AD is drawn perpendicular to base BC of an equilateral triangle ABC given BC =10 find the length of AD correct up to 1 place of decimal
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area of equilateral trianle=√3/4*a²
√3/4*10*10
=25√3=1/2base *height
25√3=1/2*10*AD
solving this AD=8.65
√3/4*10*10
=25√3=1/2base *height
25√3=1/2*10*AD
solving this AD=8.65
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