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  • Sphere

Sphere

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  • Question 1:
    1 pts
    		Which expression can be used to find the volume of the sphere?

    V=4R2πV=4R^{2}\pi

    V=43R3πV=\dfrac{4}{3}R^{3}\pi

    V=43R2πV=\dfrac{4}{3}R^{2}\pi

    V=13R2πV=\dfrac{1}{3}R^{2}\pi

  • Question 2:
    1 pts
    		The following expression can be used to find the total surface area of the sphere. A=4R2πA=4R^{2}\pi
  • Question 3:
    1 pts
    		Find the volume of a sphere with a radius of 3cm.3 cm.
    V=V=
  • Question 4:
    1 pts
    		Find the total surface area of the sphere with a radius of 7cm.7 cm.
    V=108πin.2V=108\pi in. ^{2}
    V=169πcm2V=169\pi cm ^{2}
    V=196πcm2V=196\pi cm ^{2}
  • Question 5:
    2 pts
    		Find the total surface area of the sphere shown on the picture.
    A=144πcm2A=144\pi cm ^{2}
    A=324πcm2A=324\pi cm ^{2}
    A=224πcm2A=224\pi cm ^{2}
  • Question 6:
    1 pts
    		Find the volume of a sphere shown on the picture.
    V=V=
  • Question 7:
    2 pts
    		Which of the following expression can be used to find the total surface area of the hemisphere shown on the picture?
    A=3⋅72πcm2A=3\cdot 7^{2}\pi cm^{2}
    A=2⋅72πcm2A=2\cdot 7^{2}\pi cm^{2}
    A=43⋅72πcm2A=\dfrac{4}{3}\cdot 7^{2}\pi cm^{2}
    A=3⋅73πcm2A=3\cdot 7^{3}\pi cm^{2}
  • Question 8:
    2 pts
    		Which of the following expression can be used to find the volume of the sphere shown on the picture?
    V=V=
  • Question 9:
    3 pts
    		Basketballs used in professional games must have a circumference of 81 centimeters. What is the surface area of a basketball used in a professional game?

    A=π8124⋅π2A=\pi\dfrac{81^{2}}{4\cdot \pi^{2}}

    A=4π8124⋅π2A=4\pi\dfrac{81^{2}}{4\cdot \pi^{2}}

    A=2π8124⋅πA=2\pi\dfrac{81^{2}}{4\cdot \pi}

    A=4π812π2A=4\pi\dfrac{81^{2}}{ \pi^{2}}

  • Question 10:
    3 pts
    		We can use the following expression to find a ratio comparing the volume of a sphere with radius rr to the volume of a cylinder with radiusr r and height 2r.2r. Vsphere:Vcylinder=43r3ππr2⋅2r=23V_{sphere}:V_{cylinder}=\dfrac{\dfrac{4}{3}r^{3}\pi}{\pi r^{2} \cdot 2r}=\dfrac{2}{3}
  • Question 11:
    3 pts
    		If the area of the great circle of a sphere is 32m232m^{2},what is the surface area of the sphere?

    A=64πm2A=64\pi m^{2}

    A=128m2A=128 m^{2}

    A=128π2m2A=128\pi^{2} m^{2}

    A=256π2m2A=256\pi^{2} m^{2}

  • Question 12:
    3 pts
    		If a sphere has radius rr, there exists a cone with radius rr having the same volume.

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