Question

42cm, 39cm,, 45cm find the area of a triangle if its sides are

Answer 1

Hey

a = 42 cm
b = 39cm
c = 45cm .

So ,
s = a + b + c / 2
s = 42 + 39 + 45 / 2
s = 126 / 2
s = 63 .

Now ,
$area = \sqrt{s(s - a)(s - b)(s - c)}$
$= \sqrt{63(63 - 42)(63 - 39)(63 - 45)}$
$= \sqrt{63 \times 21 \times 24 \times 18}$
$= \sqrt{9 \times 7 \times 7 \times 3 \times 6 \times 4 \times 6 \times 3}$
$= 3 \times 7 \times 3 \times 6 \times 2$
$= 108 \times 7$
$= 756$

So , area = 756 cm²

thanks :)

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Answer 2

Hey friend,
Here is the answer you were looking for:

Given,
The sides of triangle i.e. 42 cm, 39 cm and 45 cm.
$semi \: - \: circle \: = \: \frac{a + b + c}{2} \\ \\ = \frac{42 + 39 + 45}{2} \\ \\ = \frac{126}{2} \\ \\ = 63 \: cm \\ \\ area \: of \: triangle \\ (according \: to \: heron \: s \: formula) \\ = \sqrt{s(s - a)(s - b)(s - c)} \\ \\ = \sqrt{63(63 - 42)(63 - 39)(63 - 45)} \\ \\ = \sqrt{63 \times 21 \times 24 \times 18} \\ \\ = \sqrt{571536} \\ \\ = 756 {cm}^{2}$

Hope this helps!!!!

@Mahak24

Thanks...
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