How do you use Heron's formula to find the area of a triangle with sides of lengths #1 #, #2 #, and #2 #?

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To use Heron's formula to find the area of a triangle with sides of lengths 1, 2, and 2, you first calculate the semi-perimeter, ( s ), using the formula:
[ s = \frac{{a + b + c}}{2} ]
where ( a ), ( b ), and ( c ) are the lengths of the triangle's sides.
In this case, ( a = 1 ), ( b = 2 ), and ( c = 2 ).
[ s = \frac{{1 + 2 + 2}}{2} = \frac{5}{2} ]
Next, use Heron's formula to find the area, ( A ), of the triangle:
[ A = \sqrt{s(s - a)(s - b)(s - c)} ]
[ A = \sqrt{\frac{5}{2} \times \frac{5}{2} \times \frac{3}{2} \times \frac{3}{2}} ]
[ A = \sqrt{\frac{225}{16}} ]
[ A = \frac{15}{4} ]
So, the area of the triangle is ( \frac{15}{4} ) square units.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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