A triangle has two corners with angles of # pi / 4 # and # pi / 4 #. If one side of the triangle has a length of #16 #, what is the largest possible area of the triangle?

Question

A triangle has two corners with angles of #  pi  /  4  # and #  pi  /  4   #.  If one side of the triangle has a length of #16  #, what is the largest possible area of the triangle?

Avery Allen · Accepted Answer

128

The angles of a triangle have to sum up to #pi# and since two of the angles are both #pi/4#, the remaining angle must be #pi/2#. Knowing special triangles, this is a right triangle where the two legs are the same length and the hypotenuse is #sqrt2# times the legs. One of the sides of the triangle has a length of #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 16sqrt2~~22.6# instead of #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 #16^2/2=128#

Wyatt Anderson · Answer

undefined The largest possible area of the triangle can be found using the formula for the area of a triangle given two sides and the angle between them. In this case, the two sides are equal (since the two angles are equal) and one side has a length of 16. Therefore, the other side also has a length of 16. Using the formula $Area = \frac{1}{2} \times \text{base} \times \text{height}$, where the base is one of the sides and the height is the distance from the opposite corner to the base, we can calculate the area of the triangle. The height can be found using trigonometry, specifically the tangent function, since we know the angle and the adjacent side length.

$Area = \frac{1}{2} \times 16 \times 16 \times \tan(\frac{\pi}{4})$

$Area = \frac{1}{2} \times 16 \times 16 \times 1$

$Area = 128$

Therefore, the largest possible area of the triangle is 128 square units.