If an equilateral triangle has a height of $4.7 \sqrt{3}$ $4.7sqrt3$ and base of 9.4 cm, what is its area?

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Answer 1 · 2015-06-27T13:53:14

$\frac{1}{2} \times b a s e \times h e i g h t = {4.7}^{2} \sqrt{3}$ $1/2 xx base xx height = 4.7^2sqrt(3)$ $c m^{2}$ $cm^2$

$= 22.09 \sqrt{3}$ $= 22.09sqrt(3)$ $c m^{2}$ $cm^2$

$≅ 38.261$ $~= 38.261$ $c m^{2}$ $cm^2$

As with all triangles, the area is equal to half of the base multiplied by the height.

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