A rectangle parcel of land is 70 ft longer than it is wide. Each diagonal between oppsite corners is 130 ft

dimensions-of-parcel

#1

a rectangle parcel of land is 70 ft longer than it is wide. Each diagonal between oppsite corners is 130 ft. what are the dimensions of the parcel? *** NEED FOR TMO PLEASE HELP ASAP***

Answer:

Okay so the first thing I would do would be to draw a picture. Next write a set of equations. L = length and W=width. From the given information you can write L=W+70 because the length is 70 ft longer than the width or is equal to the width plus 70 feet. Next with the diagonal information you can use the Pythagorean theorem to write an equation. The diagonal is going to be the hypotenuse or c because it is opposite of the right angle of the rectangle. The width will be b and the length will be a. So L^2 + W^2 = 130^2. Next use substitution to make it all in terms of 1 variable. We know L=W+70 so we substitute that into the equation for L so (W+70)^2 + W^2 = 16900 Next foil the squared term so W^2 + 140W + 4900 + W^2 = 16900. Next add like terms so 2W^2 +140W - 12000 = 0 and then factor to solve for W 2(W+120)(W-50)= 0 So W= -120 or 50 and since it’s a distance it can’t be negative so W= 50 and then L= W+70 or 50 + 70 so L=120 so the dimensions are 120ft by 50 ft