Question 156845

Leg 1: {{{x}}}



Leg 2: {{{x+14}}} (since one leg is 14 m shorter than the other, the other leg is 14 m longer than the first one)



Hypotenuse: 26




{{{a^2+b^2=c^2}}} Start with Pythagorean's Theorem



{{{x^2+(x+14)^2=26^2}}} Plug in the given information



{{{x^2+(x+14)^2=676}}} Square 26



{{{x^2+x^2+28x+196=676}}} FOIL



{{{2x^2+28x+196=676}}} Combine like terms



{{{2x^2+28x+196-676=0}}} Subtract 676 from both sides.



{{{2x^2+28x-480=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=2}}}, {{{b=28}}}, and {{{c=-480}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(28) +- sqrt( (28)^2-4(2)(-480) ))/(2(2))}}} Plug in  {{{a=2}}}, {{{b=28}}}, and {{{c=-480}}}



{{{x = (-28 +- sqrt( 784-4(2)(-480) ))/(2(2))}}} Square {{{28}}} to get {{{784}}}. 



{{{x = (-28 +- sqrt( 784--3840 ))/(2(2))}}} Multiply {{{4(2)(-480)}}} to get {{{-3840}}}



{{{x = (-28 +- sqrt( 784+3840 ))/(2(2))}}} Rewrite {{{sqrt(784--3840)}}} as {{{sqrt(784+3840)}}}



{{{x = (-28 +- sqrt( 4624 ))/(2(2))}}} Add {{{784}}} to {{{3840}}} to get {{{4624}}}



{{{x = (-28 +- sqrt( 4624 ))/(4)}}} Multiply {{{2}}} and {{{2}}} to get {{{4}}}. 



{{{x = (-28 +- 68)/(4)}}} Take the square root of {{{4624}}} to get {{{68}}}. 



{{{x = (-28 + 68)/(4)}}} or {{{x = (-28 - 68)/(4)}}} Break up the expression. 



{{{x = (40)/(4)}}} or {{{x =  (-96)/(4)}}} Combine like terms. 



{{{x = 10}}} or {{{x = -24}}} Simplify. 



So the possible answers are {{{x = 10}}} or {{{x = -24}}} 

  

However, a negative length isn't possible. So the only answer is {{{x = 10}}}



So the first leg is 10 m, the second leg is 24 m (since {{{10+14=24}}}), and the hypotenuse is 26 m