if a b c are the sides of a triangle and s be its semiperimeter and (s-a)=90 (s-b)=50 (s-c)=10 then find value of a
chaudharyprince042:
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Given, (s - a) = 90........(i)
(s - b) = 50 ........(ii)
(s - c) = 10 .........(iii)
adding all equations (i), (ii) and (iii),
(s - a) + (s - b) + (s - c) = 90 + 50 + 10
(s + s + s) - (a + b + c) = 150
3s - (a + b + c) = 150
we know, semiperimeter , s = (a + b + c)/2
so, (a + b + c) = 2s put it above ,
3s - 2s = 150
s = 150
now, from equation (1),
150 - a = 90
150 - 90 = a
a = 60
hence, value of a = 60
(s - b) = 50 ........(ii)
(s - c) = 10 .........(iii)
adding all equations (i), (ii) and (iii),
(s - a) + (s - b) + (s - c) = 90 + 50 + 10
(s + s + s) - (a + b + c) = 150
3s - (a + b + c) = 150
we know, semiperimeter , s = (a + b + c)/2
so, (a + b + c) = 2s put it above ,
3s - 2s = 150
s = 150
now, from equation (1),
150 - a = 90
150 - 90 = a
a = 60
hence, value of a = 60
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4
In the attachment I have answered this problem.
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