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Calculate the area of a trapezium with parallel sides 0.05 m and 9 cm and height 6 cm. Also, find the total surface area and volume of a cone with slant height 13 cm and radius 5 cm.

Mathematics

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Here are the solutions for Question Four:

Question Four

a) Calculate the area of the trapezium whose parallel sides are 0.05 m0.05 \text{ m} and 9 cm9 \text{ cm} with an height of 6 cm6 \text{ cm}.

  1. Convert all units to centimeters.

    The parallel sides are a=0.05 ma = 0.05 \text{ m} and b=9 cmb = 9 \text{ cm}. The height is h=6 cmh = 6 \text{ cm}.

    a=0.05 m=0.05×100 cm=5 cma = 0.05 \text{ m} = 0.05 \times 100 \text{ cm} = 5 \text{ cm}
  2. Use the formula for the area of a trapezium.

    The area AA of a trapezium is given by A=12(a+b)hA = \frac{1}{2}(a+b)h.

    A=12(5 cm+9 cm)(6 cm)A = \frac{1}{2}(5 \text{ cm} + 9 \text{ cm})(6 \text{ cm}) A=12(14 cm)(6 cm)A = \frac{1}{2}(14 \text{ cm})(6 \text{ cm}) A=(7 cm)(6 cm)A = (7 \text{ cm})(6 \text{ cm}) A=42 cm2A = \boxed{\text{42 cm}^2}

    b) A cone has a slant height of 13 cm13 \text{ cm} with a radius of 5 cm5 \text{ cm} for its circular base. Calculate:
    i. Total surface area
    ii. Volume

    Given: Slant height l=13 cml = 13 \text{ cm}, radius r=5 cmr = 5 \text{ cm}. Use π=227\pi = \frac{22}{7}.

    i. Total surface area

  3. Use the formula for the total surface area of a cone.

    The total surface area AtotalA_{total} of a cone is given by Atotal=πr(r+l)A_{total} = \pi r (r+l).

    Atotal=227×5 cm×(5 cm+13 cm)A_{total} = \frac{22}{7} \times 5 \text{ cm} \times (5 \text{ cm} + 13 \text{ cm}) Atotal=227×5 cm×18 cmA_{total} = \frac{22}{7} \times 5 \text{ cm} \times 18 \text{ cm} Atotal=19807 cm2A_{total} = \frac{1980}{7} \text{ cm}^2 Atotal≈282.86 cm2A_{total} \approx \boxed{\text{282.86 cm}^2}

    ii. Volume

  4. Calculate the perpendicular height hh of the cone using the Pythagorean theorem.

    h2+r2=l2h^2 + r^2 = l^2 h2+(5 cm)2=(13 cm)2h^2 + (5 \text{ cm})^2 = (13 \text{ cm})^2 h2+25 cm2=169 cm2h^2 + 25 \text{ cm}^2 = 169 \text{ cm}^2 h2=169 cm2−25 cm2h^2 = 169 \text{ cm}^2 - 25 \text{ cm}^2 h2=144 cm2h^2 = 144 \text{ cm}^2 h=144 cm2h = \sqrt{144 \text{ cm}^2} h=12 cmh = 12 \text{ cm}
  5. Use the formula for the volume of a cone.

    The volume VV of a cone is given by V=13πr2hV = \frac{1}{3}\pi r^2 h.

    V=13×227×(5 cm)2×12 cmV = \frac{1}{3} \times \frac{22}{7} \times (5 \text{ cm})^2 \times 12 \text{ cm} V=13×227×25 cm2×12 cmV = \frac{1}{3} \times \frac{22}{7} \times 25 \text{ cm}^2 \times 12 \text{ cm} V=22×25×123×7 cm3V = \frac{22 \times 25 \times 12}{3 \times 7} \text{ cm}^3 V=660021 cm3V = \frac{6600}{21} \text{ cm}^3 V=22007 cm3V = \frac{2200}{7} \text{ cm}^3 V≈314.29 cm3V \approx \boxed{\text{314.29 cm}^3}

    c) Calculate the volume of the cylinder whose height is 7 cm7 \text{ cm} and a diameter of 28 cm28 \text{ cm}.

    Given: Height h=7 cmh = 7 \text{ cm}, diameter D=28 cmD = 28 \text{ cm}. Use π=227\pi = \frac{22}{7}.

  6. Calculate the radius rr of the cylinder.

    r=D2=28 cm2=14 cmr = \frac{D}{2} = \frac{28 \text{ cm}}{2} = 14 \text{ cm}
  7. Use the formula for the volume of a cylinder.

    The volume VV of a cylinder is given by V=πr2hV = \pi r^2 h.

    V=227×(14 cm)2×7 cmV = \frac{22}{7} \times (14 \text{ cm})^2 \times 7 \text{ cm} V=227×196 cm2×7 cmV = \frac{22}{7} \times 196 \text{ cm}^2 \times 7 \text{ cm} V=22×196 cm3V = 22 \times 196 \text{ cm}^3 V=4312 cm3V = \boxed{\text{4312 cm}^3}

    d) Solve the following simultaneous equation using the elimination method.

    x+y=3(1)x+y=3 \quad (1) x+y−3=9(2)x+y-3=9 \quad (2)
  8. Simplify equation (2).

    x+y−3=9x+y-3=9 x+y=9+3x+y=9+3 x+y=12(3)x+y=12 \quad (3)
  9. Use the elimination method by subtracting equation (1) from equation (3).

    (x+y)−(x+y)=12−3(x+y) - (x+y) = 12 - 3 0=90 = 9
  10. Interpret the result.

    Since 0=90=9 is a false statement, the system of equations is inconsistent. This means there is no solution that satisfies both equations simultaneously.
    The system has no solution\boxed{\text{no solution}}.

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