A triangle has sides with lengths: 1, 5, and 3. How do you find the area of the triangle using Heron's formula?
There is no such triangle, since
It's interesting to note that you will be trying to take the square root of a negative number if you try to use Heron's formula with such lengths, so there won't be any real area.
Next, the following formula defines the semiperimeter:
And the following formula yields the area:
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To find the area of a triangle using Heron's formula, follow these steps:

Calculate the semiperimeter (( s )) of the triangle, which is half the sum of the lengths of its three sides: [ s = \frac{1 + 5 + 3}{2} = 4.5 ]

Use Heron's formula to find the area (( A )) of the triangle: [ A = \sqrt{s(s  1)(s  5)(s  3)} ]

Substitute the value of ( s ) into the formula and simplify: [ A = \sqrt{4.5(4.5  1)(4.5  5)(4.5  3)} ] [ A = \sqrt{4.5 \times 3.5 \times (0.5) \times 1.5} ] [ A = \sqrt{4.5 \times 3.5 \times 1.5 \times 0.5} ] [ A = \sqrt{11.8125} ] [ A \approx 3.438 ]
Therefore, the area of the triangle is approximately ( 3.438 ) square units.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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