Dimensions of a Lot

[Mikayla2010]

A city lot has the shape of a right triangle whose hypotenuse is 7ft longer than one of the other sides. The perimeter of the lot is 392ft.
How long is each side of the lot?

Not sure how to set up the equation..

 

[Loren]

Draw a sketch of a right triangle. Label one leg x. Then label the hypotenuse x+7. Use the Pythagorean theorem to determine the length of the other leg. This length will be expressed in terms of x. Then use what you know about the perimeter to build your equation. Solve for x and answer the question stated by the problem.

 

[BigGlenntheHeavy]

\(\displaystyle (x+7)^2-x^2 \ = \ k^2, \ solve \ for \ k, \ then \ x+7+x+k \ = \ 392, \ solve \ for \ x.\)

 

[Mikayla2010]

Ok. S[sup:19qql4s6][/sup:19qql4s6]o I realize that the equation to find one side of the triangle that isn't "x" would be k[sup:19qql4s6]2[/sup:19qql4s6]+x[sup:19qql4s6]2[/sup:19qql4s6]=(x+7)[sup:19qql4s6]2[/sup:19qql4s6]
But I don't even know how do go about solving that for k...

 

[Deleted member 4993]

Ok. S[sup:2pxymilu][/sup:2pxymilu]o I realize that the equation to find one side of the triangle that isn't "x" would be k[sup:2pxymilu]2[/sup:2pxymilu]+x[sup:2pxymilu]2[/sup:2pxymilu]=(x+7)[sup:2pxymilu]2[/sup:2pxymilu]
But I don't even know how do go about solving that for k...

k[sup:2pxymilu]2[/sup:2pxymilu] = (x+7)[sup:2pxymilu]2[/sup:2pxymilu] - x[sup:2pxymilu]2[/sup:2pxymilu]

k[sup:2pxymilu]2[/sup:2pxymilu] = (x[sup:2pxymilu]2[/sup:2pxymilu]+ 14x + 49) - x[sup:2pxymilu]2[/sup:2pxymilu]

k[sup:2pxymilu]2[/sup:2pxymilu] = 14x + 49.......................................(1)

you are given the perimeter. Then

k + x + x + 7 = 392

k = 385 - 2x......................................................(2)

Using this in (1), we get

(385 - 2x)[sup:2pxymilu]2[/sup:2pxymilu] = 14x + 49

From above you'll get a quadratic equation for x - solve for x

Use (2) to solve for k