Question

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

Answer 1

Let the three sides of the triangle be a,b,c
a=18m,b=10m,c.
Perimeter=42m
Now,
a+b+c=Perimeter
$= > 18 + 10 + c = 42 \\ = > c = 42 - 28 \\ = > c = 14$
Area,
$\sqrt{s(s - a)(s - b)(s - c)} \\ = \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ = \sqrt{21 \times 3 \times 11 \times 7} \\ = \sqrt{4851} \\ = 69.649... \\ = 69.65(approx)$