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Find the altitude and the area of an equilateral triangle whose side is 8 cm long.
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in equilateral traingle altitude will be perpendicular to base and bisect it
|dw:1391379208225:dw|
from pythagoras theorm altitude=\[\sqrt{8^{2}-4^{2}}=\sqrt{64-16}=\sqrt{48}=\sqrt{16*3}=4*\sqrt{3}\]
area=(altitude*base)/2 \[\frac{ 4*\sqrt{3}*8 }{ 2 }=16*\sqrt{3}\]
@conan.195114
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sorry, used\[h=\frac{ 2\sqrt{s(s-a)(s-b)(s-c)} }{c}\] and got \[4\sqrt{3}\] is this correct?
for area i used the heron's formula and got the same answer \[16\sqrt{3}\] thanks for helping me though i found it too long to wait so i studied on my own. thanks again
okk u can also do so ....but in heron calculation will be more
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