Question

Please help with this problem

Answer 1

Answer:

The length of the short side is 14.5 units, the length of the other short side is 18.5 units, and the length of the longest side is 23.5 units.

Step-by-step explanation:

The Pythagorean Theorem

If a and b are the lengths of the legs of a right triangle and c is the length of the hypotenuse, then the sum of the squares of the lengths of the legs is equal to the square of the length of the hypotenuse.

This relationship is represented by the formula:

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

Applying the Pythagorean Theorem  to find the lengths of the three sides we get:

$(x)^2+(x+4)^2=(x+9)^2\\\\2x^2+8x+16=x^2+18x+81\\\\2x^2+8x-65=x^2+18x\\\\2x^2-10x-65=x^2\\\\x^2-10x-65=0$

Solve with the quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}$

$x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-10,\:c=-65:\quad x_{1,\:2}=\frac{-\left(-10\right)\pm \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}\\\\x_{1}=\frac{-\left(-10\right)+ \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5+3\sqrt{10}\\\\x_{2}=\frac{-\left(-10\right)- \sqrt{\left(-10\right)^2-4\cdot \:1\left(-65\right)}}{2\cdot \:1}=5-3\sqrt{10}$

Because a length can only be positive, the only solution is

$x=5+3\sqrt{10}\approx 14.5$

The length of the short side is 14.5, the length of the other short side is $14.5+4=18.5$, and the length of the longest side is $14.5+9=23.5$.