABC

Mathematics
1 answer:
MArishka [77]3 years ago
7 0

Remember Heron's formula, which states the following:

A = \sqrt{s(s - a)(s - b)(s - c)}

  • s = \dfrac{a + b + c}{2}
  • a, b, and c are the side lengths of the triangle

In this case, we can say \overline{AB} = \overline{BC} = \overline{AC} = a.


Thus,

s = \dfrac{a + a + a}{2} = \dfrac{3a}{2},


Now, let's apply this to Heron's formula:

A = \sqrt{\Big( \dfrac{3a}{2} \Big) \Big( \dfrac{a}{2} \Big) ^3}


Now, let's try to simplify this expression if possible.

A = \sqrt{\Big( \dfrac{3a}{2} \Big) \Big( \dfrac{a}{2} \Big) ^3}

A = \sqrt{\Big( \dfrac{3a}{2} \Big) \Big( \dfrac{a^3}{8} \Big)}

A = \sqrt{\dfrac{3a^4}{16}}

A = \sqrt{\dfrac{3}{16}} \cdot \sqrt{a^4}

A = a^2 \sqrt{\dfrac{3}{16}}

A = a^2 \sqrt{\dfrac{3}{4}} \cdot \sqrt{\dfrac{1}{4}}

A = a^2 \dfrac{\sqrt{3}}{2} \cdot \dfrac{1}{2}

A = \dfrac{a^2 \sqrt{3}}{4}


Our answer is \boxed{A = \dfrac{a^2 \sqrt{3}}{4}}.

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