find the percentage increase in the area of a triangle if each side is double
class 9th chapter 12 heron's formula
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let the sides of a triangle be a, b, c
s=a+b+c/2
then area of triangle =√s(s-a)(s-b) (s-c)
s=2a+2b+2c/2=2(a+b+c)/2=2s
therefore area of triangle=√2s(2s-2a)(2s-2b)(2s-2c)=√16s(s-a) (s-b) (s-c) =4s
4s-s=3
therefore percentage increase in triangle=3/s*100=300%
so this is the answer it might helped you
s=a+b+c/2
then area of triangle =√s(s-a)(s-b) (s-c)
s=2a+2b+2c/2=2(a+b+c)/2=2s
therefore area of triangle=√2s(2s-2a)(2s-2b)(2s-2c)=√16s(s-a) (s-b) (s-c) =4s
4s-s=3
therefore percentage increase in triangle=3/s*100=300%
so this is the answer it might helped you
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