Aleksandra Ciepielowska

Metoda Analityczna

$$2P_{A} = \sum_{i = 1}^{n}{x_{i}(y_{i + 1} - y_{i - 1})}$$

2P_A=93,00(52,60+42,10)+37,90(23,80+59,20)-30,90(33,80-52,60)-70,90(-78,80-23,80)-69,70(-45,90-33,80)-28,80(-42,10+78,80)+31,70(-59,20+45,90)

2P_A=23884,58m²

P_A=11942,29m²

Sprawdzenie

$$2P_{A} = \sum_{i = 1}^{n}{y_{i}(x_{i + 1} - x_{i - 1})}$$

2P_A=-59,20(37,90-31,70)+52,60(-30,90-93,00)+23,80(-70,90-37,90)+33,80(-69,70+30,90)-78,80(-28,80+70,90)-45,90(31,70+69,70)-42,10(93,00+28,80)

2P_A=I-23884,58I

P_A=11942,29m²

Metoda graficzna:

$$p_{i} = \frac{a_{i}*h_{i}}{2}$$

1^o

$$p_{1} = \frac{7,32*2,69}{2} = 9,85;\ \ \ \ \ \ \ \ \ \ p_{2} = \frac{5,88*2,00}{2} = 5,88;\ \ \ \ \ \ \ \ \ \ p_{3} = \frac{4,01*4,52}{2} = 9,06;\ \ \ \ \ \ $$

$$\text{\ \ \ \ }p_{4} = \frac{4,94*5,55}{2} = 13,71;\ \ \ \ \ \ \ \ \ \ p_{5} = \frac{8,18*3,13}{2} = 12,80;\ \ \ \ \ \ \ \ \ $$

$$P^{I} = \sum_{i = 1}^{n}p_{i}$$

P^I=51,30cm² - pole na mapie, w rzeczywistości to P=51,30*1500² = 115425000cm²=11549,25m²

2^o

$$p_{1} = \frac{7,35*2,62}{2} = 9,63;\ \ \ \ \ \ \ \ \ \ p_{2} = \frac{4,50*2,72}{2} = 6,12;\ \ \ \ \ \ \ \ \ \ p_{3} = \frac{4,50*4,61}{2} = 10,37;\ \ $$

$$\ \ p_{4} = \frac{7,79*3,18}{2} = 12,39;\ \ \ \ \ \ \ \ \ \ p_{5} = \frac{6,21*4,14}{2} = 12,56;\ \ \ \ \ \ \ \ \ $$

Aleksandra CIepielowska

$$P^{\text{II}} = \sum_{i = 1}^{n}p_{i}$$

P^II= 51,36 cm² - pole na mapie, w rzeczywistości to P=11556m²

$$P_{sr} = \frac{P^{I} + P^{\text{II}}}{2}$$

$$P_{sr} = \frac{11549,25 + 11556}{2} = 11549,25m^{2}$$

|P_sr−P_A| ≤ 500    [m²]

|11549,25−11942,29| ≤ 500 [m²]

393, 04 ≤ 500 [m²]