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What is the area of an equilateral triangle of side length 20 cm?

1 Answer

Nov 26, 2015

$100 \sqrt{3}$ $100sqrt(3)$

Explanation:

Referring to this image,
http://areeweb.polito.it/didattica/polymath/htmlS/argoment/ParoleMate/Gen_08/Img/TriangoloEquilatero%20(11)png
we know that $A B = A C = B C = 20$ $AB=AC=BC=20$ .

This means that the height cuts $A B$ $AB$ in two equals parts, $A H$ $AH$ and $H B$ $HB$ , each $10$ $10$ units long.

This means that, for example, $A H C$ $AHC$ is a right triangle with $A C = 20$ $AC=20$ and $A H = 10$ $AH=10$ , so

$C H = \sqrt{A C^{2} - A H^{2}} = \sqrt{20^{2} - 10^{2}} = \sqrt{300} = 10 \sqrt{3}$ $CH = sqrt(AC^2-AH^2) = sqrt(20^2-10^2)=sqrt(300)=10sqrt(3)$

Since we know the base and the height, then the area is

$\frac{20 \cdot 10 \sqrt{3}}{2} = 100 \sqrt{3}$ $(20*10sqrt(3))/2 = 100sqrt(3)$

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