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Calculate the area of a rectangular floor with length 18m and width 12m. Then, calculate the volume of a triangular prism roof with a triangular cross-section of sides 7m, 7m, and 12m, and a length of

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6 parts
a

216 m2\text{216 m}^2

b

389.40 m3\text{389.40 m}^3

c

K7056\text{K}7056

d(i)

27.90 m2\text{27.90 m}^2

d(ii)

11.31 m3\text{11.31 m}^3

d(iii)

11309.73 kg\text{11309.73 kg}

Step-by-step solution

16 steps

Part a· 2 steps

  1. Identify the dimensions of the rectangular floor.

    Length (LL) = 18 m, Width (WW) = 12 m.

  2. Calculate the area of the rectangular floor.

    Area=L×W\text{Area} = L \times W Area=18 m×12 m\text{Area} = 18 \text{ m} \times 12 \text{ m} Area=216 m2\text{Area} = 216 \text{ m}^2

    The area of the rectangular floor is 216 m2\boxed{\text{216 m}^2}.

Part b

389.40 m3\text{389.40 m}^3

  1. Find the semi-perimeter (ss) of the triangular cross-section.

    The sides of the triangle are a=7a = 7 m, b=7b = 7 m, and c=12c = 12 m.

    s=a+b+c2s = \frac{a+b+c}{2} s=7 m+7 m+12 m2s = \frac{7 \text{ m} + 7 \text{ m} + 12 \text{ m}}{2} s=26 m2s = \frac{26 \text{ m}}{2} s=13 ms = 13 \text{ m}

  2. Use Heron's Formula to find the cross-sectional area of the triangle.

    Areatriangle=s(s−a)(s−b)(s−c)\text{Area}_{\text{triangle}} = \sqrt{s(s-a)(s-b)(s-c)} Areatriangle=13(13−7)(13−7)(13−12)\text{Area}_{\text{triangle}} = \sqrt{13(13-7)(13-7)(13-12)} Areatriangle=13×6×6×1\text{Area}_{\text{triangle}} = \sqrt{13 \times 6 \times 6 \times 1} Areatriangle=13×36\text{Area}_{\text{triangle}} = \sqrt{13 \times 36} Areatriangle=468\text{Area}_{\text{triangle}} = \sqrt{468} Areatriangle≈21.6333 m2\text{Area}_{\text{triangle}} \approx 21.6333 \text{ m}^2

  3. Calculate the volume of the triangular prism roof.

    The length of the prism is 18 m.

    Volumeprism=cross-sectional area×length\text{Volume}_{\text{prism}} = \text{cross-sectional area} \times \text{length} Volumeprism=468 m2×18 m\text{Volume}_{\text{prism}} = \sqrt{468} \text{ m}^2 \times 18 \text{ m} Volumeprism≈21.6333 m2×18 m\text{Volume}_{\text{prism}} \approx 21.6333 \text{ m}^2 \times 18 \text{ m} Volumeprism≈389.40 m3\text{Volume}_{\text{prism}} \approx 389.40 \text{ m}^3

    The volume of the triangular prism roof is 389.40 m3\boxed{\text{389.40 m}^3}.

Part c

K7056\text{K}7056

  1. Calculate the total area of the two slant faces.

    Each slant face has an area of 7 m×18 m=126 m27 \text{ m} \times 18 \text{ m} = 126 \text{ m}^2.

    Total slant area=2×126 m2\text{Total slant area} = 2 \times 126 \text{ m}^2 Total slant area=252 m2\text{Total slant area} = 252 \text{ m}^2

  2. Calculate the total cost of the roofing iron.

    The cost is K28 per square metre.

    Total cost=Total slant area×cost per square metre\text{Total cost} = \text{Total slant area} \times \text{cost per square metre} Total cost=252 m2×K28/m2\text{Total cost} = 252 \text{ m}^2 \times \text{K}28/\text{m}^2 Total cost=K7056\text{Total cost} = \text{K}7056

    The total cost of roofing iron is K7056\boxed{\text{K}7056}.

Part d(i)

27.90 m2\text{27.90 m}^2

  1. Identify the dimensions of the cylindrical water tank.

    Radius (rr) = 1.2 m, Height (hh) = 2.5 m.

  2. Calculate the total surface area (TSA) of the closed cylinder.

    TSA=2πr2+2πrh\text{TSA} = 2\pi r^2 + 2\pi rh TSA=2π(1.2 m)2+2π(1.2 m)(2.5 m)\text{TSA} = 2\pi (1.2 \text{ m})^2 + 2\pi (1.2 \text{ m})(2.5 \text{ m}) TSA=2π(1.44 m2)+2π(3 m2)\text{TSA} = 2\pi (1.44 \text{ m}^2) + 2\pi (3 \text{ m}^2) TSA=2.88π m2+6π m2\text{TSA} = 2.88\pi \text{ m}^2 + 6\pi \text{ m}^2 TSA=8.88π m2\text{TSA} = 8.88\pi \text{ m}^2 TSA≈8.88×3.14159265... m2\text{TSA} \approx 8.88 \times 3.14159265... \text{ m}^2 TSA≈27.897 m2\text{TSA} \approx 27.897 \text{ m}^2

  3. Round the total surface area to 2 decimal places.

    TSA=27.90 m2\text{TSA} = 27.90 \text{ m}^2

    The total surface area of the water tank is 27.90 m2\boxed{\text{27.90 m}^2}.

Part d(ii)

11.31 m3\text{11.31 m}^3

  1. Identify the dimensions of the cylindrical water tank.

    Radius (rr) = 1.2 m, Height (hh) = 2.5 m.

  2. Calculate the volume (VV) of the water tank.

    V=πr2hV = \pi r^2 h V=π(1.2 m)2(2.5 m)V = \pi (1.2 \text{ m})^2 (2.5 \text{ m}) V=π(1.44 m2)(2.5 m)V = \pi (1.44 \text{ m}^2) (2.5 \text{ m}) V=3.6π m3V = 3.6\pi \text{ m}^3 V≈3.6×3.14159265... m3V \approx 3.6 \times 3.14159265... \text{ m}^3 V≈11.3097 m3V \approx 11.3097 \text{ m}^3

  3. Round the volume to 2 decimal places.

    V=11.31 m3V = 11.31 \text{ m}^3

    The volume of the water tank is 11.31 m3\boxed{\text{11.31 m}^3}.

Part d(iii)

11309.73 kg\text{11309.73 kg}

  1. Convert the volume from cubic metres to litres.

    Given that 1 m3^3 = 1000 litres.

    Volume in litres=3.6π m3×1000 litres/m3\text{Volume in litres} = 3.6\pi \text{ m}^3 \times 1000 \text{ litres/m}^3 Volume in litres=3600π litres\text{Volume in litres} = 3600\pi \text{ litres} Volume in litres≈11309.73 litres\text{Volume in litres} \approx 11309.73 \text{ litres}

  2. Calculate the mass of water in the tank.

    Given that the density of water is 1 kg per litre.

    Mass=Volume in litres×density\text{Mass} = \text{Volume in litres} \times \text{density} Mass=3600π litres×1 kg/litre\text{Mass} = 3600\pi \text{ litres} \times 1 \text{ kg/litre} Mass=3600π kg\text{Mass} = 3600\pi \text{ kg} Mass≈11309.73 kg\text{Mass} \approx 11309.73 \text{ kg}

  3. Round the mass to 2 decimal places.

    Mass=11309.73 kg\text{Mass} = 11309.73 \text{ kg}

    The mass of water in the full tank is 11309.73 kg\boxed{\text{11309.73 kg}}.

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