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You know that Area = base x height / 2, but what other ways are there to find the area of a triangle? Brace yourself for some potent formulae.

Triangle \( ABC\) has coodinates \(A= (-4, 0)\), \(B= (4 , 0)\), and \(C= (0 , 3)\).

Let \(P\) be the point in the first quadrant such that \(\triangle ABP\) has half the area of \(\triangle ABC\) but both triangles have the same perimeter.

What is the length of \(CP\)? If your solution is in a form of \(\sqrt{d}\), submit \(d\) as the answer.

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Let \(ABCD\) be a square of side length 12.

- \(E\) is the midpoint of \(CB\),
- \( FC = \frac{1}{3} DC \),
- \( GD = \frac{1}{4} DA \),
- \( AH = \frac{1}{3} AE \),
- \(J\) is the midpoint of \(FE\).

What is the area of the purple triangle?

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Which of the following triangles has a larger area:

Triangle A, with side lengths of \( 13, 13, 10 \)

\[\text{or } \]

Triangle B, with side lengths of \( 13, 13, 24 \)?

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What is the largest possible area of an isosceles triangle with two sides of length 2?

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