How would I find the area of a 45-45-90 triangle with one side of length 73?

Question

Aurora Ayala · Accepted Answer

#AB = 73 sqrt2, BC = AC = 73#

Now i hope you are aware of the Pythagoras Theorem which states;

#AB^2 = BC^2 + AC^2#

Additionally

This is the definition of a Pythagorean triplet where m>n

Now notice BC = AC in the above image

For this scenario we have special formula;

#H = Lsqrt2#

So this is a nice template for this type of a triangle;

For the side provided to be the hypotenuse it has to be a multiple of #sqrt2# which clearly it isn't;

So we can make out first assumption

#BC = AC = 73#

Now its just substitution;

#73^2 + 73^2 = x^2#

#sqrt(73^2 + 73^2) = x#

#x = sqrt(2*73^2)#

#=>AB = 73 sqrt2#

Caroline Aucoin · Answer

undefined To find the area of a 45-45-90 triangle with one side of length 73, you can use the formula for the area of a triangle, which is $ \frac{1}{2} \times \text{base} \times \text{height} $. In a 45-45-90 triangle, the legs are congruent, so if one leg has a length of 73, the other leg also has a length of 73.

Since the triangle is isosceles, when it's split along the altitude, it forms two congruent right triangles. In a 45-45-90 triangle, the legs are equal, so if one leg is 73, the other leg is also 73. To find the height (which is also the length of the other leg), you can use the Pythagorean theorem.

Let $ a $ be the length of one leg (which is also the height), and $ c $ be the length of the hypotenuse.

Using the Pythagorean theorem: $ a^2 + a^2 = c^2 $.

Substituting the given values: $ 73^2 + 73^2 = c^2 $.

Solve for $ c $ to find the length of the hypotenuse.

Once you have the length of the hypotenuse, you can then use it as the base of the triangle and one of the legs as the height to calculate the area using the formula $ \frac{1}{2} \times \text{base} \times \text{height} $.