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Bepaal die afstand op die topografiese kaart tussen die skool in blok E1 en die skool in blok J4 in kilometer.

Geography

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Answer

5,8 km

700 m

Step-by-step solution

5 steps

2.b

  1. Bepaal die oost-wes verskuiwing (Δx).

    Blok E = 4 km oos van A, blok J = 9 km oos van A.
    Beide punte is 0,1 km oos in hul blokke.

    Δx=(9+0,1)−(4+0,1)=5 km\Delta x = (9 + 0{,}1) - (4 + 0{,}1) = 5 \text{ km}

  2. Bepaal die noord-suid verskuiwing (Δy).

    Blok ry 1 = 0 km noord, blok ry 4 = 3 km noord.
    Beide punte is 0,3 km noord in hul blokke.

    Δy=(3+0,3)−(0+0,3)=3 km\Delta y = (3 + 0{,}3) - (0 + 0{,}3) = 3 \text{ km}

  3. Bereken die reguitlyn afstand met Pythagoras.

    Afstand=(Δx)2+(Δy)2=52+32=25+9=34≈5,8 km\text{Afstand} = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{5^2 + 3^2} = \sqrt{25 + 9} = \sqrt{34} \approx 5{,}8 \text{ km}

    5,8 km

    2.c

  4. Identifiseer die punthoogtes.

    Puntoogte in blok E3: 778 m
    Puntoogte in blok E5: +78+78 m = 78 m

  5. Bereken die hoogteverskil.

    Hoogteverskil=∣778−78∣=700 m\text{Hoogteverskil} = |778 - 78| = 700 \text{ m}

    700 m

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