Question

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

Answer 1

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