Question

Find the area of triangle two sides of which 18cm and 10cm and perimeter is 40cm?

Answer 1

$<b>Hey mate!!$

Third side

= (40 - 18 + 10)

= 40 - 28

= 12

Now,

S = a + b + c/2 = 40/2 = 20

$\bold { Area \: of \: triangle } \\ \\ \bold{ \sqrt{s(s - a)(s - b)(s - c)} } \\ \\ \bold{ \sqrt{20(20 - 18)(20 - 10)(20 - 12)}} \\ \\ \bold{ \sqrt{20 \times 2 \times 10 \times 8}} \\ \\ \bold{ \sqrt{4 \times 5 \times 2 \times 2\times 5 \times 2 \times 4}} \\ \\ \bold{ \sqrt{4 {}^{2} \times 5 {}^{2} \times 2 {}^{2} \times 2 }} \\ \\ \bold{ 40\sqrt{2}} \: \\ \\ \bold{ =40 \sqrt{2} \: cm {}^{2} }$

Thanks for the question!!

Answer 2

❤❤Here is your answer ✌ ✌

$\huge\boxed{\red{\bold{Answer}}}$

Third side 

= (40 - 18 + 10) 

= 40 - 28

= 12

Now,

S = a + b + c/2 = 40/2 = 20
$area = \sqrt{(s - a)(s - b)(s - c) \ } \\ = \sqrt{20(20- 18 )(20 - 10)(20 - 12) } \\ \sqrt{20\times 2 \times 10 \times 8 } \\ \sqrt{\times 4\ times 2 \times 5\times 2 \times 5 \times 2 {}^{2} \times 2 } \\ 40\sqrt{2}$