Question

find the area of the triangle with sides 40cm,41cm and 9cm​

Answer 1

Hello mate..

$\huge ⁕ \mathfrak \red{solution}⁕$

GIVEN:

The sides of triangle are 40cm, 41cm, 9cm.

$\bold{s = \frac{a + b + c}{2} } \\ \bold{ = \frac{40 + 41 + 9}{2}} \\ \bold{ = \frac{90}{2} } \\ \bold{s = 45}$

Then,

S × a = 45 × 40 => 1800

s × b = 45 × 41 => 1845

s × c = 45 × 9 => 405

$\bold \green{area \: of \: a \: triangle = \sqrt{s(s - a)(s - b)(s - c)} }$

$\bold{ = \sqrt{45(1800)(1845)(405)} } \\ \bold{ = \sqrt{45 \times1345005000 } } \\ \bold{ = \sqrt{60525225000} } \\ \bold{ = 246018.7}$

Therefore,

The area of the triangle = 246018.7

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HOPE THIS HELPS YOU..