A triangle has two corners with angles of # pi / 4 # and # pi / 2 #. If one side of the triangle has a length of #7 #, what is the largest possible area of the triangle?

Side b & c are base & height and a the hypotenuse.

To find the largest possible area of the triangle given that it has angles of ( \frac{\pi}{4} ) and ( \frac{\pi}{2} ), we need to determine the length of the third side and then calculate the area using the formula for the area of a triangle.

Since the sum of angles in a triangle is ( \pi ) radians, and we already know two angles (( \frac{\pi}{4} ) and ( \frac{\pi}{2} )), we can find the third angle:

[ \text{Third angle} = \pi - \left( \frac{\pi}{4} + \frac{\pi}{2} \right) = \frac{\pi}{4} ]

Now, we have a triangle with angles ( \frac{\pi}{4} ), ( \frac{\pi}{2} ), and ( \frac{\pi}{4} ), which forms an isosceles right triangle.

Given that one side of the triangle has a length of 7, and it's opposite one of the ( \frac{\pi}{4} ) angles, it means the other two sides are equal.

Using trigonometry in this right triangle, we can find the length of the other two sides using the known angle ( \frac{\pi}{4} ):

[ \text{Length of other two sides} = 7 \times \sqrt{2} ]

Now, we can calculate the area of the triangle using the formula:

[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} ]

Since the triangle is isosceles, we can choose any of the two equal sides as the base and the other as the height:

[ \text{Area} = \frac{1}{2} \times (7 \times \sqrt{2}) \times (7 \times \sqrt{2}) = \frac{1}{2} \times 7 \times 7 \times 2 = 49 ]

So, the largest possible area of the triangle is 49 square units.