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Find the area of the trapezium above and calculate the area of the sector of the arc.

Mathematics

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Answer

B

Step-by-step solution

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Here are the solutions to the questions:

36. Find the area of the trapezium above

  1. Identify the lengths of the parallel sides and the height from the diagram.

    The parallel sides are a=7 cma = 7 \text{ cm} and b=13 cmb = 13 \text{ cm}.
    The perpendicular height is h=8 cmh = 8 \text{ cm}.

  2. Use the formula for the area of a trapezium, A=12(a+b)hA = \frac{1}{2}(a+b)h.

    A=12(7 cm+13 cm)(8 cm)A = \frac{1}{2}(7 \text{ cm} + 13 \text{ cm})(8 \text{ cm})
  3. Calculate the area.

    A=12(20 cm)(8 cm)A = \frac{1}{2}(20 \text{ cm})(8 \text{ cm}) A=10 cm×8 cmA = 10 \text{ cm} \times 8 \text{ cm} A=80 cm2A = 80 \text{ cm}^2

    The correct option is B.
    The area of the trapezium is B. 80 cm2\boxed{\text{B. } 80 \text{ cm}^2}.

    37. A circular arc subtends angle 150∘150^\circ at the centre of a circle of radius 12 cm12 \text{ cm}, calculate the area of the sector of the arc.

  4. Identify the given values.

    Angle θ=150∘\theta = 150^\circ.
    Radius r=12 cmr = 12 \text{ cm}.

  5. Use the formula for the area of a sector, A=θ360∘πr2A = \frac{\theta}{360^\circ} \pi r^2.

    A=150∘360∘π(12 cm)2A = \frac{150^\circ}{360^\circ} \pi (12 \text{ cm})^2
  6. Calculate the area.

    A=150360π(144 cm2)A = \frac{150}{360} \pi (144 \text{ cm}^2) A=512π(144 cm2)A = \frac{5}{12} \pi (144 \text{ cm}^2) A=5π(14412) cm2A = 5 \pi \left(\frac{144}{12}\right) \text{ cm}^2 A=5π(12) cm2A = 5 \pi (12) \text{ cm}^2 A=60π cm2A = 60 \pi \text{ cm}^2

    The correct option is D.
    The area of the sector is D. 60π cm2\boxed{\text{D. } 60 \pi \text{ cm}^2}.

    38. Calculate the volume of a cuboid of length 0.76 cm0.76 \text{ cm}, breadth 2.6 cm2.6 \text{ cm} and height 0.82 cm0.82 \text{ cm}.

  7. Identify the given dimensions.

    Length l=0.76 cml = 0.76 \text{ cm}.
    Breadth b=2.6 cmb = 2.6 \text{ cm}.
    Height h=0.82 cmh = 0.82 \text{ cm}.

  8. Use the formula for the volume of a cuboid, V=l×b×hV = l \times b \times h.

    V=0.76 cm×2.6 cm×0.82 cmV = 0.76 \text{ cm} \times 2.6 \text{ cm} \times 0.82 \text{ cm}
  9. Calculate the volume.

    V=1.976 cm2×0.82 cmV = 1.976 \text{ cm}^2 \times 0.82 \text{ cm} V=1.62032 cm3V = 1.62032 \text{ cm}^3

    Rounding to two decimal places, V≈1.62 cm3V \approx 1.62 \text{ cm}^3.
    The correct option is A.
    The volume of the cuboid is A. 1.62 cm3\boxed{\text{A. } 1.62 \text{ cm}^3}.

    39. The locus of a point equidistant from the intersection of lines 3x−7y+7=03x - 7y + 7 = 0 and 4x−6y+1=04x - 6y + 1 = 0 is a

  10. Understand the geometric interpretation of the problem.

    The intersection of two distinct lines is a single point.
    The locus of a point equidistant from a single fixed point is a circle, with the fixed point as its center.
    The correct option is C.
    The locus of a point equidistant from the intersection of lines 3x−7y+7=03x - 7y + 7 = 0 and 4x−6y+1=04x - 6y + 1 = 0 is a C. Circle\boxed{\text{C. Circle}}.

    40. The gradient of the straight line joining the points P (5,−7)(5, -7) and Q (−2,−3)(-2, -3) is

  11. Identify the coordinates of the two points.

    Point P: (x1,y1)=(5,−7)(x_1, y_1) = (5, -7).
    Point Q: (x2,y2)=(−2,−3)(x_2, y_2) = (-2, -3).

  12. Use the formula for the gradient (slope) of a line, m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.

    m=−3−(−7)−2−5m = \frac{-3 - (-7)}{-2 - 5}
  13. Calculate the gradient.

    m=−3+7−7m = \frac{-3 + 7}{-7} m=4−7m = \frac{4}{-7} m=−47m = -\frac{4}{7}

    The correct option is C.
    The gradient of the straight line joining the points P (5,−7)(5, -7) and Q (−2,−3)(-2, -3) is C. −47\boxed{\text{C. } -\frac{4}{7}}.

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