Question

@Michele_Laino

Answer 1

Is it an equilateral triangle?

Answer 2

Yes. The lengths are all 14.

Answer 3

we can apply the \(Heron\)'s formula

Answer 4

here is the Heron's formula:
\[\Large area = \sqrt {p\left( {p - l} \right)\left( {p - l} \right)\left( {p - l} \right)} \]
where \(p\) is the half perimeter

Answer 5

namely:
\[\Large p = \frac{{3l}}{2} = \frac{{3 \cdot 14}}{2}...?\]

Answer 6

21

Answer 7

correct!

Answer 8

so, after a substitution, we get:
\[\Large \begin{gathered}
area = \sqrt {p\left( {p - l} \right)\left( {p - l} \right)\left( {p - l} \right)} = \hfill \\
\hfill \\
= \sqrt {21 \cdot 7 \cdot 7 \cdot 7}=...? \hfill \\
\end{gathered} \]

Answer 9

since \(p-l=21-14=7\)

Answer 10

1571.8

Answer 11

hint:
we have:
\[\Large \begin{gathered}
area = \sqrt {p\left( {p - l} \right)\left( {p - l} \right)\left( {p - l} \right)} = \hfill \\
\hfill \\
= \sqrt {21 \cdot 7 \cdot 7 \cdot 7} = \sqrt {7203} = ...? \hfill \\
\end{gathered} \]

Answer 12

84.8

Answer 13

that's right! Better is \(84.9\)
another procedure can be this:
\[\Large area = \frac{1}{2} \times 14 \times 14 \times \sin 60 = ...?\]

Answer 14

84.8---> 84.9

Answer 15

of course, both procedures give the same result

Answer 16

yes! It is the right rounded result

Answer 17

since it is:
area=84.87048

Answer 18

I think I got the correct answer for the next question.

Answer 19

The area could be 129.90

Answer 20

using the second procedure above, we get:
\[\Large area = \frac{1}{2} \times 10\sqrt 3 \times 10\sqrt 3 \times \sin 60 = ...?\]

Answer 21

130.1

Answer 22

(rounded)

Answer 23

I think it is \(130.0\)

Answer 24

Would we apply the formula if there was a apothem in the triangle?

Answer 25

no, no, I',m applying this formula:

\[\Large area = \frac{1}{2}ab\sin \gamma \]
which holds for any triangle

Answer 26

here we can note this:

as we can see, the equilateral triangle, can be subdivided into three isosceles triangles, and the area of each isosceles triangle, is:
\[\Large area = \frac{1}{2} \times 14 \times 14 \times \sin 120 = ...?\]