Question

FIND THE AREA OF TRIANGLE, TWO SIDES OF WHO ARE 18 CM AND 10 CM AND THE PERIMETEER IS 42 CM

Answer 1

Answer:

Step-by-step explanation:

Given,

Side a=18cm

Side b=10cm

Perimeter=42cm=a+b+c

:. Putting value

42=18+10+c

42=28+c

42-28=c

14=c

Now,

S=(a+b+c)/2

:. Putting value

S=42/2

S=21

Now according to Heron's formula-

Area of a triangle

=√{s(s-a)(s-b)(s-c)}

:. Putting value

=√{21(21-18)(21-10)(21-14)}

=√{21(3)(11)(7)}

=√4851

=21√11cm²

(ÆÑẞWĒRẞ)

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Answer 2

Answer:

Given,

Side a=18cm

Side b=10cm

Perimeter=42cm=a+b+c

:. Putting value

42=18+10+c

42=28+c

42-28=c

14=c

Now,

S=(a+b+c)/2

:. Putting value

S=42/2

S=21

Now according to Heron's formula-

Area of a triangle

=√{s(s-a)(s-b)(s-c)}

:. Putting value

=√{21(21-18)(21-10)(21-14)}

=√{21(3)(11)(7)}

=√4851

=21√11cm²