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

May 11, 2018

When we solve simultaneous equations

$\frac{1}{2} a b = 960$

${a}^{2} + {b}^{2} = {52}^{2}$

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

#### Explanation:

Call the legs $a$ and $b$.

$\frac{1}{2} a b = 960$

${a}^{2} + {b}^{2} = {52}^{2}$

Let's eliminate $b$ first.

$b = \frac{1920}{a}$

${a}^{2} + {\left(\frac{1920}{a}\right)}^{2} = {52}^{2}$

${a}^{4} - {52}^{2} {a}^{2} + {1920}^{2} = 0$

That's a quadratic equation in ${a}^{2}$ with a pretty negative discriminant of

${52}^{4} - 4 {\left(1920\right)}^{2} = - 7433984$

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 $A$.

$\frac{1}{2} a b = A$

$b = \frac{2 A}{a}$

${a}^{2} + {b}^{2} = {c}^{2}$

${c}^{2} = {a}^{2} + \frac{4 {A}^{2}}{a} ^ 2$

${a}^{4} - {c}^{2} {a}^{2} + 4 {A}^{2} = 0$

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 $a = b .$