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drop perpendicular from a vertex to b... 2 right are formed let height be h
h^2+((6√3)/2)^2=3√3
h^2+(3√3)/4=3√3
h^2=3√3-3√3/4
h^2=(3√3)(3/4)
h^2=(3^{1/3 + 1})/4
h^2=3^{4/3}/4
h=(3^{2/3})/2
simplify further and check calculation...but procedure will same
h^2+((6√3)/2)^2=3√3
h^2+(3√3)/4=3√3
h^2=3√3-3√3/4
h^2=(3√3)(3/4)
h^2=(3^{1/3 + 1})/4
h^2=3^{4/3}/4
h=(3^{2/3})/2
simplify further and check calculation...but procedure will same
sk700:
kya hai answer
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