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Mensuration Test - 28

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Mensuration Test - 28
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  • Question 1
    1 / -0
    Calculate the area of a trapezium whose parallel sides are 2222 and 1010 m, and a height of 1515 m.
    Solution
    Area of a trapezoid = b1+b22×h\dfrac{b_{1}+ b_{2}}{2}\times h
    = 12+102×15\dfrac{12 + 10}{2}\times 15
    = 322×15\dfrac{32}{2}\times 15
    = 240240 m2m^{2}
  • Question 2
    1 / -0
    A company packages its juice in a tetra pack shown below. How much quantity of juice does it contain.

    Solution
    Quantity of juice in the package == Volume of cuboidal package
                                                          = = l×b×hl\times b\times h
                                                           == 4×4×104\times 4\times 10
                                                           == 160160 cm3{ cm }^{ 3 }

  • Question 3
    1 / -0
    Edward bought a container of ceralac in the shape of cylinder. If the container has a radius 10m10 m and a surface area is 340π \displaystyle 340\pi . What is its height?
    Solution
    Surface area of cylinder is A=2πr(r+h)A=2πr(r+h)

    Here the cylindrical container has surface area A=340πA=340π m2^2 and radius r=10r=10 m.

    Thus,

    A=2πr(r+h)340π=2π×10(10+h)340π=20π(10+h)340π=200π+20πh20πh=340π200π20πh=140πh=140π20π=7A=2πr(r+h)\\ \Rightarrow 340π=2π\times 10(10+h)\\ \Rightarrow 340π=20π(10+h)\\ \Rightarrow 340π=200π+20πh\\ \Rightarrow 20πh=340π-200π\\ \Rightarrow 20πh=140π\\ \Rightarrow h=\dfrac { 140π }{ 20π } =7

    Hence, the height of the cylindrical container is 7m7m.
  • Question 4
    1 / -0
    The area of the trapezium is 100 m2100\ { m }^{ 2 } and height is 10 m.10\ m. Find the length of one parallel side if the length of the other parallel side is 5 m.5\ m.
    Solution
    Consider that AA is the area of the trapezium, a,ba, b are the length of the parallel sides, and hh is the height of the trapezium.

    Given:
    A=100 m2A=100\ m^2
    h=10 mh=10\ m
    a=5 ma=5\ m

    So,
    A=12×(a+b)×h100=12×(5+b)×1020=5+bb=15 m\begin{aligned}{}A &= \frac{1}{2} \times \left( {a + b} \right) \times h\\100 &= \frac{1}{2} \times \left( {5 + b} \right) \times 10\\20 &= 5 + b\\b &= 15\ m\end{aligned}

    Hence, option CC is correct.
  • Question 5
    1 / -0

    Fill in the blank: 

    ____________ is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder or pyramid.

    Solution
    Volume is the measure of the amount of space inside of a solid figure, like a cube, ball, cylinder or pyramid.
  • Question 6
    1 / -0
    A swimming pool is 100 m long and 15 m wide its deep and 24 m height respectively. Find the capacity of the pool.
    Solution
    Volume of the pool = l× b × hl \times b \times  h
    = 100 ×15 ×24100 \times 15 \times 24
    = 36000m336000 m^3
  • Question 7
    1 / -0
    A brick whose length, breadth and height are 5m,6m5m, 6m, and 7m7m respectively. Find the surface area of the brick.
    Solution

    Surface area of the cuboid having length, breadth and height as l,bl, b and hh respectively is given as  2(lb+bh+hl)2 (lb + bh + hl)

    Brick has the shape of cuboid. Let SS be the require surface area.

    S=2(5×6+6×7+7×5)S=2 (5 \times 6 + 6 \times 7 + 7 \times 5)

    2(30+42+35)2 (30 + 42 + 35)

    214m2214 m^2
  • Question 8
    1 / -0
    Fill in the blank: The volume of solid objects is measured in _____ units.
    Solution
    The volume of solid objects is measured in cubic units.
  • Question 9
    1 / -0
    Find the surface area of a cylinder with the following dimension:
    Height =10= 10 ft.
    Radius =2= 2 ft.
    Solution
    Surface area of cylinder is A=2πr(r+h)A=2πr(r+h)

    Here, radius is r=2r=2 ft and the height is h=10h=10 ft.

    Thus,

    A=2π×2(2+10)=2π×2×12=48πA=2π\times 2(2+10)=2π\times 2\times 12=48π

    Hence, the surface area of the cylinder is 48π48π ft2^2.

  • Question 10
    1 / -0
    If lateral surface area of a cylinder of height 10 cm is 100 cm 2^{ 2 }, find radius of its base.
    Solution
    Given that, the lateral surface area of a cylinder of height 10 cm10\ cm is 100 cm2100\ cm^2.
    To find out: The radius of the base of the cylinder.

    We know that, lateral surface area of a cylinder is 2πrh\displaystyle 2\pi rh
     2πrh=100\therefore \ 2\pi rh=100\\
    2πr(10)=100\Rightarrow 2\pi r (10)=100\\
    20πr=100\Rightarrow 20\pi r =100\\
    r=10020π\Rightarrow r =\dfrac{100}{20\pi}\\
     r=5π \therefore \ \displaystyle r=\dfrac { 5 }{ \pi  }

    Hence, the radius of the base of the given cylinder is 5π \dfrac { 5 }{ \pi  }.
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