Estimate the perimeter of the land to the nearest meter. Draw straight lines joining adjacent points and use Pythagorean theorem to find the lengths QA,AR,BS,BP,QP,and ,RS

Let point A be the origin $(0,0)$. The baseline AB lies along the x-axis.

• A = $(0,0)$
• B = $(56,0)$ (since the total length of the baseline is $16+28+12 = 56$ units)

The other points are defined by their offsets from the baseline:

• Point Q: 16 units from A along the baseline, 8 units to the left (negative y-direction). So, Q = $(16, -8)$.
• Point P: $16+28 = 44$ units from A along the baseline, 16 units to the left (negative y-direction). So, P = $(44, -16)$.
• Point R: 16 units from A along the baseline, 18 units to the right (positive y-direction). So, R = $(16, 18)$.
• Point S: $16+10 = 26$ units from A along the baseline, 32 units to the right (positive y-direction). So, S = $(26, 32)$.

• Length QA: A = $(0,0)$, Q = $(16, -8)$

  $$QA = \sqrt{(16-0)^2 + (-8-0)^2} = \sqrt{16^2 + (-8)^2} = \sqrt{256 + 64} = \sqrt{320} \approx 17.8885$$

• Length QP: Q = $(16, -8)$, P = $(44, -16)$

  $$QP = \sqrt{(44-16)^2 + (-16-(-8))^2} = \sqrt{28^2 + (-8)^2} = \sqrt{784 + 64} = \sqrt{848} \approx 29.1204$$

• Length PB: P = $(44, -16)$, B = $(56, 0)$

  $$PB = \sqrt{(56-44)^2 + (0-(-16))^2} = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20.0000$$

• Length BS: B = $(56, 0)$, S = $(26, 32)$

  $$BS = \sqrt{(26-56)^2 + (32-0)^2} = \sqrt{(-30)^2 + 32^2} = \sqrt{900 + 1024} = \sqrt{1924} \approx 43.8634$$

• Length SR: S = $(26, 32)$, R = $(16, 18)$

  $$SR = \sqrt{(16-26)^2 + (18-32)^2} = \sqrt{(-10)^2 + (-14)^2} = \sqrt{100 + 196} = \sqrt{296} \approx 17.2047$$

• Length RA: R = $(16, 18)$, A = $(0,0)$

  $$RA = \sqrt{(0-16)^2 + (0-18)^2} = \sqrt{(-16)^2 + (-18)^2} = \sqrt{256 + 324} = \sqrt{580} \approx 24.0832$$

The estimated perimeter of the land is 152 meters.

The perimeter of the land is $\text{152 meters}$.