Question

The Varners live on a corner lot. Often children cut across their lot to scar walking distance. The diagram to the right represents the corner lot. The children’s path is represented by dashed line. Approximate the walking distance that is saved by cutting across their property instead of walking around the lot. Hypotenuse 34. Short leg x long leg x+12

Answer 1

About 12.56 units. 
Solve for x by plugging in values into Pythagorean theorem.
$a^{2} +b^{2} = c^{2}$
$x^{2} +(x+12)^{2} = 34^{2}$
Simplify
$2x^{2} + 24x+144 = 1156$
Subtract 144 from both sides
$2x^{2} + 24x = 1012$
divide both sides by 2
$x^{2} + 12x = 506$
complete the square
$x^{2} + 12x + 36 = 506 +36 . = (x+6)^2 = 542$
take the square root of both sides.
$x+6 = 23.28$
subtract by 6 to get  about 17.28 for x.
Plug in values for x to get distance traveled on outside of the yard.
17.28 + (17.28 +12) = 46.56
subtract 34 from the length to get the difference in travel. 46.56-34=12.56

Answer 2

Distance traveled on outside of the yard is 46.56.

Difference in travel is 12.56.

Given :

Hypotenuse = 34

Short leg = x

Long leg = x + 12

Solution :

According pythagorean theorem,

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

$x^2 + (x+12)^2 = 34^2$

$x^2 +12x -506=0$ ---- (1)

We can also write above equation (1) as,

$(x+6)^2 = 542$

$x+6=23.28$

$x=17.28$

Distance traveled on outside of the yard is,

17.28 + (17.28 +12) = 46.56

Now subtract 34 from the length to get difference in travel,

46.56 - 34 = 12.56

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