Question

If 2s=a+b+c then area of triangle abc is

Answer 1

$2s = a + b + c \\ s = (a + b + c) \div 2 \\ s \: o \: area = \sqrt{s(s - a)(s - b)(s - c)}$
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Answer 2

QUESTION:

If 2s=a+b+c then area of triangle ABC is

ANSWER:

If 2s = a + b + c, then the area of ΔABC is $\sqrt{s(s-a)(s-b)(s-c)}.$

Given,

In a triangle ABC,

2s = a + b + c.

To find,

Area of ΔABC.

Solution,

In any triangle ABC with three of its sides as a, b, and c, 's' is defined as the semi-perimeter of ΔABC, and, is given by

$s=\frac{a+b+c}{2} \hfill ...(1)$

where s is the semiperimeter.

Now, when the semi-perimeter 's' is known, the area of ΔABC (say A) is determined using Heron's formula, which is given as

$A=\sqrt{s(s-a)(s-b)(s-c)}$

We can see that the given relation (2s = a + b + c) can be rearranged in the form of (1), and it will not affect the 's', as well as the area 'A'.

Therefore, if 2s = a + b + c, then the area of ΔABC is $\sqrt{s(s-a)(s-b)(s-c)}.$

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