Question

Un triángulo rectángulo tiene un área de 84 pies2 y una hipotenusa de 25 pies de largo. ¿Cuáles son las longitudes de sus otros dos lados?

Answer 1

Answer:

7 feet and 24 feet

Step-by-step explanation:

In the right triangle, the hypotenuse is 25 feet long. The area of this triangle is 84 square feet.

Let x feet and y feet be the lengths of triangle's legs.

By the Pythagorean theorem,

$x^2+y^2=25^2\\ \\x^2+y^2=625$

The area of the right triangle is half the product of its legs, thus

$84=\dfrac{1}{2}xy\\ \\xy=168$

Solve the system of two equations:

$\left\{\begin{array}{l}x^2+y^2=625\\ \\xy=168\end{array}\right.$

From the second equation:

$x=\dfrac{168}{y}$

Substitute it into the first equation:

$\left(\dfrac{168}{y}\right)^2+y^2=625\\ \\168^2+y^4=625y^2\\ \\y^4-625y^2+168^2=0\\ \\D=(-625)^2-4\cdot 168^2=(625-2\cdot 168)(625+2\cdot 168)=289\cdot 961=17^2\cdot 31\\ \\\sqrt{D}=17\cdot 31=527\\ \\y^2_{1,2}=\dfrac{-(-625)\pm 527}{2}=49,\ 576\\ \\y_1=7,\ y_2=-7,\ y_3=24,\ y_4=-24$

The length of the leg cannot be negative, so

$y_1=7\Rightarrow x_1=\dfrac{168}{7}=24\\ \\y_2=24\Rightarrow x_2=\dfrac{168}{24}=7$