Question

if Herons Formulae=
$\sqrt{s} (s - a)(s -b )(s - c)$
then solve,

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

Answer 1

$\textit{ \textbf{ Hello Mate! }}$

Perimeter of triangle = Sum of all the sides

42 cm = a + b + c

42 cm = 18 cm + 10 cm + c

14 cm = c

$herons \: formula \: \\ \sqrt{s(s - a)(s - b)(s - c)} \\ where \: s \: = \: \frac{a + b + c}{2} \\ = \frac{42}{2} cm \: = 21cm \\ \\ \sqrt{21(21 - 18)(21 - 10)(21 - 14)} \\ \sqrt{3 \times 7 \times 3 \times 11 \times 7} \\ = 21 \sqrt{11} \: {cm}^{2}$

We get, 21 × 3.32 cm^2 = 69.72 cm^2

$\boxed{ \textsf{ Hence,\:area\:of\:triangle\:is\:69.72\:sq\:cm}}$

$\texttt{ Have Great Future Ahead! }$

Answer 2

Third side = Perimeter of the △ - Sum of two sides

⇒ S3 = 42 cm - (18 cm + 10 cm)

⇒ S3 = 42 cm - 28 cm

⇒ S3 = 14 cm

Semi-perimeter = (a + b + c)½ = (42 cm)½ = 21 cm

∴ Area = √[21(21 - 14)(21 - 18)(21 - 10)]

⇒ Area = √[21 × 7 × 3 × 11]

⇒ Area = √[7 × 3 × 7 × 3 × 11]

⇒ Area = 21√11 cm²