Question

find the area of the triangle whose 2 sides are 18 and 10 and perimeter is 42

Answer 1

other side = 42 - (18 + 10)  =  14

S = 42/2 = 21

Area = √21 (21 - 18) (21 - 10) (21 - 14)

√ 21 x 3 x 11 x 7

√21 x 21 x 11

21√11

Answer 2

Let the third side of triangle be x.
perimeter = 42
18 + 10 +x = 42
28 + x = 42
x = 42-28
x = 14

S = (a+b+c)/2
S = (18+10+14)/2
S = 42/2= 21
area =
$\sqrt{s(s - a )\: (s - b )\: (s - c)}$
√21(21-18)(21-10)(21-14)
√21×3×11×7
√7×3×3×11×7
21√11


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