Question

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

Answer 1

$\mathfrak{\huge{Answer:}}$

Given are the two sides of triangle :- 18 cm, 10 cm

Perimeter is given = 42 cm

Third side will be = 42 cm - ( 18 cm + 10 cm )

Third side will be = 42 cm - 28 cm

Third side will be = 14 cm

Semiperimeter Will be = $\sf{\frac{42}{2}}$

Semiperimeter Will be (s) = 21 cm

We know by the Heron's Formula that :-

Area of a triangle whose altitude is unknown:-

$\tt{\sqrt{s(s - a)(s - b)(s - c)}}$

Solve this further

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

Solve this formed equation further

=》 $\tt{\sqrt{21 (3)(11)(7)}}$

Some more steps to go

=》 $\tt{\sqrt{4851}}$

Last one step to go

=》 $\tt{21\sqrt{11}\:cm^{2}}$

That's the answer !

Answer 2

Perimeter of triangle = 42

two sides are 18 and 10 cm

lets find third side

third side = perimeter - sum of 2 sides

= 42 - ( 18+10)

= 42 - 28

= 14 cm

So three sides are 18,10,14

Now area

Let's find semi perimeter = 42/2 = 21

So area = √( s( s-a)( s-b)( s-c))

where s is semi perimeter

a,b,c are 3 sides

Area = √( 21( 21-18)(21-10)(21-14)

= √( 21× 3 × 11 × 7)

= √4851

= 21√11

= 69.6 cm^2 approx