MHB Solve Trig Word Problem: Acre Parcel Sides 180 & 240 ft

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The discussion focuses on solving a trigonometric word problem involving a one-acre parcel with sides measuring 180 ft and 240 ft that intersect at a right angle. Using the Pythagorean theorem, the length of the fourth side is calculated to be approximately 219.64 ft. The area of triangle ABD is determined to be 21,600 ft², while triangle DBC is calculated to be 21,960 ft². Although there was mention of using Heron's theorem, the participants agree that the Pythagorean method is simpler and more effective for this problem. The calculations reinforce the accuracy of the derived fourth side length.
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A one acre parcel has 2 sides 180 ft and 240 ft intersecting at a right angle.
the other side adjacent to the 180 ft is 200 ft what is the length of the 4th side.

by Pythagorean theorem $BD = 300$

so triangle ABD = $21600 \ ft^2$
thus triangle DBC = $21960 \ ft^2$

so 21960 = (1/2)(300)(h) then h=146.4

$$\sqrt{{200}^{2}{}-146.4^2}=136.26$$

$300-136.36 =163.74$

so by Pythagorean theorem $BC$ or the 4th side $\approx$ $219.64 ft$

not sure this is correct, saw another proposed way to do this
using Heron's theorem but after trying it was ?
 
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That looks good. I agree with your answer, and I don't see any simpler method for finding it.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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