find the area of a triangle having perimeter 32cm one side of its side is equal to 11 cm and difference of the other two is 5 cm
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let the sides be a,b and c
therefore
a+b+c=32
a=11
11+b+c=32
b+c=21
and
b-c=5
on adding the equation
2b=26
b=13
and thus c=8
therefore area =√s(s-a)(s-b)(s-c)
=s=11+13+8/2
=32/2
=16
=√16*5*3*8
=√80*24
=√1920
43.8
therefore
a+b+c=32
a=11
11+b+c=32
b+c=21
and
b-c=5
on adding the equation
2b=26
b=13
and thus c=8
therefore area =√s(s-a)(s-b)(s-c)
=s=11+13+8/2
=32/2
=16
=√16*5*3*8
=√80*24
=√1920
43.8
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