How do you find the dimensions of a right triangle if a right triangle has area 960 and hypotenuse length = 52?

1 Answer
May 11, 2018

Answer:

When we solve simultaneous equations

# 1/2 a b = 960#

#a^2 + b^2 = 52 ^2#

we find no real solutions, so no such right triangle exists.

Explanation:

Call the legs #a# and #b#.

# 1/2 a b = 960#

#a^2 + b^2 = 52 ^2#

Let's eliminate #b# first.

#b = 1920/a#

#a^2 + (1920/a)^2 = 52^2 #

# a^4 - 52^2 a^2 + 1920^2 = 0 #

That's a quadratic equation in #a^2# with a pretty negative discriminant of

#52^4 - 4(1920)^2 = -7433984 #

so no real solutions for #a^2# and thus no real solutions for #a,b#.


That's the end, but we can look into it a bit deeper.

The hypotenuse is just too small to support this area. Let's find the general formula for the minimum hypotenuse for a real triangle of a given area #A#.

# 1/2 ab = A#

#b = {2A}/a#

#a^2 + b^2 = c^2#

#c^2 = a^2 + {4A^2}/a^2 #

# a^4 - c^2 a^2 + 4A^2 = 0#

The discriminant must be positive or zero:

#c^4 - 16 A^2 ge 0#

#c^4 ge 16 A^2#

Everything is positive, so

#c ge 2 sqrt{A}#

In our case

# c = 2 sqrt{960} = 16 sqrt{15} approx 61.97 # is the minimum hypotenuse than can support this area, presumably corresponding to the isosceles right triangle with #a=b.#