In ∆ABC, seg AM is perpendicular to side BC, AB =17, AC=25, BC = 28. find BM, CM and AM (hint : bm = x, cm= 28-x, AM^2= AB^2 - BM^2 = AC^2 - CM^2 get x
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In ∆ABC, seg AM is perpendicular to side BC, AB =17, AC=25, BC = 28. find BM, CM and AM (hint : bm = x, cm= 28-x, AM^2= AB^2 - BM^2 = AC^2 - CM^2 get x
Pls find attached. Is it correct? Not
BM = CM iff it would have been isosceles triangle
rather then using Hint i would use another way
to find AM height we will find first area of Triangle ABC
AB =17, AC=25, BC = 28
s = (17+25+28)/2 = 35
area = sqroot{(35)(35-17)(35-25)(35-28)}
area = sqroot(35*18×10×7)
area = 210 sqcm
area = (1/2)× 28 × AM
210 = 14 AM
AM = 15 cm
BM^2 = AB^2 - AM^2
BM^2 = 17^2 - 15^2
BM^2 = 289 - 225
BM^2 = 64
BM = 8 cm
CM = BC - BM = 28-8 = 20 cm
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