Question

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

Answer 1

let the third side of triangle =x

then

perimeter of triangle= sum of all sides of triangle

42=18+10+x

42=28x

x=14cm

s=21cm

s-a=21-18=3

s-b=21-10=11

s-c=21-14=7

area of triangle=√s-a*s-b*s-c*s

,√21*3*11*7

=21√11 ans

Answer 2

Answer

Step-by-step explanation:

a=18cm,b=10cm,c=?

a+b+c = perimeter of given triangle

18+10+(c)=42

28+(c)=42

c =14 cm

Heron's formula = $\sqrt{s (s-a)(s-b)(s-c)$

where S is semiperimeter

s = a+b+c/2

s=42/2

s=21

= $\sqrt{21(21-18)(21-10)(21-14)$

= $\sqrt{21*3*11*7$

= $21\sqrt{11$ $cm^{2}$