Question

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

Answer 1

Step-by-step explanation:

a = 18cm

b = 10cm

perimeter = 42cm

c = 42-18-10 = 14cm

$s = \frac{p}{2} = \frac{42}{2} = 21$

area of triangle

$= \sqrt s({s - a)(s - b)(s - c)} \\ = \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ = \sqrt{21 \times 3 \times 11 \times 7} \\ = \sqrt{3 \times 7 \times 3 \times 11 \times 7} \\ = 3 \times 7 \sqrt{11} \\ = 21 \sqrt{11} \\ = 61.649(approx)$

Answer 2

$Area \: of \: triangle \sqrt{s(s - a)(9 - b)(s - 0}$

Here ,S is the Semi - perimeter,

and a,b,c are the Sides of the triangle.

Given,

a = 18 cm ,b = 10 Cm

and perimeter = 42 Cm

$\: \: \: Semi \: perimeter = s \frac{perimeter}{2}$

$\: \: \: s = \frac{42}{2}$

$\: \: \: s = 21cm$

We need to find C,

Now,

Perimeter = 42 Cm.

a+b+c = 42Cm.

18cm + 10cm + c =42

28 cm + c =42 Cm

C = 42-28cm

C = 14 cm.

$Area \: of \: the \: triangle = \sqrt{s(s - a)(s - b)(s - c)}$

$putting \: a = 18cm \: ,\: b = 10cm, \: c = 14cm \: and \: s = 21cm$

$Area \: of \: the \: triangle = \sqrt{21(21 - 18)(21 - 10)(21 - 14)}$

$\: \: \: = \sqrt{21(3)(11)(7)}$

$\: \: \: = \sqrt{21(7 \times 3)(11)}$

$\: \: \: = \sqrt{21(21)(11)}$

$\: \: \: = \sqrt{21 \times 21 \sqrt{11} }$

$\: \: \: = \sqrt{ {21}^{2} \times \sqrt{11} }$

$\: \: \: \: = (21) \times \sqrt{11}$

$\: \: \: = 21 \sqrt{11}$

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