Question

A rectangular parcel of land is 70 ft longer than it is wide. Each diagonal between opposite corners is 130 ft. What are the dimensions of the parcel?

Answer 1

We have the following dimensions:
w=wide
w+70=length

We have a right triangle:
leg₁=w
leg₂=w+70
hypotenuse=130

Pythagoras theorem:
leg₁²+leg₂²=hypotenuse²
Then:

w²+(w+70)²=130²

We have to solve that equation:
w²+(w+70)²=130²
w²+w²+140w+4900=16900
2w²+140w+4900-16900=0
2w²+140w-12000=0
(2w²+140w-12000)/2=0/2
w²+70w-6000=0

w=(-70⁺₋√(4900+24000)) / 2
=(-70⁺₋170) / 2
We have two possible solutions:
Solution 1

w=(-70-170)/2=-120   (this solution is not possible because the result is negative and it have no sense in this problem).

Solution 2

w=(-70+170)/2=50  (this is the right solution)

w=50
w+70=120

ANSWER:  the length would be 120 u  (u=units of length ) and the wide would be 50 u.