Question

find the area of a triangle two sides of which are 18 cm and 10 cm and the perimeter is 42 CM

Answer 1

Given,
AB=18 cm
BC=10 cm
AB+BC+AC=42 cm
To find,
Area of the triangle
ATQ,
AB+BC+AC=42
18+10+AC=42
AC=42-28
AC=14 cm
Now,
Let,
AB=a
BC=b
AC=c
Then,
By Herons formula:-
s=(a+b+c)/2=(18+10+14)/2=42/2=21
So,
Area=√[s(s-a)(s-b)(s-c)]
=√[21(21-18)(21-10)(21-14)]
=√[21*3*11*7]
=√[231*21]
=√4851
=69.649
=70 cm²(approx)

Hope this helps you!

Answer 2

now u follow this steps

hope it will help u