Arectangular 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?

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.

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answer:
congruent segments
step-by-step explanation:
congruent segments. take a line, cut it in half, superipose one half over the other half; you'll have congruence.