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Bunuel
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When we know two of the side lengths, the greatest possible area of the triangle will occur when the two sides form a right triangle.


With \(8\) and \(10\) as the base and height of the triangle, we have an area of:

Area = 1⁄2 × Base × Height
\(Area = 1⁄2 × 8 × 10\)
\(Area = 40\)

Since the greatest possible area of the triangle with sides \(8\), \(10\), and \(z\) is \(40\),

other smaller areas ARE possible (if we change the angle to something other than 90 degrees).

Thus, it cannot have an area greater than \(40\).

The final answer is .
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Bunuel
If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle?


A. 3
B. 6
C. 12
D. 24
E. 46

Just use logic.
Think about it: given two sides and an unknown third - we can have a triangle with very very small area (almost 0), then the area increases and then finally starts decreasing again until it comes to 0 again. So there is a maximum limit to the area that the triangle can have. Hence given a range of increasing areas, the one which will not be possible is the largest one. For example, if 24 were not possible, 46 would not be possible too. But there is only one correct answer. Hence it must be 46 only.

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Bunuel
If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle?


A. 3
B. 6
C. 12
D. 24
E. 46

We know that the third side should not be greater than the sum of the other side this gives us space to accomdate even relatively small sides

This gave the spark to eleminate A since the area can be really small so A is out then for not violating the above stated fact it cannot go beyond a certain area i knew about 6-8-10 triangle which gave the area to 24 therefore the only large possibility that left was 40

Therefore IMO E
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max area = 0.5 * 8 * 10 = 40 \(units^2\)
Therefore 46 cannot be the area, so IMO Ans = E
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