A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{4}$ $pi / 4$ . If one side of the triangle has a length of $16$ $16$ , what is the largest possible area of the triangle?

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A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{4}$ $pi / 4$ . If one side of the triangle has a length of $16$ $16$ , what is the largest possible area of the triangle?

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Answer 1 · 2017-03-26T19:55:10

128

Explanation:

The angles of a triangle have to sum up to $π$ $pi$ and since two of the angles are both $\frac{π}{4}$ $pi/4$ , the remaining angle must be $\frac{π}{2}$ $pi/2$ . Knowing special triangles, this is a right triangle where the two legs are the same length and the hypotenuse is $\sqrt{2}$ $sqrt2$ times the legs.

One of the sides of the triangle has a length of $16$ $16$ . We can set that to be either the leg length or the hypotenuse length. Intuitively setting it as the leg length will give a larger triangle because the sides will be longer: $16, 16, and 16 \sqrt{2} \approx 22.6$ $16, 16, and 16sqrt2~~22.6$ instead of $\frac{16}{\sqrt{2}} \approx 11.3, \frac{16}{\sqrt{2}} \approx 11.3, and 16$ $16/sqrt2~~11.3, 16/sqrt2~~11.3, and 16$ .

Since our largest triangle has the largest area, we know that the first set is what we're looking for.

Since this is a right triangle with legs 16 and 16, the area is $\frac{16^{2}}{2} = 128$ $16^2/2=128$

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A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{4}$ $pi / 4$ . If one side of the triangle has a length of $16$ $16$ , what is the largest possible area of the triangle?

1 Answer

Explanation:

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A triangle has two corners with angles of $\frac{π}{4}$ $pi / 4$ and $\frac{π}{4}$ $pi / 4$ . If one side of the triangle has a length of $16$ $16$ , what is the largest possible area of the triangle?