# Triangle A has an area of #24 # and two sides of lengths #8 # and #12 #. Triangle B is similar to triangle A and has a side of length #5 #. What are the maximum and minimum possible areas of triangle B?

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Since triangles A and B are similar, their corresponding sides are in proportion. Let's denote the ratio of the corresponding sides of triangles A and B as ( k ).

Given that triangle A has sides of lengths 8 and 12, its area is 24. We can use Heron's formula to find the length of the third side: [ s = \frac{8 + 12 + x}{2} ] [ A = \sqrt{s(s-8)(s-12)(s-x)} = 24 ] Where ( x ) is the length of the third side.

By solving this equation, we find ( x = 4 ) or ( x = 16 ). However, since the third side must be less than the sum of the other two sides and greater than their difference, ( x ) must be 4.

Now, we can find the ratio ( k ): [ k = \frac{5}{4} ]

The maximum possible area of triangle B occurs when its side length is the maximum length it can be in proportion to triangle A, which is when ( k = \frac{5}{4} ). So, the maximum area is ( 24 \times \left(\frac{5}{4}\right)^2 = 37.5 ).

The minimum possible area of triangle B occurs when its side length is the minimum length it can be in proportion to triangle A, which is when ( k = \frac{5}{16} ). So, the minimum area is ( 24 \times \left(\frac{5}{16}\right)^2 = 2.34375 ).

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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.

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